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Hindawi Publishing CorporationJournal of Function Spaces and ApplicationsVolume 2013 Article ID 675057 9 pageshttpdxdoiorg1011552013675057
Research ArticleOn the Topological and Uniform Structure of Diversities
Andrew Poelstra
Simon Fraser University Burnaby BC Canada V5A 1S6
Correspondence should be addressed to Andrew Poelstra asp11sfuca
Received 16 July 2013 Accepted 11 September 2013
Academic Editor Chengbo Zhai
Copyright copy 2013 Andrew Poelstra This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Diversities have recently been developed as multiway metrics admitting clear and useful notions of hyperconvexity and tightspan In this note we consider the analytical properties of diversities in particular the generalizations of uniform continuityuniform convergence Cauchy sequences and completeness to diversitiesWe develop conformities a diversity analogue of uniformspaces which abstract these concepts in the metric case We show that much of the theory of uniform spaces admits a naturalanalogue in this new structure for example conformities can be defined either axiomatically or in terms of uniformly continuouspseudodiversities Just as diversities can be restricted to metrics conformities can be restricted to uniformities We find that thesetwo notions of restriction which are functors in the appropriate categories are related by a natural transformation
1 Introduction
The theory of metric spaces is well-understood and forms thebasis of much of modern analysis In 1956 Aronszajn andPanitchpakdi developed the notion of hyperconvex metricspaces [1] in order to apply the Hahn-Banach theorem ina more general setting In fact every metric space can beembedded isometrically in a minimal hyperconvex space asdiscovered by Isbell [2] (as the ldquohyperconvex hullrdquo) and laterby Dress [3] (as the ldquometric tight spanrdquo)
Theseminimal hyperconvex spaces or tight spans provedto be powerful tools for the analysis of finite metric spacesThe theory of tight spans or T-theory is overviewed in [4]Its history as well as applications to phylogeny are given in[5]
In light of these applications of T-theory Bryant andTupper developed the theory of diversities alongside anassociated tight span theory in [5] Diversities are multiwaymetrics mapping finite subsets of a ground space 119883 to thenonnegative reals The axioms were chosen based on theirspecific applications to phylogeny (where they had alreadyappeared in special cases) and their ability to admit a tightspan theory This diversity tight span theory contains themetric tight span theory as a special case (using so-calleddiameter diversities) but it also allows new behavior whichmay be useful in situations such as microbial phylogeny
where the idea of a historical ldquophylogenetic treerdquo does notmake sense Several examples along with pictures of thisphenomenon are given in [5]
A classic paper by Weil [6] developed the theory of uni-form spaces which generalize metric spaces Uniform spacesadmit notions of uniform continuity uniform convergenceand completeness which coincide with the standard notionswhen metric spaces are considered as uniform spaces Thistheory has been described in Bourbakirsquos General Topology[7] as well as Kelleyrsquos classic text [8] The metric topology canbe derived purely from properties of the uniform space (viathe so-called uniform topology) and in this sense uniformspaces lie ldquobetweenrdquo metric spaces and topologies
In this note we develop conformities which generalizediversities in analogy to Weilrsquos uniform space generalizationof metrics We will describe uniform continuity uniformconvergence Cauchy sequences and completeness for diver-sities and show that these can be characterized in termsof conformities giving an abstract framework in which toanalyze the uniform structure of diversitiesThis is motivatedby the observation that while diversities generalize metricspaces in a straightforward way (in fact they restrict to metricspaces) they can exhibit very nonsmooth behavior withrespect to these spaces (cfTheorem 1)Therefore the existingtools for metric spaces are insufficient to get a handle on thebehavior of diversities
2 Journal of Function Spaces and Applications
2 Preliminaries
Throughout this paper we will denote the finite power set ofa given set119883 by
Pfin (119883) = 119860 sube 119883 |119860| lt infin (1)
We begin with the Bryant-Tupper definition from [5]a diversity is a pair (119883 120575) where 119883 is some set and 120575
Pfin(119883) rarr R is a function satisfying
(D1) if 119860 isin Pfin(119883) 120575(119860) ge 0 and 120575(119860) = 0 iff |119860| le 1
(D2) if 119860 119861 119862 isin Pfin(119883) with 119862 = then
120575 (119860 cup 119861) le 120575 (119860 cup 119862) + 120575 (119862 cup 119861) (2)
If for some 119860 isin Pfin(119883) 120575(119860) = 0 but |119860| gt 1 wehave the weaker notion of a pseudodiversity It is shownin [5] from these axioms that if 119860 sube 119861 then 120575(119860) le
120575(119861) that is (pseudo)diversities are monotonic and that therestriction of a diversity to sets of size 2 forms a pseudometric119889(119909 119910) = 120575(119909 119910) We call this metric the induced metric ofthe diversity
For a metric space (119883 119889) there are two importantdiversities on119883 having 119889 as an induced metric as follows
(i) the diameter diversity (119883 diam119889) defined by
diam119889 (1199091 119909119899) = sup119894119895
119889 (119909119894 119909119895) (3)
when 119883 = R119899 and 119889 is the Euclidean metric we referto this diversity simply by diam
(ii) The Steiner tree diversity (119883 120575) is defined for eachfinite set 119860 sube 119883 as the infimum of the size of theminimum Steiner tree on 119860(Recall that a Steiner tree on 119860 is a tree whose vertexset 119881 satisfies 119860 sube 119881 sube 119883 with each edge (119909 119910)
weighted by 119889(119909 119910) The size of the tree is the sumof its edge weights)
In fact these examples are the extremes of diversitybehavior relative to their induced metrics in the sense thatfor any diversity (119883 120575
1015840) which induces a metric 119889 we have
diam119889 le 1205751015840le 120575 (4)
where 120575 is the Steiner tree diversity on (119883 119889) This can beshown by a straightforward argument (Bryant and Tupperupcoming)
To demonstrate the difference between the diameterand Steiner tree diversities consider the Euclidean metric(R3 119889) The induced metric of both the diameter and Steinertree diversity is the Euclidean metric For any finite set 119860
contained in an 120576-ball diam(119860) lt 120576 To contrast in any 120576-ballwe can find finite sets 119860 for which 120575(119860) is arbitrarily large
Theorem 1 The Steiner tree diversity function 120575 on R3 isunbounded on every open set of the Euclidean topology
Proof Without loss of generality we show that the result for120576-balls is about 0 For each 119899 isin N define
119866119899 = (119894
1198992119895
1198992119896
1198992) 0 le 119894 119895 119896 lt 119899 (5)
which is a grid of points contained in the cube [0 1119899]3 Sincethere are 1198993 points a minimum spanning tree connecting themembers of 119866119899 must have 119899
3minus 1 edges each of the lengths
ge 11198992 since that is the least distance between two points
Therefore the size of the minimum spanning tree on 119866119899 isat least (1198993 minus 1)119899
2 which can be taken as large as we likeby taking 119899 large enough Since the minimal Steiner tree on119866119899 has a size of at least 0615 times than that of the minimalspanning tree [9] we have 120575(119866119899) rarr infin as 119899 rarr infin eventhough diam(119866119899) rarr 0
A similar construction for the Steiner tree diversity onR2gives sets of diversity (0615 minus 120576) for every 120576 gt 0 in everyEuclidean ball On R the Steiner tree diversity and diameterdiversity are identical The dramatic difference between themany-point behavior of these two diversities in dimension 2or higher demonstrates that diversities are not characterizedby their induced metrics even up to a constant
In Section 2 and 3 we will define uniform convergenceuniform continuity and completeness explicitly in termsof an underlying diversity in Section 4 we will describeconformities which abstract these properties for diversitiesThis is in analogy to Weilrsquos uniformities which abstract thesame concepts for metric spaces
With this goal in mind we start with the followingdefinitions let (119883 120575119883) and (119884 120575119884) be diversities Given 119909 isin 119883a sequence 119909119899 sub 119883 converges to 119909 denoted 119909119899 rarr 119909 if
lim119873rarrinfin
sup11989411198942119894119899ge119873
120575119883 (119909 1199091198941
1199091198942
119909119894119899
) = 0 (6)
The sequence 119909119899 is a Cauchy sequence if
lim119873rarrinfin
sup11989411198942119894119899ge119873
120575119883 (1199091198941
1199091198942
119909119894119899
) = 0 (7)
From these definitions and the axioms (D1) and (D2) itcan be shown that limits are unique and every convergentsequence is Cauchy If every Cauchy sequence is convergentwe call the diversity complete
Finally if 119891 119883 rarr 119884 is a function such that for every 120576 gt
0 there exists some119889 gt 0 such that 120575119883(119860) lt 119889 rArr 120575119884(119891(119860)) lt
120576 for every 119860 isin Pfin(119883) we say 119891 is uniformly continuousIt is not hard to see that for diameter diversities these
definitions coincide exactly with the standard ones on theinduced metric
For the second half of the paper we will work extensivelywith filters so we state the definition here given a ground set119883 define a filter as a collection F of subsets of 119883 satisfying119860 cap 119861 whenever 119860 119861 are inF and 119861 isin F whenever 119861 supe 119860
and119860 isin F A filter base becomes a filter when all supersets ofits elements are added inwhich casewe say the base generatesthe filter
In this paper we additionally require that notin F
Journal of Function Spaces and Applications 3
3 Comparison with Metrics
In this section we contrast the convergence of sequences withrespect to diversities and their induced metrics In particularwe show that although the Cauchy property for sequencesis much stronger for diversities (we demonstrate a sequencewhich is not Cauchywith respect to a diversity even though itis Cauchy with respect to the induced metric) completenessof a diversity is equivalent to completeness of its inducedmetric This tells us that every diversity which induces aEuclidean metric (eg the Steiner tree diversity on R119899) iscomplete
Since the set of Cauchy sequences in a diversity maybe smaller than the set of Cauchy sequences of its inducedmetric this may provide a simpler way to determine com-pleteness of metric spaces
At the end of the section we construct the analogue ofcompletion for diversities
31 Completeness in Diversities and Metric Spaces
Theorem 2 Let (119883 120575) be a diversity and let 119889 be its inducedmetric If (119883 119889) is a complete metric space then (119883 120575) is acomplete diversity
Proof Suppose that (119883 119889) is complete Let 119909119899 be a Cauchysequence in (119883 120575) Then it is also Cauchy in (119883 119889) andtherefore converges to some element 119909 We claim that 119909119899 rarr
119909 in (119883 120575) To this end let 120576 gt 0 Then there exists 119873 suchthat
(i) 119889(119909119899 119909) lt 120576 for all 119899 gt 119873 (since 119909119899 rarr 119909 in (119883 119889))(ii) 120575(119909119899
1
1199091198992
119909119899119898
) lt 120576 for all 119899119894 gt 119873 (since 119909119899 isCauchy in (119883 120575))
Therefore for all 1198991 119899119898 gt 119873
120575 (119909 1199091198991
119909119899119898
) le 120575 (119909 1199091198991
) + 120575 (1199091198991
119909119899119898
)
= 119889 (119909 1199091198991
) + 120575 (1199091198991
119909119899119898
) lt 2120576
(8)
that is 119909119899 rarr 119909 in (119883 120575)
As mentioned the set of Cauchy sequences in a diversitymay be strictly smaller than the set of Cauchy sequences inthe induced metric For example let (119883 120575) be the Steinertree diversity on R3 and consider the sets 119866119899119899isinN fromTheorem 1
Order each set 119866119899 somehow and define the sequence 119909119894by concatenating them that is
119909119899 = 119866111986621198663 sdot sdot sdot (9)
which is Cauchy in the induced metric of (119883 120575) (sinceeventually every pair of points is confined to arbitrarily smallcubes [0 120576]
3) However it is not Cauchy in (119883 120575) since wesaw in the proof of Theorem 1 that 120575(119866119899) becomes arbitrarilylarge as 119899 rarr infin In other words every tail of 119909119899 hasarbitrarily large finite sets so 119909119899 is not Cauchy
In light of this example it is interesting to know that everycomplete diversity has a complete induced metric which isproved with the following lemma
Lemma 3 Let (119883 120575) be a diversity and let 119889 be its inducedmetric Let 119909119899 be Cauchy in (119883 119889)Then it has a subsequencethat is Cauchy in (119883 120575)
Proof Define the subsequence 119909119899119894
by
119899119894 = min 119899 119889 (119909119899 119909119898) lt 2minus119894
forall119898 ge 119899 (10)
Given 120576 gt 0 choose 119873 such that 21minus119873 lt 120576 Then for all 1198941 le1198942 le sdot sdot sdot le 119894119898 greater than119873
120575 (1199091198941
119909119894119898
) le 120575 (1199091198941
1199091198942
) + sdot sdot sdot + 120575 (119909119894119898minus1
119909119894119898
)
lt1
21198941+ sdot sdot sdot +
1
2119894119898lt
infin
sum
119894=119873
12119894
= 21minus119873
lt 120576
(11)
That is 119909119899119894
is Cauchy in (119883 120575)
Theorem 4 Let (119883 120575) be a diversity and let 119889 be its inducedmetric If (119883 120575) is a complete diversity then (119883 119889) is a completemetric space
Proof Let 119909119899 be a Cauchy sequence in (119883 119889) Then byLemma 3 it has a subsequence 119909119894
119899
that is Cauchy in (119883 120575)which converges to some element 119909 since the diversity iscomplete (it converges in both (119883 120575) and (119883 119889))
Then 119909119899 converges to 119909 in (119883 119889) since for any 120576 we have119889(119909119899 119909) le 119889(119909119899 119909119894
119898
) + 119889(119909119894119898
119909) lt 2120576 for 119898 119899 large enough
32 Completion In light of the equivalence between metriccompleteness and diversity completeness it is perhaps notso surprising that every diversity can be completed in acanonical way To do so we require twomore definitions from[5] an embedding 120587 1198841 rarr 1198842 is an injective map betweendiversities (1198841 1205751) and (1198842 1205752) such that 1205751(119860) = 1205752(120587(119860)) forall 119860 isin Pfin(1198841) A isomorphism is a surjective embedding
Theorem 5 Every diversity (119883 120575) can be embedded in acomplete diversity
Proof Let 119883 be the set of all Cauchy sequences in119883 Identify any two sequences 119909119894 119910119894 which satisfylim119899rarrinfin120575(119909119899 119910119899) = 0 (so 119883 is actually a set of equivalenceclasses) Define the function 120575 fromPfin(119883) rarr R by
120575 (1199091
119894 1199092
119894 119909
119899
119894)
= lim119873rarrinfin
sup1198941119894119899ge119873
120575 (1199091
119894119894
1199092
1198942
119909119899
119894119899
) (12)
It can then be shown that (119883 120575) is a complete diversityand that the map 119909 997891rarr 119909 119909 119909 from (119883 120575) is anembedding The proof is an exercise in notation
4 Journal of Function Spaces and Applications
This completion is dense in the sense that every member119909 of119883 has a sequence 119909119894 sube 119883with 119909119894 rarr 119909 in119883 (let 119910119894 be arepresentative of 119909 and define 119909119894 = 1199101 1199102 119910119894 119910119894 119910119894 )It also satisfies a universal property analogous to that formetric completion
Theorem 6 Let (119883 120575) be a diversity and let (119883 120575) be itscompletion Then for any complete diversity (119884 120574) and anyuniformly continuous function 119891 119883 rarr 119884 there is a uniqueuniformly continuous function 119891 119883 rarr 119884 which extends 119891
Proof Let 119909119894 be a representative sequence of somemembersof 119883 and define 119891(119909119894) = lim119894rarrinfin119891(119909119894) which is definedand independent of the representative since 119891 is uniformlycontinuous and 119884 is complete To show 119891 is uniformlycontinuous pick 120576 gt 0 and 119889 gt 0 such that 120574(119891(119860)) lt 120576
whenever 120575(119860) lt 119889 for all 119860 isin Pfin(119883) Then for all 119861 =
1199091
119894 1199092
119894 119909
119898
119894 isin Pfin(119883) with 120575(119861) lt 1198892 we have
120574(119891(119861)) = 120574(lim119894rarrinfin119891(119909119899
119894)119898
119899=1)) lt 120576 since for large enough
119873 120575(119891(119909119899
119873)119898
119899=1) lt 31198894
To show uniqueness of 119891 let 119892 be another uniformlycontinuous function extending 119891 to 119883 For all 119909 isin 119883 wehave 119909119894 sub 119883 with 119909119894 rarr 119909 in 119883 and by uniform continuity119892(119909) = lim119894rarrinfin119891(119909119894) = 119891(119909)
This is a universal property in the sense that for everycomplete diversity 119883
1015840 extending 119883 and having the propertythere is an isomorphism 119895 119883
1015840rarr 119883 (Specifically let 119895 be
the unique uniformly continuous extension of the identitymap 119895 119883 rarr 119883 to119883
1015840)
4 Conformities
In this section we introduce a generalization of diversitiesanalogous to uniformities which generalize metric spacesUniformities lie between metric spaces and topologies in thesense that every metric space defines a uniformity and everyuniformity defines a topology (which coincides with themetric topology when the uniformity came from a metric)Uniformities characterize uniform continuity uniform con-vergence and Cauchy sequences which are not topologicalconcepts
The carry-over from the metric case is natural butnontrivial since diversities can behave differently on sets ofdifferent cardinality Since this construction is qualitativelydifferent from metric uniformities it requires a differentname We asked ourselves ldquowhat would you call a uniformitythat came from a diversityrdquo and the answer was clear aconformity
Throughout this section we will give the analogousdefinitions and results for uniformities using the standardtreatment fromKelley [8]We begin by defining conformitiesand comparing them to uniformities we show that just likeuniformities conformities have a countable base if and onlyif they are generated by some pseudodiversity
We then briefly touch on the problem of completion forconformities
Finally we define power conformities from a conformitydefined on a set119883 we can construct a conformity onPfin(119883)
from which pseudodiversities can be considered uniformlycontinuous functions We show that every conformity isgenerated by exactly the set of pseudodiversities which areuniformly continuous from its power conformity to R Thisgives an equivalent definition of conformity in terms ofpseudodiversities
41 Conformities of Diversities Recall that for (119883 119889) a metricspace 119909119899 a sequence in119883 that 119909119899 is Cauchy if and only iffor each 120576 gt 0 there is some 119873 such that every pair of points(119909119894 119909119895) with 119894 gt 119873 119895 gt 119873 has 119889(119909119894 119909119895) lt 120576
Similarly let 119891 119883 rarr 119884 be a function between metricspaces (119883 119889) and (119884 119892) Then 119891 is uniformly continuous ifand only if each 120576 gt 0 has a 120575 gt 0 such that wheneverpairs of points (119909 119910) isin 119883 times 119883 satisfy 119889(119909 119910) lt 120575 the pairs(119891(119909) 119891(119910)) satisfy 119892(119891(119909) 119891(119910)) lt 120576
A similar characterization of uniform convergence ofsequences of functions can be given in terms of pairs ofpoints From these observations arises the theory of unifor-mities which is described in any standard text on analysis(cf [7 8]) We briefly describe the theory here For any set119883 define a uniformity on119883 as a filterU on119883 times 119883 satisfying
(U1) (119909 119909) isin 119880 for every 119909 isin 119883 119880 isin U
(U2) If 119880 isin U (119909 119910) isin 119880 then (119910 119909) isin 119880
(U3) For every119880 isin U there exists some119881 isin Uwith119881∘119881 sube
119880 where in general we define
119880 ∘ 119881 = (119909 119911) (119909 119910) isin 119880 (119910 119911) isin 119881 for some 119910 isin 119883
(13)
In particular for any pseudometric space (119883 119889) we candefine the metric uniformity as the filter on119883times119883 defined by
119880120576= (119909 119910) 119889 (119909 119910) lt 120576 (14)
for each 120576 gt 0 We see from this example that (U1) expressesthe requirement that 119889(119909 119909) = 0 for all 119909 isin 119883 (U2) expressessymmetry and (U3) expresses the triangle inequality
Uniform structure can be defined entirely with respect touniformities For example given sets 119883119884 and uniformitiesUV on 119883 and 119884 respectively we can call a function 119891
119883 rarr 119884 uniformly continuous if 119891minus1(119881) isin U for every
119881 isin V (Here 119891 acts on members of 119881 componentwise)A sequence 119909119899 sub 119883 is Cauchy if for every 119880 isin U thereis some 119873 such that pairs of elements (119909119894 119909119895) of 119909119899 are in119880 whenever 119894 119895 gt 119873 It is not hard to see that for metricuniformities these definitions coincide with the ordinaryones for metric spaces
To abstract the uniform structure of diversities unifor-mities are clearly insufficient For one thing since diversitiesmap finite sets rather than pairs we should seek a filteron Pfin(119883) rather than 119883 times 119883 Then symmetry is nolonger required but now monotonicity is Finally it is notmeaningful to compose finite sets as in (U3) so wewill need adiferent way to express an analogue of the triangle inequality
Journal of Function Spaces and Applications 5
Putting all this together we define a conformity C on 119883
as a filter onPfin(119883) satisfying
(C1) 119909 isin 119862 for every 119909 isin 119883 119862 isin C
(C2) For every 119862 isin C whenever 119860 isin 119862 and 119861 sube 119860 wehave 119861 isin 119862
(C3) For every 119862 isin C there exists some 119863 isin C with 119863 ∘
119863 sube 119862 where in general we define
119880 ∘ 119881 = 119906 cup V 119906 isin 119880 V isin 119881 and 119906 cap V = (15)
Often the term conformity is also used to refer to the pair(119883C)
An observation that will be necessary later (one whichalso holds for uniformities) is that for any119863 isin C (119863∘119863)∘119863 =
119863 ∘ (119863 ∘ 119863) so that 119863 ∘ 119863 ∘ 119863 is defined unambiguouslyTo estimate the size of this we also note that 119863 ∘ 119863 ∘ 119863 sube
(119863 ∘ 119863) ∘ (119863 ∘ 119863)As in the metric case there is a canonical way to generate
a conformity from a diversity if 120575 is a pseudodiversity on 119883we have the conformity generated by the sets
119862120576= 119860 120575 (119860) le 120576 = 120575
minus1[0 120576] (16)
for each 120576 gt 0 (This is equivalent to the one using strictinequalities but typographically nicer)
As in the metric case uniform structure can be definedon conformities in a way that generalizes that of diversitieslet (119883C) and (119884D) be conformities Then a function 119891 isuniformly continuous from 119883 to 119884 if for all 119863 isin D theset 119891
minus1(119889) 119889 isin 119863 is in C A sequence 119909119899 on 119883 is a
Cauchy sequence if for all119862 isin CPfin(119909119899119899ge119873) sube 119862 for someinteger119873 For conformities generated from diversities in theabove way these definitions coincide with those given in theprevious section
More generally given a collection of pseudodiversi-ties 120575120572120572isinA we can generate a conformity from the sets120575minus1
120572[0 120576]
120572isinA120576gt0 We therefore seek a characterization of
conformities in terms of the diversities which generate them(In a later section we will see that all conformities can bedescribed in this way so that we can define conformities interms of such sets) We begin by stating a result from Kelley[8] along with a summary of his proof
Theorem 7 A uniformity is generated by a single pseudomet-ric if and only if it has a countable base
The standard proof of this theorem goes as follows it isobvious that any uniformity generated by a pseudometric hasa countable base Conversely if there exists a countable basefor a uniformity on 119883 there exists a countable base 119880119899119899isinN
for which the following argument holds Define the function119891(119909 119910) = 2
minus119899 where 119899 = sup119894 (119909 119910) isin 119880119894 This generatesthe uniformity but does not satisfy the triangle inequality sodefine
119889 (119909 119910) = inf119898minus1
sum
119894=1
119891 (119909119894 119909119894+1) (17)
where the infimum is taken over all sequences 119909119894119898
119894=1with
1199091 = 119909 and 119909119898 = 119910 This clearly satisfies the triangleinequality so it just remains to be shown that 119889 generates theuniformity This is done by proving that 119889(119909 119910) le 119891(119909 119910) le
2119889(119909 119910) which follows from technical constraints on 119880119899Given a conformity with a countable base 119862119899 on a set119883
onemight try to translate this proof directly define a function119891(119860) Pfin(119883) rarr R by 119891(119860) = sup119894 119860 isin 119862119894 thensomehow tweak119891 to (a) satisfy the triangle inequality and (b)generate the same conformity as 119891 However it appears thatany direct analogue to the ldquoinfimum over all pathsrdquo strategyused in the metric case (there are several) cannot satisfy both(a) and (b) simultaneously
Nonetheless the result is true which is the content of thenext theorem
Lemma 8 Let (119883C) have a countable base Then it has acountable base 119862119899 satisfying 1198620 = P fin (119883)119862119894∘119862119894∘119862119894 sube 119862119894minus1
for 119894 gt 0
Proof Let 119881119899 be a countable base for C Define 1198820 =
Pfin(119883) 119882119899 = 119881119899 cap 119882119899minus1 Then 119882119899 is a nested countablebase Finally choose 119862119899 as 119862119894 = 119882119899
119894
where 119899119894 are choseninductively as 1198990 = 0 then (119882119899
119894
∘119882119899119894
)∘(119882119899119894
∘119882119899119894
) sube 119882119899119894minus1
Theorem 9 Let (119883C) be a conformity There exists apseudodiversity 120575 which generates C if and only if C has acountable base
Proof If 120575 exists the sets 1198621119899119899isinN are our baseConversely let 119862119899
infin
1be a base for C satisfying 1198620 =
Pfin(119883) and119862119894 ∘119862119894 ∘119862119894 sube 119862119894minus1 for 119894 gt 0 Define 1205751015840 onPfin(119883)
by
1205751015840(119860) =
0 119860 isin 119862119899 forall119899
2minus119896 119860 isin 119862119899 for 0 le 119899 le 119896 but 119860 notin 119862119896+1
(18)
Notice that for 119896 ge 0
1205751015840minus1
([0 2minus119896]) = 119862119896 (19)
and that 1205751015840 is monotonic by (C2) if 119860 sube 119861 then 119860 isin 119862119899
whenever 119861 isin 119862119899Define a chain as a sequence 119860 119894
119899
119894=1inPfin(119883) with119860 119894 cap
119860 119894minus1 = for 119894 = 2 119899 Define a cycle as a chain with 1198601 cap
119860119899 = Write
120575 (119860) = infchains covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
120575 (119860) = infcycles covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
(20)
Notice that 120575() = 120575() = 0We claim that 120575 is our desired pseudodiversity since the
sets (1205751015840)minus1[0 120576] generate the conformity and 120575 le 1205751015840le 4120575 We
prove this in three stages
6 Journal of Function Spaces and Applications
(S1) First of all 120575 is a pseudodiversity By (C1) for every119909 isin 119883 119899 isin N and 119909 isin 119862119899 so that 120575
1015840(119909) = 0 Also
119909 is a cycle covering itself so 120575(119909) = 0
The triangle equality also holds let 120576 gt 0 119860119862 isin
Pfin(119883) and 119861 isin Pfin(119883) be nonempty Choosecycles 119860 119894
119899
1and 119861119894
119898
1covering 119860 cup 119861 and 119861 cup 119862
respectively and for which
119899
sum
119894=1
1205751015840(119860 119894) le 120575 (119860 cup 119861) + 120576
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119861 cup 119862) + 120576
(21)
Then 119860 119894119899
1cup 119861119894
119898
1forms a cycle (after reordering)
covering 119860 cup 119862 so
120575 (119860 cup 119862) le
119899
sum
119894=1
1205751015840(119860 119894)
+
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119860 cup 119861) + 120575 (119861 cup 119862) + 2120576
(22)
(S2) Next we notice that
(i) every cycle is a chain so 120575 le 120575(ii) If 1198601 119860119899minus1 119860119899 is a chain then
1198601 119860119899minus1 119860119899 119860119899minus1 1198601 is a cyclemdashand the sum of 120575
1015840 over this cycle is less thantwice the sum of 1205751015840 over the original chain Weconclude that
120575 le 120575 le 2120575 (23)
(S3) Finally we claim that 120575 le 1205751015840le 2120575 This combined
with (23) will give the main result
Trivially 120575 le 1205751015840 For the other inequality choose
119860 isin Pfin(119883) Our strategy is to induct on the greatestinteger119873 such that 120575(119860) lt 2
minus119873
The case119873 = 0 is easy because then 1205751015840le 1 le 2120575 (this
also covers the case 120575(119860) = 1 which is not covered bythe induction) When 119873 gt 0 we can choose positive120576 less than (2
minus119873minus 120575(119860)) and a chain 119860 119894
119899
1with
119899
sum
119894=1
1205751015840(119860 119894) lt 120575 (119860) + 120576 lt 2
minus119873 (24)
If 119899 = 1 we have 1205751015840(119860) le 120575
1015840(1198601) lt 2
minus119873lt 2120575(119860)
Otherwise there is 119896 lt 119899 such that
119896minus1
sum
119894=1
1205751015840(119860 119894) le
120575 (119860)
2
119899
sum
119894=119896+1
1205751015840(119860 119894) le
120575 (119860)
2 (25)
Since 119860 119894119896minus1
1and 119860 119894
119899
119896+1are chains whose sum under 1205751015840 is
less than half that of 119860 119894119899
1 the inductive hypothesis applies
to them and we may write
1205751015840(1198601 cup sdot sdot sdot cup 119860119896minus1)
le 2120575 (1198601 cup sdot sdot sdot cup 119860119896minus1) inductive hypothesis
le 2
119896minus1
sum
119894=1
1205751015840(119860 119894) definition of 120575
le 120575 (119860) by (25)
lt 2minus119873
(26)
Similarly 1205751015840(119860119896+1 cup sdot sdot sdot cup 119860119899) lt 2minus119873 and 120575
1015840(119860119896) lt 2
minus119873 by(24) So
(1198601 cup sdot sdot sdot cup 119860119896minus1) isin 119862119873+1 119860119896 isin 119862119873+1
(119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873+1
(27)
Our double-composition hypothesis gives
(1198601 cup sdot sdot sdot cup 119860119896minus1) cup 119860119896 cup (119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873 (28)
And by monotonicity of 1205751015840
1205751015840(119860) le 120575
1015840(1198601 cup sdot sdot sdot cup 119860119899) le 2
minus119873le 2120575 (119860) (29)
This characterizes the conformities generated by singlepseudodiversities Later we will describe every conformity interms of the pseudodiversities that generate them
42 Induced Uniformities and Completeness Given a confor-mity C we define its induced uniformity as the uniformitygenerated by the sets
119880119862 = (119909 119910) 119909 119910 isin 119862 (30)
for every 119862 isin C It is straightforward to show that this isa uniformity since every singleton 119909 is in every 119862 isin Cwe have every pair (119909 119909) in every generator of the induceduniformity proving (U1) Since 119909 119910 = 119910 119909 we have (U2)Finally (U3) follows from the observation that whenever119909 119910 isin 119862 isin C and 119910 119911 isin 119863 isin C the set 119863 ∘ 119863 isin Ccontains 119909 119910 119911Then x 119911 isin 119863∘119862 by (C2) In other wordsif 119862 ∘ 119863 sube 119864 in the conformity then 119880119862 ∘ 119880119863 sube 119880119864 in theinduced uniformity Thus (U3) is implied by (C3)
Theorem 10 Let119883 be a set 120575119860119860isinA be a family of diversitieswhich generate a conformity C For each 120575119860 write 119889119860 for itsinduced metric Then the uniformity generated by the metrics119889119860119860isinA is exactly the induced uniformity ofC
Proof Denote by U119889 the uniformity generated by 119889119860119860isinAand byU119888 the uniformity induced byC A base forC is
119862120576119860 = 119865 120575119860 (119865) lt 120576 (31)
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
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2 Journal of Function Spaces and Applications
2 Preliminaries
Throughout this paper we will denote the finite power set ofa given set119883 by
Pfin (119883) = 119860 sube 119883 |119860| lt infin (1)
We begin with the Bryant-Tupper definition from [5]a diversity is a pair (119883 120575) where 119883 is some set and 120575
Pfin(119883) rarr R is a function satisfying
(D1) if 119860 isin Pfin(119883) 120575(119860) ge 0 and 120575(119860) = 0 iff |119860| le 1
(D2) if 119860 119861 119862 isin Pfin(119883) with 119862 = then
120575 (119860 cup 119861) le 120575 (119860 cup 119862) + 120575 (119862 cup 119861) (2)
If for some 119860 isin Pfin(119883) 120575(119860) = 0 but |119860| gt 1 wehave the weaker notion of a pseudodiversity It is shownin [5] from these axioms that if 119860 sube 119861 then 120575(119860) le
120575(119861) that is (pseudo)diversities are monotonic and that therestriction of a diversity to sets of size 2 forms a pseudometric119889(119909 119910) = 120575(119909 119910) We call this metric the induced metric ofthe diversity
For a metric space (119883 119889) there are two importantdiversities on119883 having 119889 as an induced metric as follows
(i) the diameter diversity (119883 diam119889) defined by
diam119889 (1199091 119909119899) = sup119894119895
119889 (119909119894 119909119895) (3)
when 119883 = R119899 and 119889 is the Euclidean metric we referto this diversity simply by diam
(ii) The Steiner tree diversity (119883 120575) is defined for eachfinite set 119860 sube 119883 as the infimum of the size of theminimum Steiner tree on 119860(Recall that a Steiner tree on 119860 is a tree whose vertexset 119881 satisfies 119860 sube 119881 sube 119883 with each edge (119909 119910)
weighted by 119889(119909 119910) The size of the tree is the sumof its edge weights)
In fact these examples are the extremes of diversitybehavior relative to their induced metrics in the sense thatfor any diversity (119883 120575
1015840) which induces a metric 119889 we have
diam119889 le 1205751015840le 120575 (4)
where 120575 is the Steiner tree diversity on (119883 119889) This can beshown by a straightforward argument (Bryant and Tupperupcoming)
To demonstrate the difference between the diameterand Steiner tree diversities consider the Euclidean metric(R3 119889) The induced metric of both the diameter and Steinertree diversity is the Euclidean metric For any finite set 119860
contained in an 120576-ball diam(119860) lt 120576 To contrast in any 120576-ballwe can find finite sets 119860 for which 120575(119860) is arbitrarily large
Theorem 1 The Steiner tree diversity function 120575 on R3 isunbounded on every open set of the Euclidean topology
Proof Without loss of generality we show that the result for120576-balls is about 0 For each 119899 isin N define
119866119899 = (119894
1198992119895
1198992119896
1198992) 0 le 119894 119895 119896 lt 119899 (5)
which is a grid of points contained in the cube [0 1119899]3 Sincethere are 1198993 points a minimum spanning tree connecting themembers of 119866119899 must have 119899
3minus 1 edges each of the lengths
ge 11198992 since that is the least distance between two points
Therefore the size of the minimum spanning tree on 119866119899 isat least (1198993 minus 1)119899
2 which can be taken as large as we likeby taking 119899 large enough Since the minimal Steiner tree on119866119899 has a size of at least 0615 times than that of the minimalspanning tree [9] we have 120575(119866119899) rarr infin as 119899 rarr infin eventhough diam(119866119899) rarr 0
A similar construction for the Steiner tree diversity onR2gives sets of diversity (0615 minus 120576) for every 120576 gt 0 in everyEuclidean ball On R the Steiner tree diversity and diameterdiversity are identical The dramatic difference between themany-point behavior of these two diversities in dimension 2or higher demonstrates that diversities are not characterizedby their induced metrics even up to a constant
In Section 2 and 3 we will define uniform convergenceuniform continuity and completeness explicitly in termsof an underlying diversity in Section 4 we will describeconformities which abstract these properties for diversitiesThis is in analogy to Weilrsquos uniformities which abstract thesame concepts for metric spaces
With this goal in mind we start with the followingdefinitions let (119883 120575119883) and (119884 120575119884) be diversities Given 119909 isin 119883a sequence 119909119899 sub 119883 converges to 119909 denoted 119909119899 rarr 119909 if
lim119873rarrinfin
sup11989411198942119894119899ge119873
120575119883 (119909 1199091198941
1199091198942
119909119894119899
) = 0 (6)
The sequence 119909119899 is a Cauchy sequence if
lim119873rarrinfin
sup11989411198942119894119899ge119873
120575119883 (1199091198941
1199091198942
119909119894119899
) = 0 (7)
From these definitions and the axioms (D1) and (D2) itcan be shown that limits are unique and every convergentsequence is Cauchy If every Cauchy sequence is convergentwe call the diversity complete
Finally if 119891 119883 rarr 119884 is a function such that for every 120576 gt
0 there exists some119889 gt 0 such that 120575119883(119860) lt 119889 rArr 120575119884(119891(119860)) lt
120576 for every 119860 isin Pfin(119883) we say 119891 is uniformly continuousIt is not hard to see that for diameter diversities these
definitions coincide exactly with the standard ones on theinduced metric
For the second half of the paper we will work extensivelywith filters so we state the definition here given a ground set119883 define a filter as a collection F of subsets of 119883 satisfying119860 cap 119861 whenever 119860 119861 are inF and 119861 isin F whenever 119861 supe 119860
and119860 isin F A filter base becomes a filter when all supersets ofits elements are added inwhich casewe say the base generatesthe filter
In this paper we additionally require that notin F
Journal of Function Spaces and Applications 3
3 Comparison with Metrics
In this section we contrast the convergence of sequences withrespect to diversities and their induced metrics In particularwe show that although the Cauchy property for sequencesis much stronger for diversities (we demonstrate a sequencewhich is not Cauchywith respect to a diversity even though itis Cauchy with respect to the induced metric) completenessof a diversity is equivalent to completeness of its inducedmetric This tells us that every diversity which induces aEuclidean metric (eg the Steiner tree diversity on R119899) iscomplete
Since the set of Cauchy sequences in a diversity maybe smaller than the set of Cauchy sequences of its inducedmetric this may provide a simpler way to determine com-pleteness of metric spaces
At the end of the section we construct the analogue ofcompletion for diversities
31 Completeness in Diversities and Metric Spaces
Theorem 2 Let (119883 120575) be a diversity and let 119889 be its inducedmetric If (119883 119889) is a complete metric space then (119883 120575) is acomplete diversity
Proof Suppose that (119883 119889) is complete Let 119909119899 be a Cauchysequence in (119883 120575) Then it is also Cauchy in (119883 119889) andtherefore converges to some element 119909 We claim that 119909119899 rarr
119909 in (119883 120575) To this end let 120576 gt 0 Then there exists 119873 suchthat
(i) 119889(119909119899 119909) lt 120576 for all 119899 gt 119873 (since 119909119899 rarr 119909 in (119883 119889))(ii) 120575(119909119899
1
1199091198992
119909119899119898
) lt 120576 for all 119899119894 gt 119873 (since 119909119899 isCauchy in (119883 120575))
Therefore for all 1198991 119899119898 gt 119873
120575 (119909 1199091198991
119909119899119898
) le 120575 (119909 1199091198991
) + 120575 (1199091198991
119909119899119898
)
= 119889 (119909 1199091198991
) + 120575 (1199091198991
119909119899119898
) lt 2120576
(8)
that is 119909119899 rarr 119909 in (119883 120575)
As mentioned the set of Cauchy sequences in a diversitymay be strictly smaller than the set of Cauchy sequences inthe induced metric For example let (119883 120575) be the Steinertree diversity on R3 and consider the sets 119866119899119899isinN fromTheorem 1
Order each set 119866119899 somehow and define the sequence 119909119894by concatenating them that is
119909119899 = 119866111986621198663 sdot sdot sdot (9)
which is Cauchy in the induced metric of (119883 120575) (sinceeventually every pair of points is confined to arbitrarily smallcubes [0 120576]
3) However it is not Cauchy in (119883 120575) since wesaw in the proof of Theorem 1 that 120575(119866119899) becomes arbitrarilylarge as 119899 rarr infin In other words every tail of 119909119899 hasarbitrarily large finite sets so 119909119899 is not Cauchy
In light of this example it is interesting to know that everycomplete diversity has a complete induced metric which isproved with the following lemma
Lemma 3 Let (119883 120575) be a diversity and let 119889 be its inducedmetric Let 119909119899 be Cauchy in (119883 119889)Then it has a subsequencethat is Cauchy in (119883 120575)
Proof Define the subsequence 119909119899119894
by
119899119894 = min 119899 119889 (119909119899 119909119898) lt 2minus119894
forall119898 ge 119899 (10)
Given 120576 gt 0 choose 119873 such that 21minus119873 lt 120576 Then for all 1198941 le1198942 le sdot sdot sdot le 119894119898 greater than119873
120575 (1199091198941
119909119894119898
) le 120575 (1199091198941
1199091198942
) + sdot sdot sdot + 120575 (119909119894119898minus1
119909119894119898
)
lt1
21198941+ sdot sdot sdot +
1
2119894119898lt
infin
sum
119894=119873
12119894
= 21minus119873
lt 120576
(11)
That is 119909119899119894
is Cauchy in (119883 120575)
Theorem 4 Let (119883 120575) be a diversity and let 119889 be its inducedmetric If (119883 120575) is a complete diversity then (119883 119889) is a completemetric space
Proof Let 119909119899 be a Cauchy sequence in (119883 119889) Then byLemma 3 it has a subsequence 119909119894
119899
that is Cauchy in (119883 120575)which converges to some element 119909 since the diversity iscomplete (it converges in both (119883 120575) and (119883 119889))
Then 119909119899 converges to 119909 in (119883 119889) since for any 120576 we have119889(119909119899 119909) le 119889(119909119899 119909119894
119898
) + 119889(119909119894119898
119909) lt 2120576 for 119898 119899 large enough
32 Completion In light of the equivalence between metriccompleteness and diversity completeness it is perhaps notso surprising that every diversity can be completed in acanonical way To do so we require twomore definitions from[5] an embedding 120587 1198841 rarr 1198842 is an injective map betweendiversities (1198841 1205751) and (1198842 1205752) such that 1205751(119860) = 1205752(120587(119860)) forall 119860 isin Pfin(1198841) A isomorphism is a surjective embedding
Theorem 5 Every diversity (119883 120575) can be embedded in acomplete diversity
Proof Let 119883 be the set of all Cauchy sequences in119883 Identify any two sequences 119909119894 119910119894 which satisfylim119899rarrinfin120575(119909119899 119910119899) = 0 (so 119883 is actually a set of equivalenceclasses) Define the function 120575 fromPfin(119883) rarr R by
120575 (1199091
119894 1199092
119894 119909
119899
119894)
= lim119873rarrinfin
sup1198941119894119899ge119873
120575 (1199091
119894119894
1199092
1198942
119909119899
119894119899
) (12)
It can then be shown that (119883 120575) is a complete diversityand that the map 119909 997891rarr 119909 119909 119909 from (119883 120575) is anembedding The proof is an exercise in notation
4 Journal of Function Spaces and Applications
This completion is dense in the sense that every member119909 of119883 has a sequence 119909119894 sube 119883with 119909119894 rarr 119909 in119883 (let 119910119894 be arepresentative of 119909 and define 119909119894 = 1199101 1199102 119910119894 119910119894 119910119894 )It also satisfies a universal property analogous to that formetric completion
Theorem 6 Let (119883 120575) be a diversity and let (119883 120575) be itscompletion Then for any complete diversity (119884 120574) and anyuniformly continuous function 119891 119883 rarr 119884 there is a uniqueuniformly continuous function 119891 119883 rarr 119884 which extends 119891
Proof Let 119909119894 be a representative sequence of somemembersof 119883 and define 119891(119909119894) = lim119894rarrinfin119891(119909119894) which is definedand independent of the representative since 119891 is uniformlycontinuous and 119884 is complete To show 119891 is uniformlycontinuous pick 120576 gt 0 and 119889 gt 0 such that 120574(119891(119860)) lt 120576
whenever 120575(119860) lt 119889 for all 119860 isin Pfin(119883) Then for all 119861 =
1199091
119894 1199092
119894 119909
119898
119894 isin Pfin(119883) with 120575(119861) lt 1198892 we have
120574(119891(119861)) = 120574(lim119894rarrinfin119891(119909119899
119894)119898
119899=1)) lt 120576 since for large enough
119873 120575(119891(119909119899
119873)119898
119899=1) lt 31198894
To show uniqueness of 119891 let 119892 be another uniformlycontinuous function extending 119891 to 119883 For all 119909 isin 119883 wehave 119909119894 sub 119883 with 119909119894 rarr 119909 in 119883 and by uniform continuity119892(119909) = lim119894rarrinfin119891(119909119894) = 119891(119909)
This is a universal property in the sense that for everycomplete diversity 119883
1015840 extending 119883 and having the propertythere is an isomorphism 119895 119883
1015840rarr 119883 (Specifically let 119895 be
the unique uniformly continuous extension of the identitymap 119895 119883 rarr 119883 to119883
1015840)
4 Conformities
In this section we introduce a generalization of diversitiesanalogous to uniformities which generalize metric spacesUniformities lie between metric spaces and topologies in thesense that every metric space defines a uniformity and everyuniformity defines a topology (which coincides with themetric topology when the uniformity came from a metric)Uniformities characterize uniform continuity uniform con-vergence and Cauchy sequences which are not topologicalconcepts
The carry-over from the metric case is natural butnontrivial since diversities can behave differently on sets ofdifferent cardinality Since this construction is qualitativelydifferent from metric uniformities it requires a differentname We asked ourselves ldquowhat would you call a uniformitythat came from a diversityrdquo and the answer was clear aconformity
Throughout this section we will give the analogousdefinitions and results for uniformities using the standardtreatment fromKelley [8]We begin by defining conformitiesand comparing them to uniformities we show that just likeuniformities conformities have a countable base if and onlyif they are generated by some pseudodiversity
We then briefly touch on the problem of completion forconformities
Finally we define power conformities from a conformitydefined on a set119883 we can construct a conformity onPfin(119883)
from which pseudodiversities can be considered uniformlycontinuous functions We show that every conformity isgenerated by exactly the set of pseudodiversities which areuniformly continuous from its power conformity to R Thisgives an equivalent definition of conformity in terms ofpseudodiversities
41 Conformities of Diversities Recall that for (119883 119889) a metricspace 119909119899 a sequence in119883 that 119909119899 is Cauchy if and only iffor each 120576 gt 0 there is some 119873 such that every pair of points(119909119894 119909119895) with 119894 gt 119873 119895 gt 119873 has 119889(119909119894 119909119895) lt 120576
Similarly let 119891 119883 rarr 119884 be a function between metricspaces (119883 119889) and (119884 119892) Then 119891 is uniformly continuous ifand only if each 120576 gt 0 has a 120575 gt 0 such that wheneverpairs of points (119909 119910) isin 119883 times 119883 satisfy 119889(119909 119910) lt 120575 the pairs(119891(119909) 119891(119910)) satisfy 119892(119891(119909) 119891(119910)) lt 120576
A similar characterization of uniform convergence ofsequences of functions can be given in terms of pairs ofpoints From these observations arises the theory of unifor-mities which is described in any standard text on analysis(cf [7 8]) We briefly describe the theory here For any set119883 define a uniformity on119883 as a filterU on119883 times 119883 satisfying
(U1) (119909 119909) isin 119880 for every 119909 isin 119883 119880 isin U
(U2) If 119880 isin U (119909 119910) isin 119880 then (119910 119909) isin 119880
(U3) For every119880 isin U there exists some119881 isin Uwith119881∘119881 sube
119880 where in general we define
119880 ∘ 119881 = (119909 119911) (119909 119910) isin 119880 (119910 119911) isin 119881 for some 119910 isin 119883
(13)
In particular for any pseudometric space (119883 119889) we candefine the metric uniformity as the filter on119883times119883 defined by
119880120576= (119909 119910) 119889 (119909 119910) lt 120576 (14)
for each 120576 gt 0 We see from this example that (U1) expressesthe requirement that 119889(119909 119909) = 0 for all 119909 isin 119883 (U2) expressessymmetry and (U3) expresses the triangle inequality
Uniform structure can be defined entirely with respect touniformities For example given sets 119883119884 and uniformitiesUV on 119883 and 119884 respectively we can call a function 119891
119883 rarr 119884 uniformly continuous if 119891minus1(119881) isin U for every
119881 isin V (Here 119891 acts on members of 119881 componentwise)A sequence 119909119899 sub 119883 is Cauchy if for every 119880 isin U thereis some 119873 such that pairs of elements (119909119894 119909119895) of 119909119899 are in119880 whenever 119894 119895 gt 119873 It is not hard to see that for metricuniformities these definitions coincide with the ordinaryones for metric spaces
To abstract the uniform structure of diversities unifor-mities are clearly insufficient For one thing since diversitiesmap finite sets rather than pairs we should seek a filteron Pfin(119883) rather than 119883 times 119883 Then symmetry is nolonger required but now monotonicity is Finally it is notmeaningful to compose finite sets as in (U3) so wewill need adiferent way to express an analogue of the triangle inequality
Journal of Function Spaces and Applications 5
Putting all this together we define a conformity C on 119883
as a filter onPfin(119883) satisfying
(C1) 119909 isin 119862 for every 119909 isin 119883 119862 isin C
(C2) For every 119862 isin C whenever 119860 isin 119862 and 119861 sube 119860 wehave 119861 isin 119862
(C3) For every 119862 isin C there exists some 119863 isin C with 119863 ∘
119863 sube 119862 where in general we define
119880 ∘ 119881 = 119906 cup V 119906 isin 119880 V isin 119881 and 119906 cap V = (15)
Often the term conformity is also used to refer to the pair(119883C)
An observation that will be necessary later (one whichalso holds for uniformities) is that for any119863 isin C (119863∘119863)∘119863 =
119863 ∘ (119863 ∘ 119863) so that 119863 ∘ 119863 ∘ 119863 is defined unambiguouslyTo estimate the size of this we also note that 119863 ∘ 119863 ∘ 119863 sube
(119863 ∘ 119863) ∘ (119863 ∘ 119863)As in the metric case there is a canonical way to generate
a conformity from a diversity if 120575 is a pseudodiversity on 119883we have the conformity generated by the sets
119862120576= 119860 120575 (119860) le 120576 = 120575
minus1[0 120576] (16)
for each 120576 gt 0 (This is equivalent to the one using strictinequalities but typographically nicer)
As in the metric case uniform structure can be definedon conformities in a way that generalizes that of diversitieslet (119883C) and (119884D) be conformities Then a function 119891 isuniformly continuous from 119883 to 119884 if for all 119863 isin D theset 119891
minus1(119889) 119889 isin 119863 is in C A sequence 119909119899 on 119883 is a
Cauchy sequence if for all119862 isin CPfin(119909119899119899ge119873) sube 119862 for someinteger119873 For conformities generated from diversities in theabove way these definitions coincide with those given in theprevious section
More generally given a collection of pseudodiversi-ties 120575120572120572isinA we can generate a conformity from the sets120575minus1
120572[0 120576]
120572isinA120576gt0 We therefore seek a characterization of
conformities in terms of the diversities which generate them(In a later section we will see that all conformities can bedescribed in this way so that we can define conformities interms of such sets) We begin by stating a result from Kelley[8] along with a summary of his proof
Theorem 7 A uniformity is generated by a single pseudomet-ric if and only if it has a countable base
The standard proof of this theorem goes as follows it isobvious that any uniformity generated by a pseudometric hasa countable base Conversely if there exists a countable basefor a uniformity on 119883 there exists a countable base 119880119899119899isinN
for which the following argument holds Define the function119891(119909 119910) = 2
minus119899 where 119899 = sup119894 (119909 119910) isin 119880119894 This generatesthe uniformity but does not satisfy the triangle inequality sodefine
119889 (119909 119910) = inf119898minus1
sum
119894=1
119891 (119909119894 119909119894+1) (17)
where the infimum is taken over all sequences 119909119894119898
119894=1with
1199091 = 119909 and 119909119898 = 119910 This clearly satisfies the triangleinequality so it just remains to be shown that 119889 generates theuniformity This is done by proving that 119889(119909 119910) le 119891(119909 119910) le
2119889(119909 119910) which follows from technical constraints on 119880119899Given a conformity with a countable base 119862119899 on a set119883
onemight try to translate this proof directly define a function119891(119860) Pfin(119883) rarr R by 119891(119860) = sup119894 119860 isin 119862119894 thensomehow tweak119891 to (a) satisfy the triangle inequality and (b)generate the same conformity as 119891 However it appears thatany direct analogue to the ldquoinfimum over all pathsrdquo strategyused in the metric case (there are several) cannot satisfy both(a) and (b) simultaneously
Nonetheless the result is true which is the content of thenext theorem
Lemma 8 Let (119883C) have a countable base Then it has acountable base 119862119899 satisfying 1198620 = P fin (119883)119862119894∘119862119894∘119862119894 sube 119862119894minus1
for 119894 gt 0
Proof Let 119881119899 be a countable base for C Define 1198820 =
Pfin(119883) 119882119899 = 119881119899 cap 119882119899minus1 Then 119882119899 is a nested countablebase Finally choose 119862119899 as 119862119894 = 119882119899
119894
where 119899119894 are choseninductively as 1198990 = 0 then (119882119899
119894
∘119882119899119894
)∘(119882119899119894
∘119882119899119894
) sube 119882119899119894minus1
Theorem 9 Let (119883C) be a conformity There exists apseudodiversity 120575 which generates C if and only if C has acountable base
Proof If 120575 exists the sets 1198621119899119899isinN are our baseConversely let 119862119899
infin
1be a base for C satisfying 1198620 =
Pfin(119883) and119862119894 ∘119862119894 ∘119862119894 sube 119862119894minus1 for 119894 gt 0 Define 1205751015840 onPfin(119883)
by
1205751015840(119860) =
0 119860 isin 119862119899 forall119899
2minus119896 119860 isin 119862119899 for 0 le 119899 le 119896 but 119860 notin 119862119896+1
(18)
Notice that for 119896 ge 0
1205751015840minus1
([0 2minus119896]) = 119862119896 (19)
and that 1205751015840 is monotonic by (C2) if 119860 sube 119861 then 119860 isin 119862119899
whenever 119861 isin 119862119899Define a chain as a sequence 119860 119894
119899
119894=1inPfin(119883) with119860 119894 cap
119860 119894minus1 = for 119894 = 2 119899 Define a cycle as a chain with 1198601 cap
119860119899 = Write
120575 (119860) = infchains covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
120575 (119860) = infcycles covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
(20)
Notice that 120575() = 120575() = 0We claim that 120575 is our desired pseudodiversity since the
sets (1205751015840)minus1[0 120576] generate the conformity and 120575 le 1205751015840le 4120575 We
prove this in three stages
6 Journal of Function Spaces and Applications
(S1) First of all 120575 is a pseudodiversity By (C1) for every119909 isin 119883 119899 isin N and 119909 isin 119862119899 so that 120575
1015840(119909) = 0 Also
119909 is a cycle covering itself so 120575(119909) = 0
The triangle equality also holds let 120576 gt 0 119860119862 isin
Pfin(119883) and 119861 isin Pfin(119883) be nonempty Choosecycles 119860 119894
119899
1and 119861119894
119898
1covering 119860 cup 119861 and 119861 cup 119862
respectively and for which
119899
sum
119894=1
1205751015840(119860 119894) le 120575 (119860 cup 119861) + 120576
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119861 cup 119862) + 120576
(21)
Then 119860 119894119899
1cup 119861119894
119898
1forms a cycle (after reordering)
covering 119860 cup 119862 so
120575 (119860 cup 119862) le
119899
sum
119894=1
1205751015840(119860 119894)
+
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119860 cup 119861) + 120575 (119861 cup 119862) + 2120576
(22)
(S2) Next we notice that
(i) every cycle is a chain so 120575 le 120575(ii) If 1198601 119860119899minus1 119860119899 is a chain then
1198601 119860119899minus1 119860119899 119860119899minus1 1198601 is a cyclemdashand the sum of 120575
1015840 over this cycle is less thantwice the sum of 1205751015840 over the original chain Weconclude that
120575 le 120575 le 2120575 (23)
(S3) Finally we claim that 120575 le 1205751015840le 2120575 This combined
with (23) will give the main result
Trivially 120575 le 1205751015840 For the other inequality choose
119860 isin Pfin(119883) Our strategy is to induct on the greatestinteger119873 such that 120575(119860) lt 2
minus119873
The case119873 = 0 is easy because then 1205751015840le 1 le 2120575 (this
also covers the case 120575(119860) = 1 which is not covered bythe induction) When 119873 gt 0 we can choose positive120576 less than (2
minus119873minus 120575(119860)) and a chain 119860 119894
119899
1with
119899
sum
119894=1
1205751015840(119860 119894) lt 120575 (119860) + 120576 lt 2
minus119873 (24)
If 119899 = 1 we have 1205751015840(119860) le 120575
1015840(1198601) lt 2
minus119873lt 2120575(119860)
Otherwise there is 119896 lt 119899 such that
119896minus1
sum
119894=1
1205751015840(119860 119894) le
120575 (119860)
2
119899
sum
119894=119896+1
1205751015840(119860 119894) le
120575 (119860)
2 (25)
Since 119860 119894119896minus1
1and 119860 119894
119899
119896+1are chains whose sum under 1205751015840 is
less than half that of 119860 119894119899
1 the inductive hypothesis applies
to them and we may write
1205751015840(1198601 cup sdot sdot sdot cup 119860119896minus1)
le 2120575 (1198601 cup sdot sdot sdot cup 119860119896minus1) inductive hypothesis
le 2
119896minus1
sum
119894=1
1205751015840(119860 119894) definition of 120575
le 120575 (119860) by (25)
lt 2minus119873
(26)
Similarly 1205751015840(119860119896+1 cup sdot sdot sdot cup 119860119899) lt 2minus119873 and 120575
1015840(119860119896) lt 2
minus119873 by(24) So
(1198601 cup sdot sdot sdot cup 119860119896minus1) isin 119862119873+1 119860119896 isin 119862119873+1
(119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873+1
(27)
Our double-composition hypothesis gives
(1198601 cup sdot sdot sdot cup 119860119896minus1) cup 119860119896 cup (119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873 (28)
And by monotonicity of 1205751015840
1205751015840(119860) le 120575
1015840(1198601 cup sdot sdot sdot cup 119860119899) le 2
minus119873le 2120575 (119860) (29)
This characterizes the conformities generated by singlepseudodiversities Later we will describe every conformity interms of the pseudodiversities that generate them
42 Induced Uniformities and Completeness Given a confor-mity C we define its induced uniformity as the uniformitygenerated by the sets
119880119862 = (119909 119910) 119909 119910 isin 119862 (30)
for every 119862 isin C It is straightforward to show that this isa uniformity since every singleton 119909 is in every 119862 isin Cwe have every pair (119909 119909) in every generator of the induceduniformity proving (U1) Since 119909 119910 = 119910 119909 we have (U2)Finally (U3) follows from the observation that whenever119909 119910 isin 119862 isin C and 119910 119911 isin 119863 isin C the set 119863 ∘ 119863 isin Ccontains 119909 119910 119911Then x 119911 isin 119863∘119862 by (C2) In other wordsif 119862 ∘ 119863 sube 119864 in the conformity then 119880119862 ∘ 119880119863 sube 119880119864 in theinduced uniformity Thus (U3) is implied by (C3)
Theorem 10 Let119883 be a set 120575119860119860isinA be a family of diversitieswhich generate a conformity C For each 120575119860 write 119889119860 for itsinduced metric Then the uniformity generated by the metrics119889119860119860isinA is exactly the induced uniformity ofC
Proof Denote by U119889 the uniformity generated by 119889119860119860isinAand byU119888 the uniformity induced byC A base forC is
119862120576119860 = 119865 120575119860 (119865) lt 120576 (31)
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Discrete Dynamics in Nature and Society
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Stochastic AnalysisInternational Journal of
Journal of Function Spaces and Applications 3
3 Comparison with Metrics
In this section we contrast the convergence of sequences withrespect to diversities and their induced metrics In particularwe show that although the Cauchy property for sequencesis much stronger for diversities (we demonstrate a sequencewhich is not Cauchywith respect to a diversity even though itis Cauchy with respect to the induced metric) completenessof a diversity is equivalent to completeness of its inducedmetric This tells us that every diversity which induces aEuclidean metric (eg the Steiner tree diversity on R119899) iscomplete
Since the set of Cauchy sequences in a diversity maybe smaller than the set of Cauchy sequences of its inducedmetric this may provide a simpler way to determine com-pleteness of metric spaces
At the end of the section we construct the analogue ofcompletion for diversities
31 Completeness in Diversities and Metric Spaces
Theorem 2 Let (119883 120575) be a diversity and let 119889 be its inducedmetric If (119883 119889) is a complete metric space then (119883 120575) is acomplete diversity
Proof Suppose that (119883 119889) is complete Let 119909119899 be a Cauchysequence in (119883 120575) Then it is also Cauchy in (119883 119889) andtherefore converges to some element 119909 We claim that 119909119899 rarr
119909 in (119883 120575) To this end let 120576 gt 0 Then there exists 119873 suchthat
(i) 119889(119909119899 119909) lt 120576 for all 119899 gt 119873 (since 119909119899 rarr 119909 in (119883 119889))(ii) 120575(119909119899
1
1199091198992
119909119899119898
) lt 120576 for all 119899119894 gt 119873 (since 119909119899 isCauchy in (119883 120575))
Therefore for all 1198991 119899119898 gt 119873
120575 (119909 1199091198991
119909119899119898
) le 120575 (119909 1199091198991
) + 120575 (1199091198991
119909119899119898
)
= 119889 (119909 1199091198991
) + 120575 (1199091198991
119909119899119898
) lt 2120576
(8)
that is 119909119899 rarr 119909 in (119883 120575)
As mentioned the set of Cauchy sequences in a diversitymay be strictly smaller than the set of Cauchy sequences inthe induced metric For example let (119883 120575) be the Steinertree diversity on R3 and consider the sets 119866119899119899isinN fromTheorem 1
Order each set 119866119899 somehow and define the sequence 119909119894by concatenating them that is
119909119899 = 119866111986621198663 sdot sdot sdot (9)
which is Cauchy in the induced metric of (119883 120575) (sinceeventually every pair of points is confined to arbitrarily smallcubes [0 120576]
3) However it is not Cauchy in (119883 120575) since wesaw in the proof of Theorem 1 that 120575(119866119899) becomes arbitrarilylarge as 119899 rarr infin In other words every tail of 119909119899 hasarbitrarily large finite sets so 119909119899 is not Cauchy
In light of this example it is interesting to know that everycomplete diversity has a complete induced metric which isproved with the following lemma
Lemma 3 Let (119883 120575) be a diversity and let 119889 be its inducedmetric Let 119909119899 be Cauchy in (119883 119889)Then it has a subsequencethat is Cauchy in (119883 120575)
Proof Define the subsequence 119909119899119894
by
119899119894 = min 119899 119889 (119909119899 119909119898) lt 2minus119894
forall119898 ge 119899 (10)
Given 120576 gt 0 choose 119873 such that 21minus119873 lt 120576 Then for all 1198941 le1198942 le sdot sdot sdot le 119894119898 greater than119873
120575 (1199091198941
119909119894119898
) le 120575 (1199091198941
1199091198942
) + sdot sdot sdot + 120575 (119909119894119898minus1
119909119894119898
)
lt1
21198941+ sdot sdot sdot +
1
2119894119898lt
infin
sum
119894=119873
12119894
= 21minus119873
lt 120576
(11)
That is 119909119899119894
is Cauchy in (119883 120575)
Theorem 4 Let (119883 120575) be a diversity and let 119889 be its inducedmetric If (119883 120575) is a complete diversity then (119883 119889) is a completemetric space
Proof Let 119909119899 be a Cauchy sequence in (119883 119889) Then byLemma 3 it has a subsequence 119909119894
119899
that is Cauchy in (119883 120575)which converges to some element 119909 since the diversity iscomplete (it converges in both (119883 120575) and (119883 119889))
Then 119909119899 converges to 119909 in (119883 119889) since for any 120576 we have119889(119909119899 119909) le 119889(119909119899 119909119894
119898
) + 119889(119909119894119898
119909) lt 2120576 for 119898 119899 large enough
32 Completion In light of the equivalence between metriccompleteness and diversity completeness it is perhaps notso surprising that every diversity can be completed in acanonical way To do so we require twomore definitions from[5] an embedding 120587 1198841 rarr 1198842 is an injective map betweendiversities (1198841 1205751) and (1198842 1205752) such that 1205751(119860) = 1205752(120587(119860)) forall 119860 isin Pfin(1198841) A isomorphism is a surjective embedding
Theorem 5 Every diversity (119883 120575) can be embedded in acomplete diversity
Proof Let 119883 be the set of all Cauchy sequences in119883 Identify any two sequences 119909119894 119910119894 which satisfylim119899rarrinfin120575(119909119899 119910119899) = 0 (so 119883 is actually a set of equivalenceclasses) Define the function 120575 fromPfin(119883) rarr R by
120575 (1199091
119894 1199092
119894 119909
119899
119894)
= lim119873rarrinfin
sup1198941119894119899ge119873
120575 (1199091
119894119894
1199092
1198942
119909119899
119894119899
) (12)
It can then be shown that (119883 120575) is a complete diversityand that the map 119909 997891rarr 119909 119909 119909 from (119883 120575) is anembedding The proof is an exercise in notation
4 Journal of Function Spaces and Applications
This completion is dense in the sense that every member119909 of119883 has a sequence 119909119894 sube 119883with 119909119894 rarr 119909 in119883 (let 119910119894 be arepresentative of 119909 and define 119909119894 = 1199101 1199102 119910119894 119910119894 119910119894 )It also satisfies a universal property analogous to that formetric completion
Theorem 6 Let (119883 120575) be a diversity and let (119883 120575) be itscompletion Then for any complete diversity (119884 120574) and anyuniformly continuous function 119891 119883 rarr 119884 there is a uniqueuniformly continuous function 119891 119883 rarr 119884 which extends 119891
Proof Let 119909119894 be a representative sequence of somemembersof 119883 and define 119891(119909119894) = lim119894rarrinfin119891(119909119894) which is definedand independent of the representative since 119891 is uniformlycontinuous and 119884 is complete To show 119891 is uniformlycontinuous pick 120576 gt 0 and 119889 gt 0 such that 120574(119891(119860)) lt 120576
whenever 120575(119860) lt 119889 for all 119860 isin Pfin(119883) Then for all 119861 =
1199091
119894 1199092
119894 119909
119898
119894 isin Pfin(119883) with 120575(119861) lt 1198892 we have
120574(119891(119861)) = 120574(lim119894rarrinfin119891(119909119899
119894)119898
119899=1)) lt 120576 since for large enough
119873 120575(119891(119909119899
119873)119898
119899=1) lt 31198894
To show uniqueness of 119891 let 119892 be another uniformlycontinuous function extending 119891 to 119883 For all 119909 isin 119883 wehave 119909119894 sub 119883 with 119909119894 rarr 119909 in 119883 and by uniform continuity119892(119909) = lim119894rarrinfin119891(119909119894) = 119891(119909)
This is a universal property in the sense that for everycomplete diversity 119883
1015840 extending 119883 and having the propertythere is an isomorphism 119895 119883
1015840rarr 119883 (Specifically let 119895 be
the unique uniformly continuous extension of the identitymap 119895 119883 rarr 119883 to119883
1015840)
4 Conformities
In this section we introduce a generalization of diversitiesanalogous to uniformities which generalize metric spacesUniformities lie between metric spaces and topologies in thesense that every metric space defines a uniformity and everyuniformity defines a topology (which coincides with themetric topology when the uniformity came from a metric)Uniformities characterize uniform continuity uniform con-vergence and Cauchy sequences which are not topologicalconcepts
The carry-over from the metric case is natural butnontrivial since diversities can behave differently on sets ofdifferent cardinality Since this construction is qualitativelydifferent from metric uniformities it requires a differentname We asked ourselves ldquowhat would you call a uniformitythat came from a diversityrdquo and the answer was clear aconformity
Throughout this section we will give the analogousdefinitions and results for uniformities using the standardtreatment fromKelley [8]We begin by defining conformitiesand comparing them to uniformities we show that just likeuniformities conformities have a countable base if and onlyif they are generated by some pseudodiversity
We then briefly touch on the problem of completion forconformities
Finally we define power conformities from a conformitydefined on a set119883 we can construct a conformity onPfin(119883)
from which pseudodiversities can be considered uniformlycontinuous functions We show that every conformity isgenerated by exactly the set of pseudodiversities which areuniformly continuous from its power conformity to R Thisgives an equivalent definition of conformity in terms ofpseudodiversities
41 Conformities of Diversities Recall that for (119883 119889) a metricspace 119909119899 a sequence in119883 that 119909119899 is Cauchy if and only iffor each 120576 gt 0 there is some 119873 such that every pair of points(119909119894 119909119895) with 119894 gt 119873 119895 gt 119873 has 119889(119909119894 119909119895) lt 120576
Similarly let 119891 119883 rarr 119884 be a function between metricspaces (119883 119889) and (119884 119892) Then 119891 is uniformly continuous ifand only if each 120576 gt 0 has a 120575 gt 0 such that wheneverpairs of points (119909 119910) isin 119883 times 119883 satisfy 119889(119909 119910) lt 120575 the pairs(119891(119909) 119891(119910)) satisfy 119892(119891(119909) 119891(119910)) lt 120576
A similar characterization of uniform convergence ofsequences of functions can be given in terms of pairs ofpoints From these observations arises the theory of unifor-mities which is described in any standard text on analysis(cf [7 8]) We briefly describe the theory here For any set119883 define a uniformity on119883 as a filterU on119883 times 119883 satisfying
(U1) (119909 119909) isin 119880 for every 119909 isin 119883 119880 isin U
(U2) If 119880 isin U (119909 119910) isin 119880 then (119910 119909) isin 119880
(U3) For every119880 isin U there exists some119881 isin Uwith119881∘119881 sube
119880 where in general we define
119880 ∘ 119881 = (119909 119911) (119909 119910) isin 119880 (119910 119911) isin 119881 for some 119910 isin 119883
(13)
In particular for any pseudometric space (119883 119889) we candefine the metric uniformity as the filter on119883times119883 defined by
119880120576= (119909 119910) 119889 (119909 119910) lt 120576 (14)
for each 120576 gt 0 We see from this example that (U1) expressesthe requirement that 119889(119909 119909) = 0 for all 119909 isin 119883 (U2) expressessymmetry and (U3) expresses the triangle inequality
Uniform structure can be defined entirely with respect touniformities For example given sets 119883119884 and uniformitiesUV on 119883 and 119884 respectively we can call a function 119891
119883 rarr 119884 uniformly continuous if 119891minus1(119881) isin U for every
119881 isin V (Here 119891 acts on members of 119881 componentwise)A sequence 119909119899 sub 119883 is Cauchy if for every 119880 isin U thereis some 119873 such that pairs of elements (119909119894 119909119895) of 119909119899 are in119880 whenever 119894 119895 gt 119873 It is not hard to see that for metricuniformities these definitions coincide with the ordinaryones for metric spaces
To abstract the uniform structure of diversities unifor-mities are clearly insufficient For one thing since diversitiesmap finite sets rather than pairs we should seek a filteron Pfin(119883) rather than 119883 times 119883 Then symmetry is nolonger required but now monotonicity is Finally it is notmeaningful to compose finite sets as in (U3) so wewill need adiferent way to express an analogue of the triangle inequality
Journal of Function Spaces and Applications 5
Putting all this together we define a conformity C on 119883
as a filter onPfin(119883) satisfying
(C1) 119909 isin 119862 for every 119909 isin 119883 119862 isin C
(C2) For every 119862 isin C whenever 119860 isin 119862 and 119861 sube 119860 wehave 119861 isin 119862
(C3) For every 119862 isin C there exists some 119863 isin C with 119863 ∘
119863 sube 119862 where in general we define
119880 ∘ 119881 = 119906 cup V 119906 isin 119880 V isin 119881 and 119906 cap V = (15)
Often the term conformity is also used to refer to the pair(119883C)
An observation that will be necessary later (one whichalso holds for uniformities) is that for any119863 isin C (119863∘119863)∘119863 =
119863 ∘ (119863 ∘ 119863) so that 119863 ∘ 119863 ∘ 119863 is defined unambiguouslyTo estimate the size of this we also note that 119863 ∘ 119863 ∘ 119863 sube
(119863 ∘ 119863) ∘ (119863 ∘ 119863)As in the metric case there is a canonical way to generate
a conformity from a diversity if 120575 is a pseudodiversity on 119883we have the conformity generated by the sets
119862120576= 119860 120575 (119860) le 120576 = 120575
minus1[0 120576] (16)
for each 120576 gt 0 (This is equivalent to the one using strictinequalities but typographically nicer)
As in the metric case uniform structure can be definedon conformities in a way that generalizes that of diversitieslet (119883C) and (119884D) be conformities Then a function 119891 isuniformly continuous from 119883 to 119884 if for all 119863 isin D theset 119891
minus1(119889) 119889 isin 119863 is in C A sequence 119909119899 on 119883 is a
Cauchy sequence if for all119862 isin CPfin(119909119899119899ge119873) sube 119862 for someinteger119873 For conformities generated from diversities in theabove way these definitions coincide with those given in theprevious section
More generally given a collection of pseudodiversi-ties 120575120572120572isinA we can generate a conformity from the sets120575minus1
120572[0 120576]
120572isinA120576gt0 We therefore seek a characterization of
conformities in terms of the diversities which generate them(In a later section we will see that all conformities can bedescribed in this way so that we can define conformities interms of such sets) We begin by stating a result from Kelley[8] along with a summary of his proof
Theorem 7 A uniformity is generated by a single pseudomet-ric if and only if it has a countable base
The standard proof of this theorem goes as follows it isobvious that any uniformity generated by a pseudometric hasa countable base Conversely if there exists a countable basefor a uniformity on 119883 there exists a countable base 119880119899119899isinN
for which the following argument holds Define the function119891(119909 119910) = 2
minus119899 where 119899 = sup119894 (119909 119910) isin 119880119894 This generatesthe uniformity but does not satisfy the triangle inequality sodefine
119889 (119909 119910) = inf119898minus1
sum
119894=1
119891 (119909119894 119909119894+1) (17)
where the infimum is taken over all sequences 119909119894119898
119894=1with
1199091 = 119909 and 119909119898 = 119910 This clearly satisfies the triangleinequality so it just remains to be shown that 119889 generates theuniformity This is done by proving that 119889(119909 119910) le 119891(119909 119910) le
2119889(119909 119910) which follows from technical constraints on 119880119899Given a conformity with a countable base 119862119899 on a set119883
onemight try to translate this proof directly define a function119891(119860) Pfin(119883) rarr R by 119891(119860) = sup119894 119860 isin 119862119894 thensomehow tweak119891 to (a) satisfy the triangle inequality and (b)generate the same conformity as 119891 However it appears thatany direct analogue to the ldquoinfimum over all pathsrdquo strategyused in the metric case (there are several) cannot satisfy both(a) and (b) simultaneously
Nonetheless the result is true which is the content of thenext theorem
Lemma 8 Let (119883C) have a countable base Then it has acountable base 119862119899 satisfying 1198620 = P fin (119883)119862119894∘119862119894∘119862119894 sube 119862119894minus1
for 119894 gt 0
Proof Let 119881119899 be a countable base for C Define 1198820 =
Pfin(119883) 119882119899 = 119881119899 cap 119882119899minus1 Then 119882119899 is a nested countablebase Finally choose 119862119899 as 119862119894 = 119882119899
119894
where 119899119894 are choseninductively as 1198990 = 0 then (119882119899
119894
∘119882119899119894
)∘(119882119899119894
∘119882119899119894
) sube 119882119899119894minus1
Theorem 9 Let (119883C) be a conformity There exists apseudodiversity 120575 which generates C if and only if C has acountable base
Proof If 120575 exists the sets 1198621119899119899isinN are our baseConversely let 119862119899
infin
1be a base for C satisfying 1198620 =
Pfin(119883) and119862119894 ∘119862119894 ∘119862119894 sube 119862119894minus1 for 119894 gt 0 Define 1205751015840 onPfin(119883)
by
1205751015840(119860) =
0 119860 isin 119862119899 forall119899
2minus119896 119860 isin 119862119899 for 0 le 119899 le 119896 but 119860 notin 119862119896+1
(18)
Notice that for 119896 ge 0
1205751015840minus1
([0 2minus119896]) = 119862119896 (19)
and that 1205751015840 is monotonic by (C2) if 119860 sube 119861 then 119860 isin 119862119899
whenever 119861 isin 119862119899Define a chain as a sequence 119860 119894
119899
119894=1inPfin(119883) with119860 119894 cap
119860 119894minus1 = for 119894 = 2 119899 Define a cycle as a chain with 1198601 cap
119860119899 = Write
120575 (119860) = infchains covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
120575 (119860) = infcycles covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
(20)
Notice that 120575() = 120575() = 0We claim that 120575 is our desired pseudodiversity since the
sets (1205751015840)minus1[0 120576] generate the conformity and 120575 le 1205751015840le 4120575 We
prove this in three stages
6 Journal of Function Spaces and Applications
(S1) First of all 120575 is a pseudodiversity By (C1) for every119909 isin 119883 119899 isin N and 119909 isin 119862119899 so that 120575
1015840(119909) = 0 Also
119909 is a cycle covering itself so 120575(119909) = 0
The triangle equality also holds let 120576 gt 0 119860119862 isin
Pfin(119883) and 119861 isin Pfin(119883) be nonempty Choosecycles 119860 119894
119899
1and 119861119894
119898
1covering 119860 cup 119861 and 119861 cup 119862
respectively and for which
119899
sum
119894=1
1205751015840(119860 119894) le 120575 (119860 cup 119861) + 120576
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119861 cup 119862) + 120576
(21)
Then 119860 119894119899
1cup 119861119894
119898
1forms a cycle (after reordering)
covering 119860 cup 119862 so
120575 (119860 cup 119862) le
119899
sum
119894=1
1205751015840(119860 119894)
+
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119860 cup 119861) + 120575 (119861 cup 119862) + 2120576
(22)
(S2) Next we notice that
(i) every cycle is a chain so 120575 le 120575(ii) If 1198601 119860119899minus1 119860119899 is a chain then
1198601 119860119899minus1 119860119899 119860119899minus1 1198601 is a cyclemdashand the sum of 120575
1015840 over this cycle is less thantwice the sum of 1205751015840 over the original chain Weconclude that
120575 le 120575 le 2120575 (23)
(S3) Finally we claim that 120575 le 1205751015840le 2120575 This combined
with (23) will give the main result
Trivially 120575 le 1205751015840 For the other inequality choose
119860 isin Pfin(119883) Our strategy is to induct on the greatestinteger119873 such that 120575(119860) lt 2
minus119873
The case119873 = 0 is easy because then 1205751015840le 1 le 2120575 (this
also covers the case 120575(119860) = 1 which is not covered bythe induction) When 119873 gt 0 we can choose positive120576 less than (2
minus119873minus 120575(119860)) and a chain 119860 119894
119899
1with
119899
sum
119894=1
1205751015840(119860 119894) lt 120575 (119860) + 120576 lt 2
minus119873 (24)
If 119899 = 1 we have 1205751015840(119860) le 120575
1015840(1198601) lt 2
minus119873lt 2120575(119860)
Otherwise there is 119896 lt 119899 such that
119896minus1
sum
119894=1
1205751015840(119860 119894) le
120575 (119860)
2
119899
sum
119894=119896+1
1205751015840(119860 119894) le
120575 (119860)
2 (25)
Since 119860 119894119896minus1
1and 119860 119894
119899
119896+1are chains whose sum under 1205751015840 is
less than half that of 119860 119894119899
1 the inductive hypothesis applies
to them and we may write
1205751015840(1198601 cup sdot sdot sdot cup 119860119896minus1)
le 2120575 (1198601 cup sdot sdot sdot cup 119860119896minus1) inductive hypothesis
le 2
119896minus1
sum
119894=1
1205751015840(119860 119894) definition of 120575
le 120575 (119860) by (25)
lt 2minus119873
(26)
Similarly 1205751015840(119860119896+1 cup sdot sdot sdot cup 119860119899) lt 2minus119873 and 120575
1015840(119860119896) lt 2
minus119873 by(24) So
(1198601 cup sdot sdot sdot cup 119860119896minus1) isin 119862119873+1 119860119896 isin 119862119873+1
(119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873+1
(27)
Our double-composition hypothesis gives
(1198601 cup sdot sdot sdot cup 119860119896minus1) cup 119860119896 cup (119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873 (28)
And by monotonicity of 1205751015840
1205751015840(119860) le 120575
1015840(1198601 cup sdot sdot sdot cup 119860119899) le 2
minus119873le 2120575 (119860) (29)
This characterizes the conformities generated by singlepseudodiversities Later we will describe every conformity interms of the pseudodiversities that generate them
42 Induced Uniformities and Completeness Given a confor-mity C we define its induced uniformity as the uniformitygenerated by the sets
119880119862 = (119909 119910) 119909 119910 isin 119862 (30)
for every 119862 isin C It is straightforward to show that this isa uniformity since every singleton 119909 is in every 119862 isin Cwe have every pair (119909 119909) in every generator of the induceduniformity proving (U1) Since 119909 119910 = 119910 119909 we have (U2)Finally (U3) follows from the observation that whenever119909 119910 isin 119862 isin C and 119910 119911 isin 119863 isin C the set 119863 ∘ 119863 isin Ccontains 119909 119910 119911Then x 119911 isin 119863∘119862 by (C2) In other wordsif 119862 ∘ 119863 sube 119864 in the conformity then 119880119862 ∘ 119880119863 sube 119880119864 in theinduced uniformity Thus (U3) is implied by (C3)
Theorem 10 Let119883 be a set 120575119860119860isinA be a family of diversitieswhich generate a conformity C For each 120575119860 write 119889119860 for itsinduced metric Then the uniformity generated by the metrics119889119860119860isinA is exactly the induced uniformity ofC
Proof Denote by U119889 the uniformity generated by 119889119860119860isinAand byU119888 the uniformity induced byC A base forC is
119862120576119860 = 119865 120575119860 (119865) lt 120576 (31)
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
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Stochastic AnalysisInternational Journal of
4 Journal of Function Spaces and Applications
This completion is dense in the sense that every member119909 of119883 has a sequence 119909119894 sube 119883with 119909119894 rarr 119909 in119883 (let 119910119894 be arepresentative of 119909 and define 119909119894 = 1199101 1199102 119910119894 119910119894 119910119894 )It also satisfies a universal property analogous to that formetric completion
Theorem 6 Let (119883 120575) be a diversity and let (119883 120575) be itscompletion Then for any complete diversity (119884 120574) and anyuniformly continuous function 119891 119883 rarr 119884 there is a uniqueuniformly continuous function 119891 119883 rarr 119884 which extends 119891
Proof Let 119909119894 be a representative sequence of somemembersof 119883 and define 119891(119909119894) = lim119894rarrinfin119891(119909119894) which is definedand independent of the representative since 119891 is uniformlycontinuous and 119884 is complete To show 119891 is uniformlycontinuous pick 120576 gt 0 and 119889 gt 0 such that 120574(119891(119860)) lt 120576
whenever 120575(119860) lt 119889 for all 119860 isin Pfin(119883) Then for all 119861 =
1199091
119894 1199092
119894 119909
119898
119894 isin Pfin(119883) with 120575(119861) lt 1198892 we have
120574(119891(119861)) = 120574(lim119894rarrinfin119891(119909119899
119894)119898
119899=1)) lt 120576 since for large enough
119873 120575(119891(119909119899
119873)119898
119899=1) lt 31198894
To show uniqueness of 119891 let 119892 be another uniformlycontinuous function extending 119891 to 119883 For all 119909 isin 119883 wehave 119909119894 sub 119883 with 119909119894 rarr 119909 in 119883 and by uniform continuity119892(119909) = lim119894rarrinfin119891(119909119894) = 119891(119909)
This is a universal property in the sense that for everycomplete diversity 119883
1015840 extending 119883 and having the propertythere is an isomorphism 119895 119883
1015840rarr 119883 (Specifically let 119895 be
the unique uniformly continuous extension of the identitymap 119895 119883 rarr 119883 to119883
1015840)
4 Conformities
In this section we introduce a generalization of diversitiesanalogous to uniformities which generalize metric spacesUniformities lie between metric spaces and topologies in thesense that every metric space defines a uniformity and everyuniformity defines a topology (which coincides with themetric topology when the uniformity came from a metric)Uniformities characterize uniform continuity uniform con-vergence and Cauchy sequences which are not topologicalconcepts
The carry-over from the metric case is natural butnontrivial since diversities can behave differently on sets ofdifferent cardinality Since this construction is qualitativelydifferent from metric uniformities it requires a differentname We asked ourselves ldquowhat would you call a uniformitythat came from a diversityrdquo and the answer was clear aconformity
Throughout this section we will give the analogousdefinitions and results for uniformities using the standardtreatment fromKelley [8]We begin by defining conformitiesand comparing them to uniformities we show that just likeuniformities conformities have a countable base if and onlyif they are generated by some pseudodiversity
We then briefly touch on the problem of completion forconformities
Finally we define power conformities from a conformitydefined on a set119883 we can construct a conformity onPfin(119883)
from which pseudodiversities can be considered uniformlycontinuous functions We show that every conformity isgenerated by exactly the set of pseudodiversities which areuniformly continuous from its power conformity to R Thisgives an equivalent definition of conformity in terms ofpseudodiversities
41 Conformities of Diversities Recall that for (119883 119889) a metricspace 119909119899 a sequence in119883 that 119909119899 is Cauchy if and only iffor each 120576 gt 0 there is some 119873 such that every pair of points(119909119894 119909119895) with 119894 gt 119873 119895 gt 119873 has 119889(119909119894 119909119895) lt 120576
Similarly let 119891 119883 rarr 119884 be a function between metricspaces (119883 119889) and (119884 119892) Then 119891 is uniformly continuous ifand only if each 120576 gt 0 has a 120575 gt 0 such that wheneverpairs of points (119909 119910) isin 119883 times 119883 satisfy 119889(119909 119910) lt 120575 the pairs(119891(119909) 119891(119910)) satisfy 119892(119891(119909) 119891(119910)) lt 120576
A similar characterization of uniform convergence ofsequences of functions can be given in terms of pairs ofpoints From these observations arises the theory of unifor-mities which is described in any standard text on analysis(cf [7 8]) We briefly describe the theory here For any set119883 define a uniformity on119883 as a filterU on119883 times 119883 satisfying
(U1) (119909 119909) isin 119880 for every 119909 isin 119883 119880 isin U
(U2) If 119880 isin U (119909 119910) isin 119880 then (119910 119909) isin 119880
(U3) For every119880 isin U there exists some119881 isin Uwith119881∘119881 sube
119880 where in general we define
119880 ∘ 119881 = (119909 119911) (119909 119910) isin 119880 (119910 119911) isin 119881 for some 119910 isin 119883
(13)
In particular for any pseudometric space (119883 119889) we candefine the metric uniformity as the filter on119883times119883 defined by
119880120576= (119909 119910) 119889 (119909 119910) lt 120576 (14)
for each 120576 gt 0 We see from this example that (U1) expressesthe requirement that 119889(119909 119909) = 0 for all 119909 isin 119883 (U2) expressessymmetry and (U3) expresses the triangle inequality
Uniform structure can be defined entirely with respect touniformities For example given sets 119883119884 and uniformitiesUV on 119883 and 119884 respectively we can call a function 119891
119883 rarr 119884 uniformly continuous if 119891minus1(119881) isin U for every
119881 isin V (Here 119891 acts on members of 119881 componentwise)A sequence 119909119899 sub 119883 is Cauchy if for every 119880 isin U thereis some 119873 such that pairs of elements (119909119894 119909119895) of 119909119899 are in119880 whenever 119894 119895 gt 119873 It is not hard to see that for metricuniformities these definitions coincide with the ordinaryones for metric spaces
To abstract the uniform structure of diversities unifor-mities are clearly insufficient For one thing since diversitiesmap finite sets rather than pairs we should seek a filteron Pfin(119883) rather than 119883 times 119883 Then symmetry is nolonger required but now monotonicity is Finally it is notmeaningful to compose finite sets as in (U3) so wewill need adiferent way to express an analogue of the triangle inequality
Journal of Function Spaces and Applications 5
Putting all this together we define a conformity C on 119883
as a filter onPfin(119883) satisfying
(C1) 119909 isin 119862 for every 119909 isin 119883 119862 isin C
(C2) For every 119862 isin C whenever 119860 isin 119862 and 119861 sube 119860 wehave 119861 isin 119862
(C3) For every 119862 isin C there exists some 119863 isin C with 119863 ∘
119863 sube 119862 where in general we define
119880 ∘ 119881 = 119906 cup V 119906 isin 119880 V isin 119881 and 119906 cap V = (15)
Often the term conformity is also used to refer to the pair(119883C)
An observation that will be necessary later (one whichalso holds for uniformities) is that for any119863 isin C (119863∘119863)∘119863 =
119863 ∘ (119863 ∘ 119863) so that 119863 ∘ 119863 ∘ 119863 is defined unambiguouslyTo estimate the size of this we also note that 119863 ∘ 119863 ∘ 119863 sube
(119863 ∘ 119863) ∘ (119863 ∘ 119863)As in the metric case there is a canonical way to generate
a conformity from a diversity if 120575 is a pseudodiversity on 119883we have the conformity generated by the sets
119862120576= 119860 120575 (119860) le 120576 = 120575
minus1[0 120576] (16)
for each 120576 gt 0 (This is equivalent to the one using strictinequalities but typographically nicer)
As in the metric case uniform structure can be definedon conformities in a way that generalizes that of diversitieslet (119883C) and (119884D) be conformities Then a function 119891 isuniformly continuous from 119883 to 119884 if for all 119863 isin D theset 119891
minus1(119889) 119889 isin 119863 is in C A sequence 119909119899 on 119883 is a
Cauchy sequence if for all119862 isin CPfin(119909119899119899ge119873) sube 119862 for someinteger119873 For conformities generated from diversities in theabove way these definitions coincide with those given in theprevious section
More generally given a collection of pseudodiversi-ties 120575120572120572isinA we can generate a conformity from the sets120575minus1
120572[0 120576]
120572isinA120576gt0 We therefore seek a characterization of
conformities in terms of the diversities which generate them(In a later section we will see that all conformities can bedescribed in this way so that we can define conformities interms of such sets) We begin by stating a result from Kelley[8] along with a summary of his proof
Theorem 7 A uniformity is generated by a single pseudomet-ric if and only if it has a countable base
The standard proof of this theorem goes as follows it isobvious that any uniformity generated by a pseudometric hasa countable base Conversely if there exists a countable basefor a uniformity on 119883 there exists a countable base 119880119899119899isinN
for which the following argument holds Define the function119891(119909 119910) = 2
minus119899 where 119899 = sup119894 (119909 119910) isin 119880119894 This generatesthe uniformity but does not satisfy the triangle inequality sodefine
119889 (119909 119910) = inf119898minus1
sum
119894=1
119891 (119909119894 119909119894+1) (17)
where the infimum is taken over all sequences 119909119894119898
119894=1with
1199091 = 119909 and 119909119898 = 119910 This clearly satisfies the triangleinequality so it just remains to be shown that 119889 generates theuniformity This is done by proving that 119889(119909 119910) le 119891(119909 119910) le
2119889(119909 119910) which follows from technical constraints on 119880119899Given a conformity with a countable base 119862119899 on a set119883
onemight try to translate this proof directly define a function119891(119860) Pfin(119883) rarr R by 119891(119860) = sup119894 119860 isin 119862119894 thensomehow tweak119891 to (a) satisfy the triangle inequality and (b)generate the same conformity as 119891 However it appears thatany direct analogue to the ldquoinfimum over all pathsrdquo strategyused in the metric case (there are several) cannot satisfy both(a) and (b) simultaneously
Nonetheless the result is true which is the content of thenext theorem
Lemma 8 Let (119883C) have a countable base Then it has acountable base 119862119899 satisfying 1198620 = P fin (119883)119862119894∘119862119894∘119862119894 sube 119862119894minus1
for 119894 gt 0
Proof Let 119881119899 be a countable base for C Define 1198820 =
Pfin(119883) 119882119899 = 119881119899 cap 119882119899minus1 Then 119882119899 is a nested countablebase Finally choose 119862119899 as 119862119894 = 119882119899
119894
where 119899119894 are choseninductively as 1198990 = 0 then (119882119899
119894
∘119882119899119894
)∘(119882119899119894
∘119882119899119894
) sube 119882119899119894minus1
Theorem 9 Let (119883C) be a conformity There exists apseudodiversity 120575 which generates C if and only if C has acountable base
Proof If 120575 exists the sets 1198621119899119899isinN are our baseConversely let 119862119899
infin
1be a base for C satisfying 1198620 =
Pfin(119883) and119862119894 ∘119862119894 ∘119862119894 sube 119862119894minus1 for 119894 gt 0 Define 1205751015840 onPfin(119883)
by
1205751015840(119860) =
0 119860 isin 119862119899 forall119899
2minus119896 119860 isin 119862119899 for 0 le 119899 le 119896 but 119860 notin 119862119896+1
(18)
Notice that for 119896 ge 0
1205751015840minus1
([0 2minus119896]) = 119862119896 (19)
and that 1205751015840 is monotonic by (C2) if 119860 sube 119861 then 119860 isin 119862119899
whenever 119861 isin 119862119899Define a chain as a sequence 119860 119894
119899
119894=1inPfin(119883) with119860 119894 cap
119860 119894minus1 = for 119894 = 2 119899 Define a cycle as a chain with 1198601 cap
119860119899 = Write
120575 (119860) = infchains covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
120575 (119860) = infcycles covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
(20)
Notice that 120575() = 120575() = 0We claim that 120575 is our desired pseudodiversity since the
sets (1205751015840)minus1[0 120576] generate the conformity and 120575 le 1205751015840le 4120575 We
prove this in three stages
6 Journal of Function Spaces and Applications
(S1) First of all 120575 is a pseudodiversity By (C1) for every119909 isin 119883 119899 isin N and 119909 isin 119862119899 so that 120575
1015840(119909) = 0 Also
119909 is a cycle covering itself so 120575(119909) = 0
The triangle equality also holds let 120576 gt 0 119860119862 isin
Pfin(119883) and 119861 isin Pfin(119883) be nonempty Choosecycles 119860 119894
119899
1and 119861119894
119898
1covering 119860 cup 119861 and 119861 cup 119862
respectively and for which
119899
sum
119894=1
1205751015840(119860 119894) le 120575 (119860 cup 119861) + 120576
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119861 cup 119862) + 120576
(21)
Then 119860 119894119899
1cup 119861119894
119898
1forms a cycle (after reordering)
covering 119860 cup 119862 so
120575 (119860 cup 119862) le
119899
sum
119894=1
1205751015840(119860 119894)
+
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119860 cup 119861) + 120575 (119861 cup 119862) + 2120576
(22)
(S2) Next we notice that
(i) every cycle is a chain so 120575 le 120575(ii) If 1198601 119860119899minus1 119860119899 is a chain then
1198601 119860119899minus1 119860119899 119860119899minus1 1198601 is a cyclemdashand the sum of 120575
1015840 over this cycle is less thantwice the sum of 1205751015840 over the original chain Weconclude that
120575 le 120575 le 2120575 (23)
(S3) Finally we claim that 120575 le 1205751015840le 2120575 This combined
with (23) will give the main result
Trivially 120575 le 1205751015840 For the other inequality choose
119860 isin Pfin(119883) Our strategy is to induct on the greatestinteger119873 such that 120575(119860) lt 2
minus119873
The case119873 = 0 is easy because then 1205751015840le 1 le 2120575 (this
also covers the case 120575(119860) = 1 which is not covered bythe induction) When 119873 gt 0 we can choose positive120576 less than (2
minus119873minus 120575(119860)) and a chain 119860 119894
119899
1with
119899
sum
119894=1
1205751015840(119860 119894) lt 120575 (119860) + 120576 lt 2
minus119873 (24)
If 119899 = 1 we have 1205751015840(119860) le 120575
1015840(1198601) lt 2
minus119873lt 2120575(119860)
Otherwise there is 119896 lt 119899 such that
119896minus1
sum
119894=1
1205751015840(119860 119894) le
120575 (119860)
2
119899
sum
119894=119896+1
1205751015840(119860 119894) le
120575 (119860)
2 (25)
Since 119860 119894119896minus1
1and 119860 119894
119899
119896+1are chains whose sum under 1205751015840 is
less than half that of 119860 119894119899
1 the inductive hypothesis applies
to them and we may write
1205751015840(1198601 cup sdot sdot sdot cup 119860119896minus1)
le 2120575 (1198601 cup sdot sdot sdot cup 119860119896minus1) inductive hypothesis
le 2
119896minus1
sum
119894=1
1205751015840(119860 119894) definition of 120575
le 120575 (119860) by (25)
lt 2minus119873
(26)
Similarly 1205751015840(119860119896+1 cup sdot sdot sdot cup 119860119899) lt 2minus119873 and 120575
1015840(119860119896) lt 2
minus119873 by(24) So
(1198601 cup sdot sdot sdot cup 119860119896minus1) isin 119862119873+1 119860119896 isin 119862119873+1
(119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873+1
(27)
Our double-composition hypothesis gives
(1198601 cup sdot sdot sdot cup 119860119896minus1) cup 119860119896 cup (119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873 (28)
And by monotonicity of 1205751015840
1205751015840(119860) le 120575
1015840(1198601 cup sdot sdot sdot cup 119860119899) le 2
minus119873le 2120575 (119860) (29)
This characterizes the conformities generated by singlepseudodiversities Later we will describe every conformity interms of the pseudodiversities that generate them
42 Induced Uniformities and Completeness Given a confor-mity C we define its induced uniformity as the uniformitygenerated by the sets
119880119862 = (119909 119910) 119909 119910 isin 119862 (30)
for every 119862 isin C It is straightforward to show that this isa uniformity since every singleton 119909 is in every 119862 isin Cwe have every pair (119909 119909) in every generator of the induceduniformity proving (U1) Since 119909 119910 = 119910 119909 we have (U2)Finally (U3) follows from the observation that whenever119909 119910 isin 119862 isin C and 119910 119911 isin 119863 isin C the set 119863 ∘ 119863 isin Ccontains 119909 119910 119911Then x 119911 isin 119863∘119862 by (C2) In other wordsif 119862 ∘ 119863 sube 119864 in the conformity then 119880119862 ∘ 119880119863 sube 119880119864 in theinduced uniformity Thus (U3) is implied by (C3)
Theorem 10 Let119883 be a set 120575119860119860isinA be a family of diversitieswhich generate a conformity C For each 120575119860 write 119889119860 for itsinduced metric Then the uniformity generated by the metrics119889119860119860isinA is exactly the induced uniformity ofC
Proof Denote by U119889 the uniformity generated by 119889119860119860isinAand byU119888 the uniformity induced byC A base forC is
119862120576119860 = 119865 120575119860 (119865) lt 120576 (31)
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
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Journal of Function Spaces and Applications 5
Putting all this together we define a conformity C on 119883
as a filter onPfin(119883) satisfying
(C1) 119909 isin 119862 for every 119909 isin 119883 119862 isin C
(C2) For every 119862 isin C whenever 119860 isin 119862 and 119861 sube 119860 wehave 119861 isin 119862
(C3) For every 119862 isin C there exists some 119863 isin C with 119863 ∘
119863 sube 119862 where in general we define
119880 ∘ 119881 = 119906 cup V 119906 isin 119880 V isin 119881 and 119906 cap V = (15)
Often the term conformity is also used to refer to the pair(119883C)
An observation that will be necessary later (one whichalso holds for uniformities) is that for any119863 isin C (119863∘119863)∘119863 =
119863 ∘ (119863 ∘ 119863) so that 119863 ∘ 119863 ∘ 119863 is defined unambiguouslyTo estimate the size of this we also note that 119863 ∘ 119863 ∘ 119863 sube
(119863 ∘ 119863) ∘ (119863 ∘ 119863)As in the metric case there is a canonical way to generate
a conformity from a diversity if 120575 is a pseudodiversity on 119883we have the conformity generated by the sets
119862120576= 119860 120575 (119860) le 120576 = 120575
minus1[0 120576] (16)
for each 120576 gt 0 (This is equivalent to the one using strictinequalities but typographically nicer)
As in the metric case uniform structure can be definedon conformities in a way that generalizes that of diversitieslet (119883C) and (119884D) be conformities Then a function 119891 isuniformly continuous from 119883 to 119884 if for all 119863 isin D theset 119891
minus1(119889) 119889 isin 119863 is in C A sequence 119909119899 on 119883 is a
Cauchy sequence if for all119862 isin CPfin(119909119899119899ge119873) sube 119862 for someinteger119873 For conformities generated from diversities in theabove way these definitions coincide with those given in theprevious section
More generally given a collection of pseudodiversi-ties 120575120572120572isinA we can generate a conformity from the sets120575minus1
120572[0 120576]
120572isinA120576gt0 We therefore seek a characterization of
conformities in terms of the diversities which generate them(In a later section we will see that all conformities can bedescribed in this way so that we can define conformities interms of such sets) We begin by stating a result from Kelley[8] along with a summary of his proof
Theorem 7 A uniformity is generated by a single pseudomet-ric if and only if it has a countable base
The standard proof of this theorem goes as follows it isobvious that any uniformity generated by a pseudometric hasa countable base Conversely if there exists a countable basefor a uniformity on 119883 there exists a countable base 119880119899119899isinN
for which the following argument holds Define the function119891(119909 119910) = 2
minus119899 where 119899 = sup119894 (119909 119910) isin 119880119894 This generatesthe uniformity but does not satisfy the triangle inequality sodefine
119889 (119909 119910) = inf119898minus1
sum
119894=1
119891 (119909119894 119909119894+1) (17)
where the infimum is taken over all sequences 119909119894119898
119894=1with
1199091 = 119909 and 119909119898 = 119910 This clearly satisfies the triangleinequality so it just remains to be shown that 119889 generates theuniformity This is done by proving that 119889(119909 119910) le 119891(119909 119910) le
2119889(119909 119910) which follows from technical constraints on 119880119899Given a conformity with a countable base 119862119899 on a set119883
onemight try to translate this proof directly define a function119891(119860) Pfin(119883) rarr R by 119891(119860) = sup119894 119860 isin 119862119894 thensomehow tweak119891 to (a) satisfy the triangle inequality and (b)generate the same conformity as 119891 However it appears thatany direct analogue to the ldquoinfimum over all pathsrdquo strategyused in the metric case (there are several) cannot satisfy both(a) and (b) simultaneously
Nonetheless the result is true which is the content of thenext theorem
Lemma 8 Let (119883C) have a countable base Then it has acountable base 119862119899 satisfying 1198620 = P fin (119883)119862119894∘119862119894∘119862119894 sube 119862119894minus1
for 119894 gt 0
Proof Let 119881119899 be a countable base for C Define 1198820 =
Pfin(119883) 119882119899 = 119881119899 cap 119882119899minus1 Then 119882119899 is a nested countablebase Finally choose 119862119899 as 119862119894 = 119882119899
119894
where 119899119894 are choseninductively as 1198990 = 0 then (119882119899
119894
∘119882119899119894
)∘(119882119899119894
∘119882119899119894
) sube 119882119899119894minus1
Theorem 9 Let (119883C) be a conformity There exists apseudodiversity 120575 which generates C if and only if C has acountable base
Proof If 120575 exists the sets 1198621119899119899isinN are our baseConversely let 119862119899
infin
1be a base for C satisfying 1198620 =
Pfin(119883) and119862119894 ∘119862119894 ∘119862119894 sube 119862119894minus1 for 119894 gt 0 Define 1205751015840 onPfin(119883)
by
1205751015840(119860) =
0 119860 isin 119862119899 forall119899
2minus119896 119860 isin 119862119899 for 0 le 119899 le 119896 but 119860 notin 119862119896+1
(18)
Notice that for 119896 ge 0
1205751015840minus1
([0 2minus119896]) = 119862119896 (19)
and that 1205751015840 is monotonic by (C2) if 119860 sube 119861 then 119860 isin 119862119899
whenever 119861 isin 119862119899Define a chain as a sequence 119860 119894
119899
119894=1inPfin(119883) with119860 119894 cap
119860 119894minus1 = for 119894 = 2 119899 Define a cycle as a chain with 1198601 cap
119860119899 = Write
120575 (119860) = infchains covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
120575 (119860) = infcycles covering 119860
119899
sum
119894=1
1205751015840(119860 119894)
(20)
Notice that 120575() = 120575() = 0We claim that 120575 is our desired pseudodiversity since the
sets (1205751015840)minus1[0 120576] generate the conformity and 120575 le 1205751015840le 4120575 We
prove this in three stages
6 Journal of Function Spaces and Applications
(S1) First of all 120575 is a pseudodiversity By (C1) for every119909 isin 119883 119899 isin N and 119909 isin 119862119899 so that 120575
1015840(119909) = 0 Also
119909 is a cycle covering itself so 120575(119909) = 0
The triangle equality also holds let 120576 gt 0 119860119862 isin
Pfin(119883) and 119861 isin Pfin(119883) be nonempty Choosecycles 119860 119894
119899
1and 119861119894
119898
1covering 119860 cup 119861 and 119861 cup 119862
respectively and for which
119899
sum
119894=1
1205751015840(119860 119894) le 120575 (119860 cup 119861) + 120576
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119861 cup 119862) + 120576
(21)
Then 119860 119894119899
1cup 119861119894
119898
1forms a cycle (after reordering)
covering 119860 cup 119862 so
120575 (119860 cup 119862) le
119899
sum
119894=1
1205751015840(119860 119894)
+
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119860 cup 119861) + 120575 (119861 cup 119862) + 2120576
(22)
(S2) Next we notice that
(i) every cycle is a chain so 120575 le 120575(ii) If 1198601 119860119899minus1 119860119899 is a chain then
1198601 119860119899minus1 119860119899 119860119899minus1 1198601 is a cyclemdashand the sum of 120575
1015840 over this cycle is less thantwice the sum of 1205751015840 over the original chain Weconclude that
120575 le 120575 le 2120575 (23)
(S3) Finally we claim that 120575 le 1205751015840le 2120575 This combined
with (23) will give the main result
Trivially 120575 le 1205751015840 For the other inequality choose
119860 isin Pfin(119883) Our strategy is to induct on the greatestinteger119873 such that 120575(119860) lt 2
minus119873
The case119873 = 0 is easy because then 1205751015840le 1 le 2120575 (this
also covers the case 120575(119860) = 1 which is not covered bythe induction) When 119873 gt 0 we can choose positive120576 less than (2
minus119873minus 120575(119860)) and a chain 119860 119894
119899
1with
119899
sum
119894=1
1205751015840(119860 119894) lt 120575 (119860) + 120576 lt 2
minus119873 (24)
If 119899 = 1 we have 1205751015840(119860) le 120575
1015840(1198601) lt 2
minus119873lt 2120575(119860)
Otherwise there is 119896 lt 119899 such that
119896minus1
sum
119894=1
1205751015840(119860 119894) le
120575 (119860)
2
119899
sum
119894=119896+1
1205751015840(119860 119894) le
120575 (119860)
2 (25)
Since 119860 119894119896minus1
1and 119860 119894
119899
119896+1are chains whose sum under 1205751015840 is
less than half that of 119860 119894119899
1 the inductive hypothesis applies
to them and we may write
1205751015840(1198601 cup sdot sdot sdot cup 119860119896minus1)
le 2120575 (1198601 cup sdot sdot sdot cup 119860119896minus1) inductive hypothesis
le 2
119896minus1
sum
119894=1
1205751015840(119860 119894) definition of 120575
le 120575 (119860) by (25)
lt 2minus119873
(26)
Similarly 1205751015840(119860119896+1 cup sdot sdot sdot cup 119860119899) lt 2minus119873 and 120575
1015840(119860119896) lt 2
minus119873 by(24) So
(1198601 cup sdot sdot sdot cup 119860119896minus1) isin 119862119873+1 119860119896 isin 119862119873+1
(119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873+1
(27)
Our double-composition hypothesis gives
(1198601 cup sdot sdot sdot cup 119860119896minus1) cup 119860119896 cup (119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873 (28)
And by monotonicity of 1205751015840
1205751015840(119860) le 120575
1015840(1198601 cup sdot sdot sdot cup 119860119899) le 2
minus119873le 2120575 (119860) (29)
This characterizes the conformities generated by singlepseudodiversities Later we will describe every conformity interms of the pseudodiversities that generate them
42 Induced Uniformities and Completeness Given a confor-mity C we define its induced uniformity as the uniformitygenerated by the sets
119880119862 = (119909 119910) 119909 119910 isin 119862 (30)
for every 119862 isin C It is straightforward to show that this isa uniformity since every singleton 119909 is in every 119862 isin Cwe have every pair (119909 119909) in every generator of the induceduniformity proving (U1) Since 119909 119910 = 119910 119909 we have (U2)Finally (U3) follows from the observation that whenever119909 119910 isin 119862 isin C and 119910 119911 isin 119863 isin C the set 119863 ∘ 119863 isin Ccontains 119909 119910 119911Then x 119911 isin 119863∘119862 by (C2) In other wordsif 119862 ∘ 119863 sube 119864 in the conformity then 119880119862 ∘ 119880119863 sube 119880119864 in theinduced uniformity Thus (U3) is implied by (C3)
Theorem 10 Let119883 be a set 120575119860119860isinA be a family of diversitieswhich generate a conformity C For each 120575119860 write 119889119860 for itsinduced metric Then the uniformity generated by the metrics119889119860119860isinA is exactly the induced uniformity ofC
Proof Denote by U119889 the uniformity generated by 119889119860119860isinAand byU119888 the uniformity induced byC A base forC is
119862120576119860 = 119865 120575119860 (119865) lt 120576 (31)
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Journal of Function Spaces and Applications
(S1) First of all 120575 is a pseudodiversity By (C1) for every119909 isin 119883 119899 isin N and 119909 isin 119862119899 so that 120575
1015840(119909) = 0 Also
119909 is a cycle covering itself so 120575(119909) = 0
The triangle equality also holds let 120576 gt 0 119860119862 isin
Pfin(119883) and 119861 isin Pfin(119883) be nonempty Choosecycles 119860 119894
119899
1and 119861119894
119898
1covering 119860 cup 119861 and 119861 cup 119862
respectively and for which
119899
sum
119894=1
1205751015840(119860 119894) le 120575 (119860 cup 119861) + 120576
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119861 cup 119862) + 120576
(21)
Then 119860 119894119899
1cup 119861119894
119898
1forms a cycle (after reordering)
covering 119860 cup 119862 so
120575 (119860 cup 119862) le
119899
sum
119894=1
1205751015840(119860 119894)
+
119898
sum
119894=1
1205751015840(119861119894) le 120575 (119860 cup 119861) + 120575 (119861 cup 119862) + 2120576
(22)
(S2) Next we notice that
(i) every cycle is a chain so 120575 le 120575(ii) If 1198601 119860119899minus1 119860119899 is a chain then
1198601 119860119899minus1 119860119899 119860119899minus1 1198601 is a cyclemdashand the sum of 120575
1015840 over this cycle is less thantwice the sum of 1205751015840 over the original chain Weconclude that
120575 le 120575 le 2120575 (23)
(S3) Finally we claim that 120575 le 1205751015840le 2120575 This combined
with (23) will give the main result
Trivially 120575 le 1205751015840 For the other inequality choose
119860 isin Pfin(119883) Our strategy is to induct on the greatestinteger119873 such that 120575(119860) lt 2
minus119873
The case119873 = 0 is easy because then 1205751015840le 1 le 2120575 (this
also covers the case 120575(119860) = 1 which is not covered bythe induction) When 119873 gt 0 we can choose positive120576 less than (2
minus119873minus 120575(119860)) and a chain 119860 119894
119899
1with
119899
sum
119894=1
1205751015840(119860 119894) lt 120575 (119860) + 120576 lt 2
minus119873 (24)
If 119899 = 1 we have 1205751015840(119860) le 120575
1015840(1198601) lt 2
minus119873lt 2120575(119860)
Otherwise there is 119896 lt 119899 such that
119896minus1
sum
119894=1
1205751015840(119860 119894) le
120575 (119860)
2
119899
sum
119894=119896+1
1205751015840(119860 119894) le
120575 (119860)
2 (25)
Since 119860 119894119896minus1
1and 119860 119894
119899
119896+1are chains whose sum under 1205751015840 is
less than half that of 119860 119894119899
1 the inductive hypothesis applies
to them and we may write
1205751015840(1198601 cup sdot sdot sdot cup 119860119896minus1)
le 2120575 (1198601 cup sdot sdot sdot cup 119860119896minus1) inductive hypothesis
le 2
119896minus1
sum
119894=1
1205751015840(119860 119894) definition of 120575
le 120575 (119860) by (25)
lt 2minus119873
(26)
Similarly 1205751015840(119860119896+1 cup sdot sdot sdot cup 119860119899) lt 2minus119873 and 120575
1015840(119860119896) lt 2
minus119873 by(24) So
(1198601 cup sdot sdot sdot cup 119860119896minus1) isin 119862119873+1 119860119896 isin 119862119873+1
(119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873+1
(27)
Our double-composition hypothesis gives
(1198601 cup sdot sdot sdot cup 119860119896minus1) cup 119860119896 cup (119860119896+1 cup sdot sdot sdot cup 119860119899) isin 119862119873 (28)
And by monotonicity of 1205751015840
1205751015840(119860) le 120575
1015840(1198601 cup sdot sdot sdot cup 119860119899) le 2
minus119873le 2120575 (119860) (29)
This characterizes the conformities generated by singlepseudodiversities Later we will describe every conformity interms of the pseudodiversities that generate them
42 Induced Uniformities and Completeness Given a confor-mity C we define its induced uniformity as the uniformitygenerated by the sets
119880119862 = (119909 119910) 119909 119910 isin 119862 (30)
for every 119862 isin C It is straightforward to show that this isa uniformity since every singleton 119909 is in every 119862 isin Cwe have every pair (119909 119909) in every generator of the induceduniformity proving (U1) Since 119909 119910 = 119910 119909 we have (U2)Finally (U3) follows from the observation that whenever119909 119910 isin 119862 isin C and 119910 119911 isin 119863 isin C the set 119863 ∘ 119863 isin Ccontains 119909 119910 119911Then x 119911 isin 119863∘119862 by (C2) In other wordsif 119862 ∘ 119863 sube 119864 in the conformity then 119880119862 ∘ 119880119863 sube 119880119864 in theinduced uniformity Thus (U3) is implied by (C3)
Theorem 10 Let119883 be a set 120575119860119860isinA be a family of diversitieswhich generate a conformity C For each 120575119860 write 119889119860 for itsinduced metric Then the uniformity generated by the metrics119889119860119860isinA is exactly the induced uniformity ofC
Proof Denote by U119889 the uniformity generated by 119889119860119860isinAand byU119888 the uniformity induced byC A base forC is
119862120576119860 = 119865 120575119860 (119865) lt 120576 (31)
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Function Spaces and Applications 7
where 120576 ranges overR+ and119860 ranges overAThen a base forC is
119880119862120576119860
= (119909 119910) 120575119860 (119909 119910) lt 120576 = (119909 119910) 119889119860 (119909 119910) lt 120576
(32)
But this is just the canonical base forU119889
Corollary 11 Let (119883C) be a conformity Then C has acountable base if and only if its induced uniformity does
Proof By Theorem 9 C has a countable base if and only ifit is generated by a single pseudodiversity by Theorem 10this occurs if and only if the induced uniformity is generatedby a single pseudometric A standard result [7 8] shows thatuniformities with countable bases are exactly those generatedby single pseudometrics
Next we give some standard definitions For a uniformspace (119883U) the uniform topology ofU on119883 is the smallesttopology containing the sets
119873(119909119880) = 119910 (119909 119910) isin 119880 (33)
for all 119909 isin 119883 119880 isin U Notice that if U is generatedby a pseudometric this coincides with the pseudometrictopology
With the same space (119883U) we call a filter F on 119883
Cauchy if for every 119880 isin U there is some 119865 isin F with119865 times 119865 sube 119880 We say thatF converges to some 119909 isin 119883 if everyneighborhoodof119909 (in the uniform topology) is inFWe thencall a uniformity complete if every Cauchy filter convergesIt can be shown that a metric space is complete if and onlyif its generated uniformity is and that every uniformity canbe embedded minimally (ie satisfying a universal propertywith respect to uniformly continuous maps) in a completeuniformity [7 8]
The analogous definitions for conformities are as followsLet 119865 be a filter on 119883 If for all 119862 isin C there exists 119891 isin 119865
with Pfin(119891) sube 119862 then 119865 is a Cauchy filter If 119909 isin 119883 and forall 119862 isin C there exist 119891 isin 119865 withPfin(119891) sube 119860 119860 cup 119909 isin 119862then119865 converges to119909 Finally if every Cauchy filter convergesto some point in119883 we sayC is complete
Theorem 12 A pseudodiversity (119883 120575) is complete if and onlyif its conformityC is
Proof Suppose (119883 120575) is complete and let 119865 be a Cauchy filteron 119883 Then for every 120576 gt 0 there is some 119891
120576isin 119865 so that
Pfin(119891120576) sube 119860 120575(119860) lt 120576 Take some sequence 120576119899 rarr 0 and
define the sets 119892119899 sube 119883 by 1198921= 1198911205761 119892119899 = 119891
120576119899 cap 119892120576119899minus1 for 119899 gt 1
Choose 119909119899
isin 119892119899 for each 119899 to form a Cauchy sequence
119909119899 with some limit 119909 For any 120576 gt 0 find an integer 119873 sothat 120576119899 lt 120576 and 120575(119909119899 119909) lt 120576 for all 119899 ge 119873 Then if 119886 isin
Pfin(119891120576119899) so is 119886 cup 119909119899 so that 120575(119886 cup 119909) le 120575(119886 cup 119909119899) +
120575(119909119899 119909) lt 2120576 We conclude that 119865 converges to 119909Conversely suppose that every Cauchy filter converges in
C and let 119909119899 be a Cauchy sequence in (119883 120575) Choose thesets 119865119873 = 119909119899
infin
119873 These sets generate a Cauchy filter with
some limit 119909 It is clear that 119909119899 rarr 119909
For any conformity (119883C) generated by a diversity theconformity is complete if and only if the diversity is Thediversity is complete if and only if its inducedmetric is whichin turn is complete if and only if its uniformity is [7 8] thuscompleteness of the conformity is equivalent to completenessof its induced uniformity In fact this is true in general as thenext theorem shows
Theorem 13 Let (119883C) be a conformity with completeinduced uniformityU ThenC is complete
Proof Suppose that U is complete and let F be a Cauchyfilter with respect to C ThenF is also Cauchy with respectto U since for all 119862 isin C we have 119909 119910 119909 119910 isin 119865 sube
Pfin(119865) sube 119862 for some 119865 isin F then 119865 times 119865 sube 119880119862 Thus Fconverges in U to some element 119909 and we claim that it alsoconverges to 119909 in C To this end fix 119862 isin C Choose 119863 isin Cso that 119863 ∘ 119863 sube 119862 and 119865 isin F so that (a) 119910 isin 119865 whenever(119909 119910) isin 119880119863 and (b) Pfin(119865) sube 119863 Then for all 119860 isin Pfin(119865)119860 cup 119909 isin 119862 (If 119860 = 119860 cup 119909 isin 119862 trivially Otherwisepick 119910 isin 119860 and we will have 119860 isin 119863 and 119909 119910 isin 119863 so that119860 cup 119909 119910 = 119860 cup 119909 isin 119862)
We end this section with two open questions as follows
(1) Does the converse to Theorem 13 holds that is ifa conformity (119883C) is complete must its induceduniformity be
(2) We saw in Section 32 that for any diversity (119883 120575)it is possible to embed 119883 in a complete diversitywhich was universal meaning that any uniformlycontinuous map from 119883 to a complete diversity isfactored through the embedding It is shown in [8]that every uniformity can be embedded in a completeuniformity This embedding is also universalIs there a notion of universal completion for confor-mities
43 Diversities of Conformities Not every conformity has acountable base For example let 119883 be the space of functions119891 [0 1] rarr [0 1] and consider the ldquopointwise convergencerdquoconformity generated by the sets
119862119909
120576= 1198911 119891119899 diam (1198911 (119909) 119891119899 (119909)) lt 120576 (34)
for every 120576 gt 0 119909 isin [0 1] This conformity has no countablebase by Corollary 11 since its induced uniformity does nothave a countable base [10] Thus by Theorem 9 it is notgenerated by any pseudodiversity
In this section we will show that every conformity isgenerated by the collection of pseudodiversities which areuniformly continuous with respect to it in an appropriatesense In the case of uniformities this is done by constructinga so-called product uniformity given a uniformity on a set119883the product uniformity is constructed on119883times119883Then a givenpseudometric 119889 may or may not be uniformly continuousfrom the product uniformity to the Euclidean uniformity onR It can be proven [7 8] that a uniformity U is exactlythe uniformity generated by all pseudometrics which areuniformly continuous from its product uniformity
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Journal of Function Spaces and Applications
Since pseudodiversities are functions on finite sets ratherthan pairs given a conformity on a set119883we seek a conformityon Pfin(119883) from which to judge uniform continuity ofpseudodiversities
In fact such a conformity exists for which we can provethe same result given a conformity (119883C) define the powerconformity C119875 as the conformity onPfin(119883) generated by thesets
119862119906 = 1198601 119860119899 119899 le 1 or119899
⋃
119894=1
119860 119894 isin 119906 (35)
where 119906 ranges over all members ofC
Lemma 14 A power conformity is a conformity
Proof First the 119862119906s form a filter base since 119862119906 cap119862V = 119862119906capV isin
C119875 for any 119862119906 119862V isin C119875 For all 119860 isin Pfin(119883) 119860 is in every119862119906 by definition It is immediate that whenever 119860 119894 is in 119862119906so is every subset of 119860 119894
Finally every 119862119906 has a 119862V with 119862V ∘ 119862V sube 119862119906 choose Vwith V ∘ V sube 119906 inC If 119860 119894
119899
119894=1 119861119894119898
119894=1are in 119862V with some 119860 119894
equal to some119861119895 then (a)119898 le 1 and 119899 le 1 so their union hasat most one element and therefore must lie in 119862119906 (b) exactlyone of119898 le 1 or 119899 le 1 in which case one of the sets is a subsetof the other so their union lies in V (and therefore 119906) or (c)119898 gt 1 and 119899 gt 1 so the sets ⋃119899
119894=1119860 119894 and ⋃
119898
119894=1119861119894 are sets in V
with nonempty intersectionThen since V∘V sube 119906 their unionlies in 119906 In every case we have 119860 119894
119899
119894=1cup 119861119894
119898
119894=1isin 119862119906
Theorem 15 Let (119883C) be a conformity A pseudodiversity 120575
is uniformly continuous fromC119875 to (R diam) if and only if theset 119881120576 = 119860 120575(119860) lt 120576 is inC for each 120576 gt 0
Proof First suppose that every119881120576 is inC For each 120576 gt 0 theset
119862119906 = 1198601 119860119899 119899 le 1 or 120575(
119899
⋃
119894=1
119860 119894) lt 120576 (36)
is inC119875 (notice that it has the form of (35) with 119906 = 119881120576) Let119860 119861 isin 119862120576 then 120575(119860) le 120575(119860cup119861) lt 120576 and similarly 120575(119861) lt 120576Thus |120575(119860) minus 120575(119861)| lt 120576 so 120575 is uniformly continuous
Conversely suppose that 120575 is uniformly continuousThen for any 120576 gt 0 there exists some 119906 isin C such that every1198601 119860119899 isin 119862119906 satisfies sup119894119895|120575(119860 119894)minus120575(119860119895)| lt 120576 Since forany119860 isin 119906 the set 119860 lies in119862119906 this implies that 120575(119860) lt 120576which in turn implies that 119906 sube 119881120576 which finally implies that119881120576 is inC
Corollary 16 Every conformity is generated by the pseu-dodiversities which are uniformly continuous from its powerconformity to (R diam)
Proof LetC be a conformityD be the conformity generatedby the pseudodiversities which are uniformly continuousfrom the power conformity to (R diam) By Theorem 9 wehave C sube D since every member 119906 of C is in a countably-based subconformity ofC (Take 1199060 = 119906 119906119894 such that 119906119894 ∘ 119906119894 sube119906119894minus1 119894 gt 0 as a base)
Then by Theorem 15 every pseudodiversity which isuniformly continuous generates a subset of C that is D sube
C
We saw at the beginning of this section that some confor-mities can be generated by sets of the form 120575
minus1
120572[0 120576]
120572isinA120576gt0
whereA is some collection of pseudodiversities 120576 gt 0 Whatwe have just shown is that all conformities are generatedin this way so that we may define a conformity as a filtergenerated in this way by some collection of diversities
5 Category Theory
In [5] Bryant andTupper introduced the categoryDvywhoseobjects are diversities and morphisms nonexpansive maps(functions 119891 between diversities (119883 120575) and (119884 120588) such that120588(119891(119860)) le 120575(119860) for all finite 119860 sube 119883) This compares withMet [11] whose objects are metric spaces and morphismsnonexpansive maps (functions 119891 between metric spaces(119883 119889) and (119884 119901) such that 119901(119891(119909) 119891(119910)) le 119889(119909 119910) for all119909 119910 isin 119883)
It is not hard to see that for both metric spaces anddiversities nonexpansive maps are uniformly continuous Inthe metric case they are also continuous
We introduce the category Conf whose objects are con-formities and morphisms uniformly continuous functionsThis compares with Unif [11] whose objects are uniformitiesand morphisms uniformly continuous functions
We also recall Top whose objects are topological spacesand morphisms continuous maps and CAT whose objectsare categories and morphisms are functors (maps betweencategories which preserve composition)
With these categories in hand we can summarize therelationships between diversities conformities and metricspaces by observing that themaps in the following diagram inCAT are functors and that the diagram as a whole commutes
rd
tm
tu
r120575
ud
u120575
Met Div
Top
Unif Conf
(37)
where
(i) 119903120575 maps conformities to their induced uniformspaces
(ii) 119903119889 maps diversities to their induced metric spaces
(iii) 119906120575 maps diversities to the conformities that theygenerate
(iv) 119906119889mapsmetric spaces to the uniform spaces that theygenerate
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Function Spaces and Applications 9
(v) 119905119898 maps metric spaces to their metric topologies(vi) and 119906119898maps uniform spaces to their uniform topolo-
gies
Notice that each functor leaves the underlying sets unchangedfor example 119906119889maps ametric space (119883 119889) to a uniform space(119883U) The morphisms are also unchanged as functionsfor example a nonexpansive map 119891 119883 rarr 119884 in Metis considered a continuous map in Top under 119905119898 and auniformly continous map in Unif under 119906119889 but it is the samefunction from the set119883 to the set 119884 in all cases
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
Research funded in part by NSERC
References
[1] N Aronszajn and P Panitchpakdi ldquoExtension of uniformlycontinuous transformations and hyperconvex metric spacesrdquoPacific Journal of Mathematics vol 6 pp 405ndash439 1956
[2] J R Isbell ldquoSix theorems about injective metric spacesrdquo Com-mentarii Mathematici Helvetici vol 39 pp 65ndash76 1964
[3] A W M Dress ldquoTrees tight extensions of metric spacesand the cohomological dimension of certain groups a noteon combinatorial properties of metric spacesrdquo Advances inMathematics vol 53 no 3 pp 321ndash402 1984
[4] A W M Dress V Moulton and W Terhalle ldquo119879-theory anoverviewrdquo European Journal of Combinatorics vol 17 no 2-3pp 161ndash175 1996
[5] D Bryant and P F Tupper ldquoHyperconvexity and tight-spantheory for diversitiesrdquo Advances in Mathematics vol 231 no 6pp 3172ndash3198 2012
[6] AWeil ldquoSur les espaces a structure uniforme et sur la topologiegeneralerdquo Actuarial Science of India vol 551 p 162 1937
[7] N Bourbaki General Topology Springer New York NY USA1989 Translation of Topologie Generale
[8] J L Kelley General Topology Springer New York NY USA1975
[9] D Z Du ldquoOn Steiner ratio conjecturesrdquo Annals of OperationsResearch vol 33 no 1ndash4 pp 437ndash449 1991
[10] S Henry ldquoReference uniformity of pointwise convergence hasno countable baserdquo MathOverflow June 2013 httpmathover-flownetquestions134421
[11] J Adamek H Herrlich and G E Strecker Abstract and Con-crete Categories The Joy of Cats Pure and Applied MathematicsJohn Wiley amp Sons New York NY USA 1990
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of