Where is it at? Brian Davey GAIA 2013 - La Trobe …Boolean algebras Boolean spaces (i.e., compact,...

The full versus strong problemWhere is it at?

Brian Davey

GAIA 2013La Trobe University

15 July, 2013

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Outline

I Examples of natural dualities

I Dualities, full dualities and strong dualities

I Full vs strong

I The lattice of alter egos

I Full dualities and universal Horn sentences

I So where is it at?

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Stone duality

Boolean algebras Boolean spaces(i.e., compact, Hausdorff anda basis of clopen sets)

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Stone duality

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Stone duality

Boolean algebra of allfinite or cofinite subsets of N

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Stone duality

Countable atomlessBoolean algebraFB(!)

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Stone duality

Boolean algebras Boolean spaces

(i.e., compact, Hausdorff anda basis of clopen sets)

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Stone duality

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Stone duality

B = ISP(B), where

B = h{0, 1};_,^, 0, 0, 1i d 0

d 1IScP+(B), where

B = h{0, 1};Tid

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Priestley duality

Bounded distributive lattices Priestley spaces

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Priestley duality

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Priestley duality

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Priestley duality

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Priestley duality

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Priestley duality

D = ISP(D), where

D = h{0, 1};_,^, 0, 1i d 0

d 1 IScP+(D), where

D = h{0, 1};6,Ti d 0

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Natural dualities: alter egos

Generalizing our examples, we start with a finite algebra M andwish to find a dual category for the quasivariety A := ISP(M).

An alter ego of a finite algebraA structure M = hM;G,H,R,Ti is an alter ego of M if

I G is a set of operations on M, each of which is ahomomorphism with respect to M,

I H is a set of partial operations on M, each of which is ahomomorphism with respect to M,

I R is a set of relations on M, each of which is asubuniverse of the appropriate power of M, and

I T is the discrete topology on M.

An alter ego of M is often denoted by M⇠.

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Natural dualities: categories and functorsLet M = hM;G,H,R,Ti be an alter ego of M.

The categories A and X

I Define A := ISP(M): the algebraic category of interest.

I Define X := IScP+(M): the potential dual category for A.

The contravariant functors D and E

I There are natural hom-functors D : A ! X and E : X ! A.

I For each algebra A in A, the underlying set of D(A) is theset hom(A,M) of all homomorphisms from A into M, andD(A) is a topologically closed substructure of MA.

I For each structure X in X, the underlying set of E(X) is theset hom(X,M) of all continuous homomorphisms from Xinto M, and E(X) is a subalgebra of MX .

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The contravariant functors D and E

I There are natural hom-functors D : A ! X and E : X ! A.

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The contravariant functors D and EI There are natural hom-functors D : A ! X and E : X ! A.

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Natural dualities: embeddings

Natural embeddings

For every A 2 A and X 2 X, there are naturally definedembeddings

eA : A ! ED(A) and "X : X ! DE(X).

These embeddings yield natural transformations

e : idA ! ED and " : idX ! DE ,

and hD,E , e, "i is a dual adjunction between A and X.

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Natural dualities: embeddings

Natural embeddings

For every A 2 A and X 2 X, there are naturally definedembeddings

eA : A ! ED(A) and "X : X ! DE(X).

These embeddings yield natural transformations

e : idA ! ED and " : idX ! DE ,

and hD,E , e, "i is a dual adjunction between A and X.

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Duality

If eA : A ! ED(A) is an isomorphism, for all A in A, then wesay that M yields a duality on A (or that M dualises M).

Equivalently, M yields a duality on A if the dual adjunctionhD,E , e, "i is a dual category equivalence between A anda full subcategory of X.

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Duality

If eA : A ! ED(A) is an isomorphism, for all A in A, then wesay that M yields a duality on A (or that M dualises M).

Equivalently, M yields a duality on A if the dual adjunctionhD,E , e, "i is a dual category equivalence between A anda full subcategory of X.

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Full duality

If, in addition, "X : X ! DE(X) is an isomorphism, for all X in X,then M yields a full duality on A (or that M fully dualises M).

Equivalently, M yields a full duality on A if the dual adjunctionhD,E , e, "i is a dual category equivalence between A and X.

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Embeddings, injectivity and strong duality

Let M be any alter ego of a finite algebra M, and let

D : A ! X and E : X ! A

be the induced hom-functors.

It is easy to see that:

I D and E send surjections to embeddings,I D sends embeddings in A to surjections in X if

and only if M is injective in A, andI E sends embeddings in X to surjections in A if

and only if M is injective in X.

Strong dualityIf M fully dualises M and M is injective in X (so that surjectionsin A correspond to embeddings in X), we say that M yields astrong duality on A (or that M strongly dualises M).

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D : A ! X and E : X ! A

be the induced hom-functors.It is easy to see that:

I D and E send surjections to embeddings,

I D sends embeddings in A to surjections in X ifand only if M is injective in A, and

I E sends embeddings in X to surjections in A ifand only if M is injective in X.

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D : A ! X and E : X ! A

and only if M is injective in A, and

I E sends embeddings in X to surjections in A ifand only if M is injective in X.

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D : A ! X and E : X ! A

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D : A ! X and E : X ! A

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Examples

I Both of our original examples — Stone duality andPriestley duality — are examples of strong dualities.

I Every finite lattice-based algebra admits a strong duality.[Clark, Davey 95]

I The unary algebra

admits a duality, but not a full duality.[Hyndman, Willard 00]

I The two-element implication algebra I := h{0, 1};!i doesnot admit a duality.[Davey, Werner 80]

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Examples

I The unary algebra

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Examples

I The unary algebra

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Examples

I The unary algebra

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Full versus strong

The Full versus Strong Problem, which dates back to thebeginnings of the theory in 1980, asks:

I Does there exist a finite algebra M and an alter ego Mof M that yields a duality on A = ISP(M) that is full butnot strong?

Why did this problem take 27 years to be solved?

I We have a very good understanding of strong dualities,and until recently, the only technique we had for proving aduality was full was by proving it was strong.

I While there are many examples of strong (and thereforefull) dualities with M not injective in A, for 27 years everyfull duality that was found had the property that M wasinjective in X.

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Full versus strong

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Full versus strong

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Full versus strong

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Full but not strong at the finite level

In 2005, Davey, Haviar and Willard gave a duality for boundeddistributive lattices, based on 3 = h{0, a, 1};_,^, 0, 1i, that isfull at the finite level but not strong at the finite level.

(1) The alter ego 3 := h{0, a, 1}; f , g,Ti dualises 3. [DHPr 95]

(2) The alter ego 3h := h{0, a, 1}; f , g, h,Ti fully dualises 3 atthe finite level but does not strongly dualise 3 at the finitelevel. [DHW 05]

(3) The alter ego 3h does not fully dualise 3. [DHW 05]

(4) In fact, every full duality for D is strong. [DHN 07]

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Full but not strong at the finite level

In 2005, Davey, Haviar and Willard gave a duality for boundeddistributive lattices, based on 3 = h{0, a, 1};_,^, 0, 1i, that isfull at the finite level but not strong at the finite level.

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Full but not strong at the finite levelIn 2005, Davey, Haviar and Willard gave a duality for boundeddistributive lattices, based on 3 = h{0, a, 1};_,^, 0, 1i, that isfull at the finite level but not strong at the finite level.

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Full but not strong!In 2007, Clark, Davey and Willard gave a full but not strongduality for the quasi-primal algebra

Q := h{0, a, b, 1}; t ,_,^, 0, 1i,

where t is the ternary discriminator.

(1) The alter ego Q> := h{0, a, b, 1}; u, u�1,Ti stronglydualises Q. [DWe 80]

(2) The alter ego Q? := h{0, a, b, 1}; graph(u),Ti fullydualises Q but does not strongly dualise Q. [CDW 07]

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Q := h{0, a, b, 1}; t ,_,^, 0, 1i,

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Q := h{0, a, b, 1}; t ,_,^, 0, 1i,

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Full and never strong!This left the question: does there exists a finite algebra M that isfully dualised by some alter ego but strongly dualised by none?

In 2009, Pitkethly gave a three-element example.

I Define the three-element topological structureS := h{0, 1, 2};_,Ti, where _ is the partial semilatticeoperation given below.

_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

I Now define the set of operations F := {f : Sn ! S | n � 1}.I Then S is an alter ego of the algebra S = h{0, 1, 2};F i.I The alter ego S fully dualises S. [Pi 09]I No alter ego strongly dualises S. [Pi 09]

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In 2009, Pitkethly gave a three-element example.

I Define the three-element topological structureS := h{0, 1, 2};_,Ti, where _ is the partial semilatticeoperation given below.

_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

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In 2009, Pitkethly gave a three-element example.I Define the three-element topological structure

S := h{0, 1, 2};_,Ti, where _ is the partial semilatticeoperation given below.

_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

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_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

I Now define the set of operations F := {f : Sn ! S | n � 1}.

I Then S is an alter ego of the algebra S = h{0, 1, 2};F i.I The alter ego S fully dualises S. [Pi 09]I No alter ego strongly dualises S. [Pi 09]

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_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

I Now define the set of operations F := {f : Sn ! S | n � 1}.I Then S is an alter ego of the algebra S = h{0, 1, 2};F i.

I The alter ego S fully dualises S. [Pi 09]I No alter ego strongly dualises S. [Pi 09]

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_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

I Now define the set of operations F := {f : Sn ! S | n � 1}.I Then S is an alter ego of the algebra S = h{0, 1, 2};F i.I The alter ego S fully dualises S. [Pi 09]

I No alter ego strongly dualises S. [Pi 09]

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_ 0 1 20 0 1 ·1 1 1 12 · 1 ·

1c0 c⌦⌦

⌦⌦

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What does it mean for two full dualities based on thesame algebra to be different?

We have seen that Q has two different fully dualising alter egos:

Q> := hQ; u, u�1,Ti and Q? := hQ; graph(u),Ti.

But in what sense are the two full dualities really different?

I The two dual categories X> and X? are both duallyequivalent to A := ISP(Q).

I So the categories X> and X? are equivalent.I In fact, they are concretely isomorphic.

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I So the categories X> and X? are equivalent.

I In fact, they are concretely isomorphic.

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The difference

The two categories have different embeddings!

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The difference

The two categories have different embeddings!

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The difference

The two categories have different embeddings!17 / 32

Comparing alter egos

Let M be any finite algebra.

There is a natural quasi-order on the alter egos of M:

M1 = hM;G1,H1,R1,Ti is a structural reduct ofM2 = hM;G2,H2,R2,Ti, written M1 v M2, if

I every total operation in G1 is a term function of M2,I every partial operation in H1 extends to a term function

of M2, andI every relation in R1 [ dom(H1) is conjunct-atomic definable

from M2.

Under this quasi-order, the collection of all alter egos of Mforms a complete lattice, AM — the lattice of alter egos of M.

(In fact, AM is doubly algebraic. [DPiW 12])

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Comparing alter egos

Let M be any finite algebra.

There is a natural quasi-order on the alter egos of M:

M1 = hM;G1,H1,R1,Ti is a structural reduct ofM2 = hM;G2,H2,R2,Ti, written M1 v M2, if

I every total operation in G1 is a term function of M2,I every partial operation in H1 extends to a term function

of M2, andI every relation in R1 [ dom(H1) is conjunct-atomic definable

from M2.

Under this quasi-order, the collection of all alter egos of Mforms a complete lattice, AM — the lattice of alter egos of M.

(In fact, AM is doubly algebraic. [DPiW 12])

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Comparing alter egos, continued

The structural reduct quasi-order v is “natural” for at leasttwo reasons:

I M1 v M2 precisely when there is a “nice” forgetful functorfrom IScP+(M2) to IScP+(M1). [DHW 052]That is,

I every closed substructure of (M2)S is also aclosed substructure of (M1)S, and

I every M2-morphism ' : X ! Y, where X 6 (M2)S andY 6 (M2)T , is also an M1-morphism.

I The relation v comes from a Galois connection on partialoperations, analogous to the one between operations andrelations used in clone theory. [DPiW 12]

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Lattice of alter egos: two examples

B = h{0, 1};_,^, 0, 0, 1i

2-elementBoolean algebra

D = h{0, 1};_,^, 0, 1i

2-elementbounded lattice

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The lattice of alter egos

How do the various flavours of duality occur within thelattice of alter egos?

The following are true (not always obviously so). [DPiW 12]

I The alter egos that yield a finite-level duality form aprincipal filter, DM, of the lattice AM.If any of these alter egos yield a duality, then they all do.

I The alter egos that yield a finite-level strong duality formthe top element of the lattice AM.So there is essentially only one candidate alter ego forstrong duality.

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How do the various flavours of duality occur within thelattice of alter egos?

The following are true (not always obviously so). [DPiW 12]

I The alter egos that yield a finite-level duality form aprincipal filter, DM, of the lattice AM.If any of these alter egos yield a duality, then they all do.

I The alter egos that yield a finite-level strong duality formthe top element of the lattice AM.So there is essentially only one candidate alter ego forstrong duality.

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I The alter egos that yield a finite-level full duality form acomplete sublattice, FM, of the lattice of alter egos AM.Thus FM is a doubly algebraic lattice.

I The alter egos that yield a full duality form an increasingsubset of FM.

I We do not know in general whether the alter egos thatyield a full duality must form a filter of FM, i.e., whetherthey are closed under meet.

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The lattice of finite-level dualities

The lattice DM of alter egos that dualise M at the finite level

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The lattice of finite-level full dualities

What is known about the lattice FM of alter egos that fullydualise M at the finite level?

I Whenever full implies strong at the finite level, we have|FM| = 1. For example, if M is a finite abelian group,semilattice or relative Stone Heyting algebra. [DHN 07]

I If M is a quasi-primal algebra, then FM is finite, sinceAM is finite. [DPiW 12]

I The lattice FQ has been calculated explicitly for the4-element quasi-primal algebra Q. [DPiW 12]

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The lattice of full dualities for Q

r = graph(u)r0 = fix(u)r1 = dom(u)r2 = ran(u)

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The lattice of finite-level full dualities for 3

Davey, Haviar and Pitkethly [DHPi 10] use Priestley duality tostudy the lattice F3 of all alter egos that fully dualise the3-element bounded lattice 3 at the finite level.

They prove that:

I the lattice F3 has size 2@0 by order-embedding thepowerset }(!) into F3;

I F3 is non-modular;

I there is an infinite ascending chain

31 @ 32 @ 33 @ 34 @ 35 @ · · ·

of alter egos, in the lattice A3, that alternately do anddo not fully dualise 3 at the finite level.

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They prove that:

31 @ 32 @ 33 @ 34 @ 35 @ · · ·

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They prove that:

31 @ 32 @ 33 @ 34 @ 35 @ · · ·

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They prove that:

31 @ 32 @ 33 @ 34 @ 35 @ · · ·

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Moving upwards through the lattice FM

The following theorem shows how the alternating chain

31 @ 32 @ 33 @ 34 @ 35 @ · · ·

might be constructed.

Theorem (Full enrichment [DPiW 12])Assume that M1 fully dualises M [at the finite level ], andM1 is a structural reduct of M2. The following are equivalent:

I M2 fully dualises M [at the finite level ];I for each relation r in R2\R1 [ {dom(h) | h 2 H2\H1},

every homomorphism h : r ! M extends to a term functionof M2.

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31 @ 32 @ 33 @ 34 @ 35 @ · · ·

It follows that the set of alter egos that fully dualise at the finitelevel is closed under arbitrary unions!

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31 @ 32 @ 33 @ 34 @ 35 @ · · ·

How do we move downwards, or even sideways, through thelattice FM?

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uH sentences

A universal Horn sentence (uH sentence, for short) in thelanguage (G,H,R) is a first-order sentence of the form

8~vh⇣ k

↵i(~v)⌘! �(~v)

where each ↵i(~v) is an atomic formula, and �(~v) is either anatomic formula or ‘false’.

I For each finite structure M, we have

ISP+(M) = Mod(ThuH(M)).

I In general, IScP+(M) ✓ ModBT(ThuH(M)).I A finite structure M = hM;G,H,R,Ti is called standard if

IScP+(M) = ModBT(ThuH(M)).

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uH sentences

8~vh⇣ k

↵i(~v)⌘! �(~v)

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uH sentences

8~vh⇣ k

↵i(~v)⌘! �(~v)

I In general, IScP+(M) ✓ ModBT(ThuH(M)).

I A finite structure M = hM;G,H,R,Ti is called standard ifIScP+(M) = ModBT(ThuH(M)).

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uH sentences

8~vh⇣ k

↵i(~v)⌘! �(~v)

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Standard structuresA finite structure M = hM;G,H,R,Ti is called standard ifIScP+(M) = ModBT(ThuH(M)).

Examples

I The cyclic group Cn = hCn; ·,�1 , 1i is strongly dualised bythe standard alter ego Cn = hCn; ·,�1 , 1,Ti.

I The semilattice S = h{0, 1};^i is strongly dualised by thestandard alter ego S = h{0, 1};^, 0, 1,Ti.

I Every quasi-primal algebra M is strongly dualised by thestandard alter ego M = hM;K [ Inv(M),Ti, where

I K is the set of nullaries corresponding to one-elementsubalgebras of M, and

I Inv(M) is the inverse semigroup of partial automorphismsof M.

I The bounded lattice 2 = h{0, 1};_,^, 0, 1i is stronglydualised by the non-standard alter ego 2 = h{0, 1};6,Ti.

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Standard structuresA finite structure M = hM;G,H,R,Ti is called standard ifIScP+(M) = ModBT(ThuH(M)).

Examples

I The cyclic group Cn = hCn; ·,�1 , 1i is strongly dualised bythe standard alter ego Cn = hCn; ·,�1 , 1,Ti.

I The semilattice S = h{0, 1};^i is strongly dualised by thestandard alter ego S = h{0, 1};^, 0, 1,Ti.

I Every quasi-primal algebra M is strongly dualised by thestandard alter ego M = hM;K [ Inv(M),Ti, where

I K is the set of nullaries corresponding to one-elementsubalgebras of M, and

I Inv(M) is the inverse semigroup of partial automorphismsof M.

I The bounded lattice 2 = h{0, 1};_,^, 0, 1i is stronglydualised by the non-standard alter ego 2 = h{0, 1};6,Ti.

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Full dualities and uH axiomatisations [DPiW 13]

Assume we have an alter ego M1 that fully dualises M.

I We show how to create new fully dualising alter egos M2using an axiomatisation of the universal Horn theory of M1.

Using this result we can prove the folowing.

Theorem (1)Assume M is fully dualised by some standard alter ego.If an alter ego M fully dualises M at the finite level, thenM is standard and fully dualises M.

Theorem (2)Assume M is fully dualised at the finite level by an alter ego Mof finite type and that we have a finite basis for ThuH(M).Then there is an algorithm for producing the smallestfull-at-the-finite-level alter ego M?.

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By combining these two results we can give a new proofthat Q? fully dualises Q.

I The alter ego Q> := h{0, a, b, 1}; u, u�1,Ti stronglydualises Q and is standard. [DWe 80]

I The algorithm from Theorem (2) yields the alter egoQ? := h{0, a, b, 1}; graph(u),Ti, which therefore fullydualises at the finite level.

I By Theorem (1), Q? is standard and fully dualises Q.

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So where is it at?

The study of full dualities is where it is at!

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So where is it at?

The study of full dualities is where it is at!

32 / 32

D. M. Clark and B. A. Davey, The quest for strong dualities,J. Austral. Math. Soc. Ser. A 58 (1995), 248–280.

D. M. Clark, B. A. Davey and R. Willard, Not every fullduality is strong! Algebra Universalis, 57 (2007), 375–381.

B. A. Davey, M. Haviar and T. Niven, When is a full dualitystrong? Houston J. Math. 33 (2007), 1–22.

B. A. Davey, M. Haviar, T. Niven and N. Perkal, Full but notstrong dualities: extending the realm, Algebra Universalis56 (2007), 37–56.

B. A. Davey, M. Haviar and J. G. Pitkethly, Using colouredordered sets to study finite-level full dualities, AlgebraUniversalis 64 (2010), 69–100.

B. A. Davey, M. Haviar and R. Willard, Structuralentailment, Algebra Universalis 54 (2005), 397–416.

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B. A. Davey, M. Haviar and R. Willard, Full does not implystrong, does it?, Algebra Universalis 54 (2005), 1–22.

B. A. Davey, J. G. Pitkethly and R. Willard, The lattice ofalter egos, Internat. J. Algebra Comput. 22 (2012),1250007, 36 pp.

B. A. Davey, J. G. Pitkethly and R. Willard, New-from-oldfull dualities via axiomatisation. About to be submitted.B. A. Davey and H. Werner, Dualities and equivalences forvarieties of algebras, Contributions to Lattice Theory(Szeged, 1980), Colloquia Mathematica Societatis JánosBolyai 33, North-Holland, 1983, pp. 101–275.

J. Hyndman and R. Willard, An algebra that is dualizablebut not fully dualizable, J. Pure Appl. Algebra 151 (2000),31–42.J. G. Pitkethly, A full duality that cannot be upgraded to astrong duality, Houston J. Math. 35 (2009), 757–774.

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Where is it at? Brian Davey GAIA 2013 - La Trobe …Boolean algebras Boolean spaces (i.e., compact,...

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