Lecture Notes

Chapter 1

Preliminaries

These are lecture notes, not a textbook. They are not meant to replace going tothe lectures. Specifically, they do not contain any motivation for the conceptsnor an intuitive explanation. They also may be out of order with the lectures.They (together with the example sheets) should contain the examinable book-work. These notes were prepared before actually giving the lectures and henceshould not be considered the finished lecture notes.

You need to become confident in set algebra, i.e. establishing equalities(and inequalities) of sets like X \

A =

X \A. There will be plenty of

opportunity for that. If you have difficulty with any of these, please contactme or your class tutor! In the classes, I dont expect class tutors to prove anyof these basic facts but rather concentrate on the topological matter.

1.1 A short summary of the Part A Topology course

I assume you know about the following concepts and can prove the lemmasbelow. If not, talk to me (or write me an e-mail at [email protected])as soon as possible!

Definition 1.1.1. A topology on a set X is a non-empty collection of subsetsof X closed under finite intersections and arbitrary unions and containing both and X, i.e.

(O0) 6= P (X);

(O1) and X ;

(O2) U, V = U V ;

(O3) U =U .

The pair (X, ) is called a topological space. When is clear, we simply referto it as the topological space X.

Elements of are called open sets, their complements are called closed sets.The collection of all closed sets is denoted by C (X).

1

2 CHAPTER 1. PRELIMINARIES

Lemma 1.1.2. Suppose X is a topological space.

(C0) 6= C (X) P (X);

(C1) C (X) and X C (X);

(C2) C,D C (X) = C D C (X);

(C3) C C (X) =C C (X);

Definition 1.1.3. The closure of a subset A of a topological space is given byA = {x X : open U x. U A 6= }.

Lemma 1.1.4. Suppose X is a topological space.

(Cl0) = ;

(Cl1) for all A X, A A;

(Cl2) for all A X, A = A;

(Cl3) for all A,B X, A B = A B.

Lemma 1.1.5. Suppose X is a topological space and A X.A =

{C C (X) : A C}.

Definition 1.1.6. The interior of a subset A of a topological space is given byint (A) = {x X : open U x. U A}.

Lemma 1.1.7. Suppose X is a topological space and A X.int (A) = X \X \A and A = X \ int (X \A)

Definition 1.1.8. A subset D of a topoological space X is dense in X if andonly if D = X. It is nowhere dense if and only if int

(D)= .

Definition 1.1.9. Suppose X and Y are topological spaces. A functionf : X Y is continuous if and only if for every open U Y , f1 (U) isopen (in X).

Lemma 1.1.10. Suppose X and Y are topological spaces and f : X Y afunction. The following are equivalent:

1. f is continuous;

2. C C (Y ) . f1 (C) C (X);

3. A X. f(A) f(A);

Lemma 1.1.11. The composition of continuous functions is continuous, i.e.if X,Y, Z are topological spaces and f : X Y and g : Y Z are continuousthen g f : X Z is continuous.

1.1. A SHORT SUMMARY OF THE PART A TOPOLOGY COURSE 3

Definition 1.1.12. A function d : X X R+0 is a metric on a set X if andonly if

(M0) for all x, y X, d(x, y) = 0 if and only if x = y;

(M1) for all x, y X, d(x, y) = d(y, x);

(M2) for all x, y, z X, d(x, z) d(x, y) + d(y, z);

The pair (X, d) is called a metric space. When d is clear, we simply re-fer to the metric space as X. For R+ and x X, we write B (x) ={y X : d(x, y) < }.

Lemma 1.1.13. Suppose X is a metric space.Then {

U X : x U R+s.t.B (x) U}

is a topology on X, called the induced topology on X.

Definition 1.1.14. A topological space X is Hausdorff if and only if for allx, y X such that x 6= y there exists disjoint open sets U, V such that x Uand y V .

Lemma 1.1.15. Every topology induced by a metric is Hausdorff.

Definition 1.1.16. A topological space X is compact if and only if every opencover of X has a finite subcover.

Lemma 1.1.17. A topological space X is compact if and only if for everyfamily C C (X) of closed sets such that for every finite subfamily D CD 6= , we have

C 6= .

Definition 1.1.18. The usual Euclidean topology on (any subset of Rn) is theone induced by the metric d(x, y) = x y. Its open sets are unions of openballs.

Chapter 2

Bases, Subbases, Neighbourhood

Bases

2.1 Bases

Definition 2.1.1. Suppose X is a topological space.A basis for X is a collection B of open subsets of X such that {

U : U B}

is the topology on X.If a basis has been fixed, its elements are called basic open sets and their

complements basic closed sets.

Example 2.1.2. Suppose X is a metric space.{B (x) : R+, x X} is a basis for the topology on X.

Lemma 2.1.3. Suppose X is a topological space.A collection B of open subsets of X is a basis for X if and only if for every

x U open X there is B B with x B U .

Lemma 2.1.4. Suppose X is a topological space and B a basis for X.Then

(B0) B 6= ,B = X;

(B1) if A,B B and x A B then there is C B with x C A B.

Lemma 2.1.5. Suppose X is a set and B a family of subsets of X satisfyingB0 and B1.

Then there is a unique topology on X, namely {U : U B}, such that B

is a basis for this topology.

Remark 2.1.6. Condition (B1) is satisfied in particular when A,B B =A B B.

Example 2.1.7. The collection of singletons is a bases for the discrete topo-logical space.

5

6 CHAPTER 2. BASES, SUBBASES, NEIGHBOURHOOD BASES

Example 2.1.8. Suppose X is a set and X. The co-finite particular pointtopology with particular point on X has is generated by the basis

{{x} : x X \ {}} {U X : U,X \ U is finite} .

Remark 2.1.9. Note that the above example can be adapted in various ways.Firstly, the family F = {U X : U,X \ U is finite} . can be replacedby any family of sets containing and being closed under finite intersec-tions (e.g. co-compact sets, co-countable sets, etc). Secondly the familyF\ = {{X} : x X \ {}} can be replaced by any basis for a topology onX \ {}.

Remark 2.1.10. This class of examples above might seem very abstract, but agood number of well known spaces can be generated this way:

Convergent Sequence Let X = {}, F the collection of co-finite sets containing , F\ thesingletons. Then X with the topology generated is homeomorphic to aconvergent sequence {2n : n } {0}.

Unit Circle Let X = R {}, F the collection of co-compact (w.r.t. the usualtopology on R) sets containing i.e. {U X : inX,X \ U compact},F\ any basis for R. ThenX with the topology generated is homeomorphicto the unit circle. The (Riemann) sphere can be generated in the sameway by replacing R with R2.

The 1-point compactifications of the rationals Let X = Q {}, F the collection of co-compact (w.r.t. the usualtopology on Q) sets containing , F\ a basis for the usual topology on Q.Then X with the topology generated is a compact, non-Hausdorff spacewhich contains Q as a dense subspace.

Lemma 2.1.11. Suppose X,Y are topological spaces, B is a basis for Y andf : X Y is a function.

f is continuous if and only if the preimage of every basic open subset of Yis open in X.

2.2 Subbases

Definition 2.2.1. Suppose X is a topological space.A subbasis for X is a collection S of open subsets of X such that{

F : F finite S}

is a basis for X.If a subbasis has been fixed, its elements are called subbasic open sets.

Lemma 2.2.2. Suppose X is a set and S a family of subsets of X.Then there is a unique topology on X, namely the one generated by {

F : F finite S}

with S a subbasis.

2.3. NEIGHBOURHOOD BASES 7

Example 2.2.3. The co-finite topology on a setX is generated by the subbasis{X \ {x} : x X}.

Lemma 2.2.4. Suppose X,Y are topological spaces, S is a subbasis for Y andf : X Y is a function.

f is continuous if and only if the preimage of every subbasic open subset ofY is open in X.

2.3 Neighbourhood Bases

Definition 2.3.1. Suppose X is a topological space.A neighbourhood basis for X is a family (Nx)xX of open subsets of X such

that:

N0 for all x X, Nx 6= ;

N1 for all x X,N Nx, x N ;

N2 for all x X,N1, N2 Nx there is N3 Nx such that x N3 N1N2;

N3 for all x X and for all open U X with x U there is N Nx withx N U ;

Lemma 2.3.2. Suppose X is a topological space, (Nx)xX a neighbourhoodbasis for X, A X and x A.

Then x int (A) if and only if there is N Nx with x N A.

Lemma 2.3.3. Suppose X is a set and (Nx)xX a family of subsets of Xsatisfying N0, N1 and N2. Then there is a unique topology on X such that(Nx)xX is a neighbourhood basis for (X, ).

Lemma 2.3.4. Suppose X is a toplogical space.If (Nx)xX is a neighbourhood basis for X then

xX Nx is a basis for X.

Conversely if B is a basis for X then ({B B : x B})xX is a neighbourhoodbasis for X.

Chapter 3

Products

Definition 3.0.5. Suppose X is a set and and are topologies on X. is coarser than (or equivalently that is finer than ) if and only if

.

Lemma 3.0.6. Suppose X is a set and and are topologies on X. is coarser than if and only if id : (X,) (X, ) is continuous.

Definition 3.0.7. Suppose X is a set, Y = (Y) a family of topologicalspaces and F = (f : X Y) a family of functions.

The initial topology on X wrt F is the coarsest topology such that for every the function f is continuous.

Lemma 3.0.8. Suppose X is a set, Y = (Y) a family of topological spacesand F = (f : X Y) a family of functions.

The initial topology on X is generated by the subbasis{f1 (U) : , U open Y

}.

Theorem 3.0.9. Suppose X is a set, Y = (Y) a family of topologicalspaces, F = (f : X Y) a family of functions and Z a topological space.

A function f : Z X is continuous wrt to the initial topology on X if andonly if for every the map f f is continuous.

Theorem 3.0.10. Suppose X is a topological space, Y = (Y) a family oftopological spaces and F = (f : X Y) a family of functions.

If for every topological space Z a function f : Z X is continuous if andonly if for every the map f f is continuous, then the topology on X isthe initial topology wrt F .

Proof. Let be the topology on X and let be the initial topology wrt F .Letting Z = (X, ) and f = id we see that each f is continuous and hencethat .

Now let Z = (X,) and again f = id. Then f f : (X,) Y iscontinuous for every so that f must be continuous. But this implies that , as required.

9

10 CHAPTER 3. PRODUCTS

Definition 3.0.11. Suppose (X) is a family of topological spaces.The (Tychonoff or product) topology on X =

X is the initial topol-

ogy with respect to the family () of projections : X Xf 7 f().

Definition 3.0.12. Suppose X and Y are topological spaces.Y X is the set of functions from X to Y .C (X,Y ) =

{f Y X : f is continuous

}is the set of continuous functions

from X to Y .The topology of C(X,Y ) inherits as a subspace from the product Y X =

xX Y is called the topology of pointwise convergence.

Chapter 4

Separation

4.1 Low Separation Axioms

Definition 4.1.1. Recall that a topological space X is Hausdorff (T2) if andonly if for every distinct x, y X there are disjoint open U, V with x U andy V .

Example 4.1.2. Suppose X is an infinite set. The co-finite topology on X isgiven by {U X : X \ U is finite} {}. Note that the co-finite topology onX is not Hausdorff (in fact it is anti-Hausdorff as any two non-empty open setsintersect). Note as well that every singleton {x} (for x X) is closed.

Example 4.1.3. Let X = {0, 1} and let = {, {1} , X}. Note that is atopology on X. The topological space S = (X, ) is called the Sierpinski space.Note that S is not Hausdorff and that {1} = X.

Theorem 4.1.4. Suppose X is a topological space and C X.C is closed in X if and only if there is a continuous function fC : X S

such that f1 (0) = C.

Lemma 4.1.5 (Embedding Lemma). Suppose X is a topological space, (f : X Y)a family of continuous functions that separates points, i.e. that if x, y X andx 6= y then there is such that f(x) 6= f(y).

If the set{f1 (V ) : , V open Y

}is a basis for X then the diago-

nal f : X

Y given by (f) (x)() = f(x) is an embeddingof X into

Y, i.e. a continuous map which is a homeomorphism onto its

image.

Proof. In this proof, we will abbreviate f to .Observe that is continuous, by noting that = F.Since the family (f) separates points, is injective.We claim that for U open X the set (U) is open in (U): suppose

y (U) and fix x U with (x) = y. Find and open V Y such thatx f

1 (V ) U and set V = 1 (V ). By construction we have y V and

11

12 CHAPTER 4. SEPARATION

if z V (X) then there is x X such that ((x)) = f(x

) V . Thusx U and therefore z (U). Hence y

1 (V )(X) (U)(X)and we have shown the claim.

Definition 4.1.6. SupposeX is a topological space. A family (f : X Y)of continuous maps separates points (separates points from closed sets) [sep-arates closed sets] if and only if for every distinct x, y X (for every x Xand C C (X) with x / C) [for every disjoint closed C,D X] there is and distinct z1, z2 Y such that f(x) = z1 and f(y) = z2 (f(x) = z1 andf (C) {z2}) [f (C) z1 and f (D) {z2}].

Definition 4.1.7. A topological space X is a T0 topological space if and onlyif C (X, S) separates points.

Example 4.1.8. Every T0 topological space can be embedded into a productof Sierpinski spaces.

Lemma 4.1.9. A topological space X is a T0 topological space if and only iffor every distinct x, y X there is an open set U X which contains exactlyone of x and y.

Definition 4.1.10. Suppose X is a topological space.X is a T1 space if and only if for every x X the singleton {x} is closed.X is a T2 or Hausdorff space if and only if for every distinct x, y X there

are disjoint open U x, V y.X is a Urysohn space if and only ifX is T1 and C (X, [0, 1]) separates points.X is a Tychonoff (T3.5) space if and only ifX is T1 and C (X, [0, 1]) separates

points from closed sets.X is a functionally normal space if and only if X is T1 and C (X, [0, 1])

separates closed sets.X is a regular space if and only if X is T1 and for every x X and closed

C X with x / C there are disjoint open U x and V C.X is a normal space if and only if X is T1 and for every disjoint closed

C,D X there are disjoint open U C and V D.

4.2 High Separation Axioms

Theorem 4.2.1. X is a Tychonoff space if and only if X is homeomorphic toa subspace of [0, 1] for some .

Proof. If X is Tychonoff the by the Embedding Lemma it is homeomorphic toa subspace of a power of [0, 1].

For the converse, let x U open X [0, 1]. We can find a basic [0, 1]-open V =

ni=0 i

1 (Ui) with Ui open [0, 1] such that x V X U .For each i note that xi Ui open [0, 1] and (by considering D = X \Ui)

we can find a continuous function (e.g. y 7 d(y,D)/d(x,D)) fi : [0, 1] [0, 1]with fi(x) = 1 and fi(X \ Ui) {0}.

4.2. HIGH SEPARATION AXIOMS 13

Let F =n

i=0 fi i . This is a continuous function from [0, 1] into

[0, 1] which maps x to 1 and [0, 1] \ V to 0. Its restriction to X is thus asrequired.

Definition 4.2.2. Suppose X is a topological space.X is regular (T3) if and only if X is T1 and for every x X and C C (X)

with x / C there are disjoint open sets U, V such that x U and C V .X is normal (T4) if and only ifX is T1 and for every disjoint closed C,D X

there are disjoin open U, V X such that C U and D V .

Lemma 4.2.3. Suppose X is a topological space.If X is Tychonoff then for every x X and C C (X) with x / C there

are disjoint open sets U, V such that x U and C V .

Lemma 4.2.4. If X is functionally normal then it is normal.

Theorem 4.2.5 (Urysohns Lemma). If X is normal, then it is functionallynormal.

Proof. Let C,D X be disjoint closed sets.By induction on any countable dense subset D of (0, 1) (e.g. the dyadic

rationals {m2n : n N,m {1, . . . , 2n 1}} or the rationals), we will defineopen sets Uq for q D such that for q, r D with q < r we have

C Uq Uq Ur Ur X \D.

Let U, V X be disjoint open such that C U and D V . Note that

C U U X \ V X \D

, so letting U1/2 = U satisfies our condition (for n = 1).Suppose we have defined Uq for all q D with denominator 2

n for n kand letm

{1, . . . , 2k+1 1

}. Ifm is even, we have already defined Um2(k+1) .

If m is odd, say m = 2l + 1 we let C = Ul2k (or C if l = 0) and D =

X \ U(l+1)2k (or X \D if l + 1 = 2k) and note that these are disjoint closed

sets by inductive hypothesis. Hence we can find disjoint open U , V such that

C U U X \ V X \D

and define Um2(k+1) = U, preserving the inductive hypothesis.

In the remainder of the proof, we interpret inf as 1.

Claim: for every x X we have

inf {r D : x Ur} = inf{r D : x Ur

}.

Since Ur Ur we clearly have . Now suppose s > inf{r D : x Ur

}. Since

D is dense we may assume wlog that s D and can find t {r D : x Ur

}


with t < s. Hence x Ut Us and thus s {r D : x Ur}, giving theresult.

Finally we will define our function fC,D : X [0, 1] by

f(x) = inf{q D : x Uq

}.

Now, if x C then x Uq C for all q D so that f(x) = 0. Thusf (C) {0}. Next, if x D then {q D : x Uq} = so that f(x) = 1.Hence f (D) {1}.

To check continuity, we observe that

f1 ([0, a)) = {x X : r D [0, a)s.t.x Ur} =

rD,ra

X \ Ur

both of which are open sets.

Theorem 4.2.6.

T4 = T3.5 = T3 = T2 = T1 = T0

andT3.5 = Urysohn = T2

Theorem 4.2.7 (Tietzes Extension Theorem). Suppose X is a topologicalspace.

X is normal if and only if for every closed C X and continuous mapf : C [0, 1] there is a continuous map F : X [0, 1] such that F |C= f , i.e.an extension.

Proof. Suppose every continuous map has an extension. Let C,D be disjointclosed and observe that D : C D [0, 1] is continuous. Hence the extensionis a function F : X [0, 1] such that F (C) {0} and F (D) {1}.

Now suppose X is normal, and fix a closed set C X and a continuousmap f : C [1, 1]. We will define a sequence Fn : X [3

n, 3n] of

continuous maps so that supxC

f(x)Nn=0 Fn(x) ( 23)n (i.e. such thatFn approximates f on C). We let F0 = 0. Having defined Fk, note that

gk = f

nk Fk|C is a continuous map from C to [ak, ak] =[(23

)k,(23

)k].

Hence Ck = gk1 ((,ak/3]) and Dk = gk

1 ([ak/3,)) are disjoint closedsets in C and hence in X. (Intuitively, Ck and Dk are the sets where ourapproximation is still pretty bad in one or the other direction.) Since X isnormal, we can find a continuous map Fk+1 : X [ak/3, ak/3] such thatFk+1 (Ck) {ak/3} and Fk+1 (Dk) {ak/3}. Now, if x Ck we haveg(x) Fk+1(x) [

23ak, 0] and similarly for x Dk. If on the other hand x

4.2. HIGH SEPARATION AXIOMS 15

C \ (Ck Dk), i.e. g(x) (ak/3, ak/3) then g(x) Fk+1(x) ( 23ak,

23ak

).

So supxC

f(x)nk+1 Fn(x) = supxC |g(x) Fk+1(x)| 23ak = ( 23)k+1as required.

Now note that by the Weierstrass M-test, F =

n=0 Fn is a well-defined,continuous function from X to [1, 1] and clearly F extends f .

Theorem 4.2.8. For i 3.5 Ti is a productive property, i.e. if all X, are Ti then so is

X.

Proof. Let P =

X.

T0 (T1) [T2]: Let x, y P and suppose x 6= y. Then there is withx 6= y and hence open U [disjoint open U, V] containing exactly one ofx, y (containing x but not y) [with x U and y V]. Hence

1 (U)contains exactly one of x, y (contains x but not y).[Hence x 1 (U) andy

1 (V) are disjoint open.]

T3: Let x P , and U P be open such that x U . Find a basic openV =

kk

1 (Vk) with x V U . For each k find Wk with xk

Wk Wk Vk and observe that

x k

k1 (Wk)

k

k1 (Wk)

k

k1 (Wk) V U.

T3.5: Let x P , U P open with x U . Find a basic open V =kk

1 (Vk) (k = 1, . . . , N) with x V U and for each k find acontinuous fk : Xk [0, 1] with fk(xk) = 0 and fk(X \ Vk) {1}.Let f(x) =

k(fk k)(x)/N and note that f is continuous (the sum of

continuous functions) and that f(x) = 0 and f(P \ U) {1} as required.

Theorem 4.2.9. For each i 3.5, Ti is a hereditary property, i.e. if X is Tiand A X then A is Ti in the subspace topology.

Proof. Immediate.

Theorem 4.2.10 (Jones Lemma). Suppose X is a normal space with a count-able dense subset D. Then X does not contain a closed, discrete subspace Awith |A| 20 .

Proof. SupposeX has a countable dense subsetD and f, g C(X, [0, 1]). Thenf = g if and only if f |D= g|D (for a quick proof, use a question on sheet 3),i.e. there is an injection C(X, [0, 1]) C(D, [0, 1]). Now

|C(D, [0, 1])| [0, 1]D = |[0, 1]||D| = (20)0 = 20.0 = 20

so that |C(X, [0, 1])| 20 .


Now suppose A is a closed, discrete subspace. Then [0, 1]A = C(A, [0, 1])(every function from a discrete space is continuous) and hence |C(A, [0, 1])| =(20)|A| = 20.|A|. Moreover, since A is closed, every element of C(A, [0, 1]) canbe extended to an element of C(X, [0, 1]) by Tietzes extension theorem, i.e.|C(A, [0, 1])| |C(X, [0, 1])|. Hence we must have that 20.|A| 20 , whichimplies that |A| < 20 . (Note that it is consistent [though pretty unlikely] withZFC that for every < 20 we have 2 = 20 , so that this is the best ZFCbound available.)

Chapter 5

Filters

5.1 Abstract Filters

Definition 5.1.1. Suppose X is a set.A filter F is a non-empty collection of pairwise intersecting subsets of X

which is closed under finite intersections and supersets, i.e. which satisfies:

F0 6= F P (X);

F1 / F ;

F2 F,G F = F G F ;

F3 F F F G X = G F .

Example 5.1.2. 1. {X} is a filter on X, called the indiscrete filter.

2. If A X then PA = {B X : A B} is a filter on X called the principalfilter at A. If A = {x} is a singleton, we also write Px.

3. SupposeX is a topological space and x X. ThenNx = {A X : x int (A)}is a filter on X called the neighbourhoodfilter at x.

4. LetX be any infinite set. Then the collection of co-finite sets {A X : X \A is finite}is a filter on X called the co-finite or Frechet filter.

Definition 5.1.3. Suppose X is a set, F , G are filters on X and A X.We say that F meshes with G, F#G, if and only if for every F F and

G G we have F G 6= . If F meshes with G, we also write F#G ={F G : F F , G G}.

We write A#F as shorthand for PA#F .

Lemma 5.1.4. Suppose X is a set and F , G are filters on X.Then F#G is a filter on X.

17

18 CHAPTER 5. FILTERS

Definition 5.1.5. Suppose X is a set and F and G are filters on X.G refines F (or G is finer than F or G extends F), written F G, if and

only if F G.This defines a partial order on the collection of all filters on X with least

element the indiscrete filter on X.

Lemma 5.1.6. Suppose X is a set and F , G are filters on X.Then F#G is the smallest filter on X containing both F and G.

Lemma 5.1.7. Suppose X is a set and {F : } a collection of filters onX.

If {F : } is totally ordered wrt , i.e. is a chain, then{F : }

is a filter on X.Hence every filter on X can be extended to a maximal filter wrt (under

the Axiom of Choice).

Definition 5.1.8. Suppose X is a set.An ultrafilter U on X is a filter on X which is maximal wrt , i.e. if G is

any filter on X and U G then U = G.

Example 5.1.9. Suppose X is a set and x X. Then Px is an ultrafilter onX.

In fact, it is consistent with ZF (i.e. set-theory without the Axiom ofChoice) that the only ultrafilters on X are the principal filters.

Lemma 5.1.10. Every non-principal ultrafilter contains the Frechet filter.

Lemma 5.1.11. Suppose X is a set and U a filter on X.U is an ultrafilter on X if and only if for every A X A U or X \A U .Moreover, if U is an ultrafilter on X and A,B X with A B U then

A U or B U .

Proof. Suppose U is a filter on X and A X such that A / U then no subsetof A belongs to U . But this means that (X \ A)#U and hence by maximalityof U we have that X \A U .

Conversely, suppose that V is a strictly finer filter than U . Find V V \U .Then X \ V U V a contradiction.

If U is and ultrafilter and A B U but A / U then X \A U and hence(A B) (X \A) = B \A U . Thus B U as required.

Definition 5.1.12. Suppose X is a set.A non-empty collection B of subsets of X is a filter-base if and only if

1. / B;

2. A,B B = C Bs.t.C A B.

A filter-base B generates the filter {A X : B Bs.t.B A}, the smallestfilter containing B. We write limB for the set of limit points of the generatedfilter.

5.2. FILTERS AND TOPOLOGY 19

Example 5.1.13. Then {[n,) : n } is a filter-base on generating theFrechet filter on .

More generally, suppose X is any infinite set and x = xn : n a se-quence in X. Then the collection of tails of x, {{xm : m n} : n }, is afilter-base. The generated filter will be called the filter generated by the se-quence x.

Lemma 5.1.14. Suppose X,Y are sets, f : X Y is a function and F is afilter on X. We write f (F) = {f (F ) : F F}.

1. f (F) is a filter-base on Y ; we will write G for the filter generated by it.

2. If f is surjective, then f (F) is a filter on Y ;

3. A G if and only if f1 (A) F ;

4. If F is an ultrafilter, then the filter generated by f (F) is an ultrafilter onY .

5.2 Filters and Topology

Definition 5.2.1. Suppose X is a topological space and F a filter on X.The set of limit points of F is limF = {x X : Nx F}. If x limF we

say that F converges to x, written F x.The set of cluster points of F is {x X : Nx#F}.

Lemma 5.2.2. Suppose X is a topological space, x X, A X and F andG are filters on X.

1. Px x and PA clusters at every point of A;

2. if F G then limF G;

3. if F x then F clusters at x;

4. the set of cluster points of F isFF F ;

5. x A if and only if A#NF if and only if there is a filter containing Awhich converges to x;

Lemma 5.2.3. Suppose X is a topological space and U an ultrafilter on X.If U clusters at x, then U converges to x.Thus, if F is a filter which clusters at x then there is a filter finer than F

that converges to x.

Theorem 5.2.4. Suppose X,Y are topological spaces and f : X Y a func-tion.

The following are equivalent:

1. f is continuous;

20 CHAPTER 5. FILTERS

2. for every filter F on X we have F x = f (F) f(x);

3. for every ultrafilter F on X we have F x = f (F) f(x);

Lemma 5.2.5. Suppose (X, d) is a metric space, xn : n a sequence in Xand x X.

The sequence converges to x (in the metric sense that d(xn, x) 0 asn) if and only if the filter generated by the sequence converges to x.

Definition 5.2.6. Recall that if X is a topological space and xn : n is asequence in X then it converges to some point x X if and only if for everyopen set U x there ism with n m = xn U , i.e. {xn : n m} U ,i.e. U contains a tail of the sequence.

Example 5.2.7. Let X be an uncountable set and fix a point X. Then

c(A) =

{A; if A is countable

A {} ; if A is uncountable

defines a closure operator and hence a topology on X.Note that the only sequences converging to are those which are eventually

constant with value equal to . Hence, there is no sequence in X \ {} whichconverges to . However, the collection of co-countable subsets (i.e. thoseA X such that X \A is countable) is a filter converging to .

Example 5.2.8. Let F be the Frechet filter on .F has no cluster points. Thus no filter finer than F converges and there

are non-convergent ultrafilters on (using AC).

Example 5.2.9. Let X = [0, 1] with its usual topology and let U be any ultra-filter on X. Then

FU F is non-empty and in fact a singleton. U converges

to the element of this singleton.

Chapter 6

Compactness

Definition 6.0.10. A topological space X is compact if and only if every opencover has a finite subcover, i.e. for every collection U of open sets such thatU = X, there exists a finite V U such that

V = X.

Theorem 6.0.11. A space is compact if and only if every family C of closedsets with the finite intersection property (i.e. if F finite C then

F 6= )

has non-empty intersection (i.e.C 6= ).

Proof. Duality.

6.1 Tychonoffs Theorem

Theorem 6.1.1 (Tychonoffs Theorem). The product of compact spaces iscompact.

Lemma 6.1.2. Suppose X is a compact space and U is an ultrafilter on X.Then U converges, i.e. limU 6= .

Proof. Assume not: let U be a non-convergent ultrafilter on X. Hence for everyx X we can find Nx Nx such that Nx / U and thus int (Nx) / U so thatwe may assume wlog that Nx is open.

Since U is an ultrafilter we must have X \Nx U for every x X.Then {Nx : x X} is an open cover of X and hence has a finite subcover V.

That meansV = X so = X \

V =

V V X \ V U , a contradiction.

Lemma 6.1.3. Suppose X is a topological space such that every ultrafilter onX converges.

Then X is compact.

Proof. Suppose X is not compact and let C be a non-empty collection ofclosed sets with the finite intersection property but empty intersection. Then{F : F finite C} is a filter-base for some filter F .

21

22 CHAPTER 6. COMPACTNESS

By Zorns Lemma (AC) we may extend F to an ultrafilter G which convergesto some x X (by assumption). Since

C = we can find C C G such

that x / C. But then X \ C Nx G and hence = C (X \ C) G, acontradiciton.

Theorem 6.1.4. The product of compact spaces is compact.

Proof of Tychonoffs Theorem. Suppose X, are compact topologicalspaces. Let U be an ultrafilter on X =

X. Then for each , (U)

is an ultrafilter on X and by compactness and Lemma 6.1.2 thus converges tosome x X.

We claim that U converges to x = (x): for suppose Nx Nx. Let V =

1 (U) be a basic open set such that x V U (i.e. finite and for , x U open X).

For , U () x implies U U () and thus 1 (U) U . But

U is closed under finite intersection so that V U . Finally U is closed undersupersets so that U U as required.

Hence every ultrafilter on X converges, so that by Lemma 6.1.3 X is com-pact.

Theorem 6.1.5 (Non-examinable). Tychonoffs theorem (together with ZF)implies the Axiom of Choice.

Proof. SupposeX, are non-empty sets and for each find some x / X(e.g. x = X. Set Y = X {x} with the topology generated by thesubbasis {A X : X \A finite } {x}. Clearly Y is compact and henceY is compact.Now, X is closed in Y and thus

{

1 (X) : }

Y is a family of closedsubsets of

Y. Now for any finite subset F , the set

F

1 (X) isnon-empty: for F , fix y X (finitely many choices) and observe that

f() =

{y; F

x; / Fgives an element of it.

By compactness

1 (X) is non-empty; choose some element f andobserve that f() X, so that f (or more precisely its graph) gives an elementof

X. Hence the latter is non-empty.Since the sets X were arbitrary non-empty sets, we have shown that any

product of non-empty sets is non-empty, which is equivalent to the Axiom ofChoice.

We now give a different, short, proof of the result:

Theorem 6.1.6. A topological space is Tychonoff if it is a subspace of a powerof [0, 1].

Proof. Note that Y = [0, 1] is compact Hausdorff by Tychonoffs theorem andproductivity of Hausdorffness. Hence Y is normal and thus Tychonoff (as Y isalso T1). Also note that being Tychonoff is hereditary. Thus subspaces of Yare Tychonoff as required.

6.2. COMPACTIFICATIONS 23

6.2 Compactifications

Definition 6.2.1. Suppose X is a topological space.A compactification of X is a homeomorphic embedding h : X Y where

Y is compact and h(X) is dense in Y . The compactification is a Hausdorffcompactification if and only if Y is Hausdorff.

We frequently identify X with its image under h and denote the compacti-fication simply by Y .

We denote Y \ X (or more precisely Y \ h(X)) as the remainder of thecompactification.

Theorem 6.2.2. A space X has a Hausdorff compactification if and only if Xis T3.5.

Proof. Suppose Y is compact Hausdorff and contains (a homeomorphic copy of)X. Then Y is normal, hence T3.5 and thus any subspace is T3.5. Conversely, ifXis T3.5 then by the embedding theorem X can be homeomorphically embeddedinto some Y = [0, 1] (some product of closed intervals). Identifying X with

its image, we note that YX

is compact Hausdorff and contains X as a densesubspace.

Example 6.2.3. S1 and [0, 1] are Hausdorff compactifications of (0, 1).

Definition 6.2.4. Suppose h1 : X Y1 and h2 : X Y2 are Haudorff com-pactifications of X. We write (h1, Y1) c (h2, Y2) if and only if there is acontinuous g : Y2 Y1 such that h1 = g h2.

Lemma 6.2.5. If X is Hausdorff, D a dense subset of X and f : X Y acontinuous map such that f |D is a homeomorphic embedding, then f (X \A)f (A) = .

Proof. If x X \ D and y D are such that f(x) = f(y) then wlog letX = D {x} and Y = f(D). Find disjoint open U x, V y and notethat D \ V is closed in D and hence f(D \ V ) is closed in f(D) = Y and thusf1(f(D \ V )) = D \ V is closed in X. But x / V gives that x / D.

Lemma 6.2.6. If g is a continuous map witnessing that (h1, Y1) c (h2, Y2)then g maps remainders to remainders.

Lemma 6.2.7. Suppose that hi : X Yi, i = 1, 2 are Hausdorff compactifica-tions of X such that (h1, Y1) c (h2, Y2) and (h2, Y2) c (h1, Y1). Then thereis a homeomorphism g : Y2 Y1 such that h1 = g h2.

Proof. By definition there is a continuous g2,1 : Y2 Y1 such that h1 = g2,1h2and a continuous g1,2 : Y1 Y2 such that h2 = g1,2 h1. Note that g1,2 g2,1 h2 = h2 so that g1,2 g2,1 is the identity on h2(X). As Y2 is Hausdorff it follows

that g1,2 g2,1 is the identity on h2(X) = Y2. Similarly g2,1 g1,2 is the identityon Y1, i.e. g2,1 is a homeomorphism as required (and g2,1 its inverse).


Definition 6.2.8. Suppose X is a T3.5 topological space. On the collectionof Hausdorff compactifications of X the relation (h1, Y1) (h2, Y2) if and onlyif (h1, Y1) c (h2, Y2) and (h2, Y2) c (h1, Y1) is an equivalence relation. cinduces a partial order on the set of equivalence classes.

Lemma 6.2.9. Every equivalence class of Hausdorff compactifications of X is

represented by a pair (h, Y ) where Y is a subset of [0, 1](2|X|).

Hence the collection of equivalence classes of Hausdorff compactifications isa set.

Proof. If D is dense in a regular space Y , then{int(A): A D

}is a basis of

Y of size 2|D|. Since h(X) is dense in Y and Y is completely regular it can be

embedded in [0, 1](2|X|) as required.

Theorem 6.2.10. The collection of equivalence classes of Hausdorff compact-ifications has suprema.

Proof. Let (h, Y) be Hausdorff compactifications ofX. LetH = Y. Then(H,H(X)) is a compactifcation of X and the witness that it is an upperbound.

If (c, Z) is another compactification which is an upper bound witnessed bysome f then f witnesses that (c, Z) (H,H(X)).

Theorem 6.2.11 (Stone-Cech-property). Suppose X is a T3.5 Hausdorff space.Then there exists a unique (equivalence class of) Hausdorff compacticfication : X X such that for every compact Hausdorff space Y and every contin-uous h : X Y there is a continuous h : X Y such that h = (h) .

This (equivalence class of) compactifications is called the Stone-Cechcompactification.

Proof. We first prove uniqueness: Suppose (h, Y ) were another such compact-ification. Then h : X Y has a continuous extension h : X Y withh = h , so that (, X) c (h, Y ). But symmetrically (h, Y ) c (i, X)yielding uniqueness of the equivalence class.

For existence, let (, X) be the greatest compactification of X (whichexists by the previous two lemmas). If f : X Z is a continuous map and Zis compact Hausdorff, then h = f : X XZ is an embedding of X intoa compact Hausdorff space. So there is a continuous map g : h(X) X suchthat g = h (by maximality of (, X)). Now let f = Z g and observethat f = Z g = Z h = f as required.

Theorem 6.2.12. Suppose (h, Y ) is a compactification of X such that everycontinuous map f : X [0, 1] can be extended to Y . Then (h, Y ) is equivalentto the Stone-Cechcompactification.

Proof. Identify X with h(X).Suppose g : X K is a continuous function from X into a compact Haus-

dorff space K and consider K as the subset of some [0, 1]. Then each fcan be extended to some continuous G : Y [0, 1]. Now let G = G, so


that G is continuous. Firstly, if x X then G(x)() = g(x)() so G extendsg. But then

G(Y ) = G(XY) G(X) = g(X) K = K

so that G is a continuous extension of g into K as required.

Remark 6.2.13. Apart from establishing the existence of the Stone-Cech-compactifications,it should in this course never be necessary to consider the concrete construc-tion of it. You should only ever have to use the Stone-Cech-property or thelast theorem.

Of particular interest is N. Note in this context that any function f : NK is continuous since N has the discrete topology.

Local Compactness

Definition 6.2.14. A Hausdorff topological space X is locally compact if andonly if for every open U and every x U there is an open V and a compact Ksuch that x V K U , i.e. such that {K X : K compact , x int (K)}generates the neighbourhoodfilter at x.

Lemma 6.2.15. Every Hausdorff locally compact space is Tychonoff.

Proof. Let x U open X where X is locally compact Hausdorff. Find anopen V and compact K such that x V K U . Since K is compact

Hausdorff, it is normal, so we can find a K-open W such that x W KW

V . Note that W is then an open subset of V which is open in X so open. Also

note that KW

is closed in K so compact so closed in the Hausdorff space X.As K is normal we can find a continuous f : K [0, 1] with f(x) = 1 and

f(K \W ) {0}. Let F : X [0, 1] extend f by defining it identically 0 onX \K. Then F |W= f |W and F |X\W= 0 are both continuous. Thus F is easilyseen to be continuous. But F (x) = f(x) = 0 and F (X \U) F (X \W ) {0}as required.

Lemma 6.2.16. If A is dense in X and U open in X then U A = U .

Proof. is obvious. So assume x U and let W x be open. Then W Uis open and non-empty so that W U A is non-empty. As W was arbitraryx inU A.

Lemma 6.2.17. If X is Hausdorff and A X is locally compact then A isthe intersection of an open and a closed set, i.e. A is open in A.

Conversely if X is locally compact Hausdorff and A is the intersection ofan open and a closed set, then A is locally compact.

Proof. SupposeX is Hausdorff and A locally compact. WlogX = A. For a Afind A-open Va and compact Ka with a Va Ka A and write Va = UaAwhere Ua is open in X. Since A is dense in X we have Va = Ua. But Ka is


compact so closed in the Hausdorff space X. Hence Ua Uq = Va Ka Aand thus a is in the interior of A. But a A was arbitrary, so A is open in A.

For the converse we first note that open subsets of locally compact spacesare locally compact as compactness is absolute and open subsets of open sets areopen. We will now show that closed subsets of locally compact Hausdorff spacesare locally compact: Let A closed X which is lcoally compact Hausdorff, leta A and let U a be open in A. Then U = V A for some V open in X andwe can find X-open W and compact K such that a W K V . But K is acompact subset of a Hausdorff space so closed in X and thus K A is a closedsubset of the compact space K hence compact. Thus a W A K A V A = U and W A is open in A and K A is compact as required.

Theorem 6.2.18. Suppose X is a locally compact Tychonoff space. Then thereis a compactification (h, Y ) of X such that Y \ h(X) is a singleton. This com-pactification is called the Alexandroff one-point compactification and is some-times denoted by X.

Proof. Let / X and define Y = X {} and let i : X Y be the inclusionmap. Equip Y with the topology

{U : U open X} {V : V and X \ V compact} .

Then Y is a compact Hausdorff topological space (compactness follows from thefact that covering by an open set leaves a compact subset of X to be covered;Hausdorffness follows from X being Hausdorff and locally compact). ClearlyX is dense in Y so that (i, Y ) is a compactification with Y \X = {}.

Lemma 6.2.19. Suppose X is a Tychonoff space and there is a compactifica-tion (h, Y ) of X such that h(X) is open in Y . Then X is locally compact.

Proof. Hausdorff compact spaces are locally compact so that h(X) is an opensubset of a locally compact space, hence locally compact. But X and h(X) arehomeomorphic, so X is locally compact.

Theorem 6.2.20. A Tychonoff space X is locally compact if and only if chas a least element.

Proof. The one-point compactification X is the c least element: if (h, Y ) isa compactification of X then h(X) is open in Y (since h(X) is an open subsetof h(X) = Y ). So consider the map f : Y X given by f(h(x)) = (x) forx X and f(y) = otherwise. If U is open in X then clearly f1(U) = h(U)is open in h(X) so open in Y . Otherwise, if U contains then X \U is compactso h(X \ U) is compact and hence f1(U) is open in Y (as it automaticallycontains Y \ h(X)).

Conversely, if (h, Y ) is c least we claim that Y \ h(X) is a singleton. Ifnot, let y1, y2 Y \ h(X) be distinct and consider X = Y \ {x1, x2} which isopen in Y , so locally compact. Hence X exists and is some compactificationof X: (h, X ). So we can find f : X Y by leastness and note that f = hon X and hence on all of X , a contradiction.


Theorem 6.2.21. For a Tychonoff topological space X the following are equiv-alent:

(i) X is locally compact.

(ii) In every compactification of X its image is open.

(iii) The image of X in X is open.

(iv) In some compactfification of X its image is open.

Proof. By constructing the one-point compactification of X we have shownthat (i) = (iv).

So suppose (iv) holds for some compactification (h, Y ) of X. Note that Xexists as X is Tychonoff and let H : X Y witness maximality of X. As Hmaps remainders onto remainders and the image of X onto the image of X wemust have X \ (X) = H1(Y \ h(X)) and the latter is closed by continuityof H and our assumption.

Now suppose that (iii) holds, let (h, Y ) be some Hausdorff compactificationof X and let H : X Y witness maximality of X. As before we haveH(X \ (X)) = Y \ h(X). But X \ (X) is closed (by assumption) in thecompact space X thus compact. Hence its image under the continuous mapH is compact in the Hausdorff space Y giving that Y \h(X) is closed, i.e. h(X)is open in Y as required.

Finally note that if (ii) holds then X is homeomorphic to an open subsetof a compact, hence locally compact, space (since X is Tychonoff it has someHausdorff compactification) so that X is locally compact.

1 - a worked example [Not examinable]

A set is well-ordered if and only if it is totally ordered and every non-emptysubset has a least element. is the typical example of a well-ordered set, butthere are others (e.g. a convergent increasing sequence with its limit, or twosuch sequences together or in fact an increasing sequence of limits of increasingsequences, the increasing sequences and the overall limit etc).

Two such set are equivalent if and only if there is a bijection which preservesthe order in both directions (i.e. x y f(x) f(y), so they areisomorphic as ordered sets).

A countable ordinal is (an equivalence class) of a well-ordered countableset.

Let 1 be the set of all countable ordinals. It is a well-ordered, uncountableset. Its topology (like that of all ordered sets) is the one generated by thesubbasis

{(, ) : 1} {(,) : 1}

where (, ) = { 1 : < } and (,) = { 1 : < }.If we write + 1 for the least element of { 1 : < } (and this

set is always non-empty).A longer introduction to 1 is given in the Set Theory courses. We need:


Lemma 6.2.22. 1 is an ordered topological space closed under countablesuprema (i.e. if C 1 is countable then supC 1) where A = {supC : C countable A}.Moreover (, ] = (, +1) is compact, so that 1 is locally compact. Fi-nally 1 is not compact and (thus) does not have a maximum.

A club subset of 1 is a closed (in the topological sense) unbounded (in theorder sense, i.e. there is not 1 which is an upper bound) subset of 1.

Here is the crucial property of 1:

Lemma 6.2.23. Any two club subsets of 1 meet (and their intersection is aclub). In fact, the countable intersection of club subsets is a club.

Proof. Let Cn, n be club subsets of 1 and let 1. By induction on nand m we will define increasing sequences (nm)m,mn so that

nm Cn for

each m n, nm for each m n, nm

n+1m

0m+1 for all m n + 1,

n .As base step observe that C0 is unbounded so choose

00 + 1 in C0.

There are two kinds of inductive steps: if we have just defined nm andn m + 1 then we observe that Cn+1 is unbounded so that we can findn+1m

nm in Cn+1. If n 6 m + 1 then we observe that C0 is unbounded so

that we can find 0m+1 nm in C0.

Now note that {nm : m n, n } is a countable set in 1 and hence hasa supremum . But by construction we must have for each fixed n that = sup {nm : m n} so that Cn. Hence

n Cn and clearly

+ 1 > . Thusn Cn is unbounded.

As an intersection of closed sets is closed, this proves the claim.

Theorem 6.2.24. Every (bounded) continuous f : 1 [0, 1] is eventuallyconstant, i.e. there is 1 such that f is constant on (,).

Proof. Let Cn,m = [(m 1)/2n, (m + 1)/2n] for m = 1, . . . , 2n 1, n .

Clearly Cn,m is a closed subset of [0, 1] so that Dn,m = f1 (Cn,m) is closed

(as f is continuous).By the previous lemma we have that ifDn,m andDn,m are both unbounded

then they meet so that

C = {Cn,m : Dn,m is unbounded}

is a collection of closed sets with the finite intersection property, and hencenon-empty intersection.

If x C then as C is countable and by the previous lemma, we must have

f1(x) club and thus the intersection contains at most thus exactly one pointx0.

LetO = {Cn,m : x0 / Cn,m} and for each C O find C such that f1 (C)

(, C) which is possible as this is bounded. As O is countable, 0 =supCO C + 1 1. Now let > 0: We must have f() Cn,m wheren,m is such that x0 Cn,m by construction of 0. But as that is true for anyfixed n, we must have f() = x0 (since the diameters of the Cn,m tend to 0 asn).


Theorem 6.2.25. 1 1.

Proof. As 1 is locally compact 1 exists. If f : 1 [0, 1] is continuous itis eventually constant with value x0 so can be extended continuously to 1by simply defining it to be x0 on the added point. Hence 1 satisfies theStone-Cech property and is thus aequivalent ot 1.

.

Chapter 7

Connectedness

7.1 Basics

Definition 7.1.1. Suppose X is a topological space. X is disconnected if andonly if there are disjoint, non-empty clopen (closed and open) sets C,D Xsuch that C D = X. X is connected if and only if it is not disconnected.

Lemma 7.1.2. Suppose X is a topological space. X is connected if and only ifevery continuous map f : X {0, 1} is constant where {0, 1} is equipped withthe discrete topology {, {0} , {1} , {0, 1}}. Equivalently X is disconnected ifand only if there is a continuous surjection from X onto the discrete two-pointspace {0, 1}.

Proof. Suppose X is disconnected and let C,D be a clopen partition of non-empty sets of X. Then C (the indicator function on C) is a continuoussurjection onto {0, 1}. Conversely if f : X {0, 1} is a continuous surjection,then f1 ({0}) , f1 ({1}) are a clopen partition of non-empty sets of X givingthat X is disconnected.

We now have a few lemmas from Part A, all of which are easily provedusing the previous characterization of connectedness:

Lemma 7.1.3. Suppose X is a topological space. If A, is a familyof connected subsets of X such that

A is non-empty, then

A is

connected.

Lemma 7.1.4. Suppose X is a topological space. If A0, A, are con-nected subsets of X such that for each we have A0 A 6= , thenA is connected.

Lemma 7.1.5. Suppose X is a topological space and An, n a countablefamily of connected subsets of X such that for each n we have AnAn+1 6=. Then

n An is connected.

These lemmas will be extensively used in the following.

31

32 CHAPTER 7. CONNECTEDNESS

Definition 7.1.6. Suppose X is a topological space and x X. The com-ponent of x in X is C(x) =

{C X : x C,C connected} and is called a

component of X.

Lemma 7.1.7. The components of X are the equivalence classes under theequivalence relation x y if and only if there is there is C X which isconnected and contains x and y. Hence they are the maximal non-empty con-nected subsets of X, i.e. C X is a component of X if and only if C 6= ,C is connected and for every D X, which is a proper superset of C, D isdisconnected.

Proof. Clearly is an equivalence relation (if x y z witnessed by Cx,y andCy,z then Cx,y Cy,z is connected and contains x and z) and the equivalenceclass of x X equals C(x). Also, given x X we have that C(x) is connected.Hence the equivalence classes are connected and non-empty. Finally if D isa proper superset of C(x) then there is y D \ C(x). Hence D cannot beconnected as otherwise x y is witnessed by D and hence y C(x) a con-tradiction. Thus the equivalence classes are the maximal connected subsets ofX.

Definition 7.1.8. Suppose X is a topological space and x X. The qua-sicomponent of x in X is Q(x) =

{C X : x C,C clopen in X} and is

called a quasicomponent of X.

Lemma 7.1.9. Suppose X is a topological space and x X. Then C(x) Q(x).

Proof. Suppose C X is connected and x C and that D is clopen in Xand x D. If C 6 D then find y C \ D and note that C D, C \ D arenon-empty clopen subsets of C which partition C. Hence C is disconnected acontradiction. Thus C D and hence the result follows.

Theorem 7.1.10 (Sura-Bura Lemma). If X is compact Hausdorff then thecomponents and quasi-components of X are the same.

Proof. By the last lemma we have that C(x) Q(x) for x X. It is thus suffi-cient to show that every quasi-component is connected as then the maximalityof C(x) ensures that Q(x) C(x).

Let Q be a quasi-component of X and assume Q is disconnected: finddisjoint non-empty Q-closed C,D with Q = C D and note that C,D are Xclosed since Q is closed.

Since X is normal there are disjoint X-open U C, V D. Then Q ={F : Q F clopen X} U V so that by compactness there is a clopen

F with Q F U V .Now

U F U F = U (U V ) F = U F

so that U F is clopen and similarly V F is clopen. Now wlog U F Q 6= then Q U F so that D Q = , a contradiction.

7.2. DISCONNECTED SPACES 33

Example 7.1.11. We cannot omit compactness in the Sura-Bura-Lemma: LetX = ([0, 1] {1/2n : n }) {(0, 0), (1, 0)}. The components of X are easilyseen to be [0, 1]{1/2n} for n , {(0, 0)} and {(1, 0)}. The quasicomponentsof X are [0, 1] {1/2n} for n and {(0, 0), (1, 0)}: every clopen set Ccontaining (0, 0) contains (0, 1/2n) for infinitely many n. As [0, 1] {1/2n} isconnected C must therefore contain (1, 1/2n) for infinitely many n and as C isclosed it must hence also contain (1, 0).

7.2 Disconnected Spaces

Definition 7.2.1. A topological space X is totally disconnected if and only ifeach component is a singleton.

A topological space X is zero-dimensional if and only if it has a basis ofclopen sets.

For a topological space X we write Cl (X) = {C X : C clopen}.

Lemma 7.2.2. A zero-dimensional Hausdorff space is totally disconnected.

Proof. In a zero-dimensional Hausdorff space we have C(x) Q(x) = {x}.

Lemma 7.2.3. A totally disconnected, compact Hausdorff space is zero-dimensional.

Proof. By the Sura-Bura Lemma we have {x} =CCl (X)x C for x X.

So if x U open X then by the compactness principle used in the proof ofthe Sura-Bura Lemma there is clopen C with x C U . Thus Cl (X) is abasis for X.

Definition 7.2.4. A Boolean Algebra is a tuple (B,,,,, 0, 1) where Bis a set, a partial order on B, : B B B; (a, b) 7 a b = inf {a, b}, : B B B; (a, b) 7 a b = sup {a, b}, : B B;B 7 B is uniquelydefined by B B = 0, 0 the least element of B and 1 the greates element ofB and ,, behave as expected (e.g. distributivity and (a b) = a band a = a). We call the meet and the join of two elements of B and bthe negation of b.

A function f : B B between Boolean Algebras is a morphism if and onlyif it is , , , , 0 and 1 preserving.

A filter F on B is a subset of B such that 0 / F , a, b F = a b Fand a F , a b = b F . A maximal filter called an ultrafilter on B.

Lemma 7.2.5. A filter U is an ultrafilter on a Boolean Algebra B iff for eachb B exactly one of b U and b U holds.

Also, the usual terminology and results about filters and ultrafilters holdwhen is replaced by and by .

Theorem 7.2.6. Suppose X is a zero-dimensional Hausdorff space then BX =(BX = Cl (X) ,,,, X \ ., , X) is a Boolean Algebra. Moreover, if f : X Y is a continuous function between such spaces then f : BY BX ;C 7f1 (C) is a morphism.

34 CHAPTER 7. CONNECTEDNESS

Proof. Simply check the various requirements, none of which are difficult toverify.

Theorem 7.2.7. Suppose B = (B,,,,, 0, 1) is a Boolean Algebra. ThenXB = {U : U is an ultrafilter on B} with topology generated by the basis

{Ub = {U : b U} : b B}

is a compact Hausdorff zero-dimensional topological space. Moreover if f : B B is a morphism between boolean algebras then f : XB XB ; f(U) =f1 (U) is a continuous map.

Proof. First note that it is easy to see that the set given is indeed a basis:UaUb = Uab and U1 = XB . Since Ub = XB \UB it is a basis of clopen sets.

Next, let us check Hausdorffness. Let U ,V be distinct ultrafilters on B.Then there is b B which is in exactly one of them, wlog U . Hence b Vand thus Ub and Ub are the required disjoint open sets.

Next, compactness: If C is a collection of basic closed sets (i.e. closedsets whose complements are basic open) then C is in fact a collection of basicopen sets. If it has the finite intersection property, then it is a filter base andcan hence be extended to an ultrafilter U . Clearly U

C so the latter is

non-empty as required.Finally assume that U is an ultrafilter of on B, f : B B a continuous

map. We claim that f1 (U) = {b B : f(b) U} is the basis for an ultrafilteron B: If f(b), f(b) U then f(b b) = f(b) f(b) U . If f(b) U andb b then f(b) f(b) so that f(b) U . Hence U is a filter. Also if f(b) / Uthen f(b) = f(b) U , hence we have an ultrafilter. Thus f is well defined.

Now, let Ub be basic open in XB . But f(U) Ub iff f1 (U) Ub iff

b f1 (U) iff f(b) U so that f1 (Ub) = Uf(b), so f is continuous.

Theorem 7.2.8 (Stone Duality). If X is compact Hausdorff zero-dimensionalthen X is homeomorphic to XBX (and for a map f : X Y we have (f

) = funder idenitfying X and Y with XBX and XBY respectively).

If B is a Boolean Algebra then B is isomorphic to BXB (and for a morphismf : B B we have (f) = f under the obvious identifications).

Proof. We contend ourselves with showing that X is homeomorphic to XBX :define a function f : X XBX by x 7 {C Cl (X) : x C}.

Firstly this is well defined: clearly f(x) is a filter on BX and as for anyclopen set we have either x C or x X \ C = C it is in fact an ultrafilter.

Next it is injective as X is Hausdorff: if x 6= y there is some clopen C x, y / C and then C f(x) but C / f(y).

f is continuous: if UC is basic open in XBX (so C Cl (B)) then f(x) UCif and only if x C, giving that f1 (UC) = C is open in X.

f is open onto f(X): if C is clopen in X then f(C) = UC f(X) is openin f(X). As images and unions commute, f is open onto its image.

We have shown that f is an embedding.

7.2. DISCONNECTED SPACES 35

To show that f is surjective, we need compactness of X: Suppose U is anultrafilter on BX . Then it is a collection of closed (in fact clopen) sets with thefinite intersection property, so has non-empty intersection. As X is Hausdorffand U is an ultrafilter, this intersection will be a single point x and it is theneasy to verify that f(x) = U .

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