METRICANDTOPOLOGICALSPACES

Part IB of the Mathematical Tripos of Cambridge

This course, consisting of 12 hours of lectures, was given by Prof. Pelham Wilson in Easter Term 2012. This LATEX version of the notes was prepared by Henry Mak, last revised in June 2012, and is available online at http://people.pwf.cam.ac.uk/hwhm3/. Comments and corrections to [email protected]. No part of this document may be used for profit.

Course schedules Metrics: Definition and examples. Limits and continuity. Open sets and neighbourhoods. Characterizing limits and continuity using neighbourhoods and open sets. [3]

Topology: Definition of a topology. Metric topologies. Further examples. Neighbourhoods, closed sets, convergence and continuity. Hausdorff spaces. Homeomorphisms. Topological and non-topological properties. Completeness. Subspace, quotient and product topologies. [3]

Connectedness: Definition using open sets and integer-valued functions. Examples, in- cluding intervals. Components. The continuous image of a connected space is connected. Path-connectedness. Path-connected spaces are connected but not conversely. Connected open sets in Euclidean space are path-connected. [3]

Compactness: Definition using open covers. Examples: finite sets and Œ0; 1. Closed subsets of compact spaces are compact. Compact subsets of a Hausdorff space must be closed. The compact subsets of the real line. Continuous images of compact sets are compact. Quotient spaces. Continuous real-valued functions on a compact space are bounded and attain their bounds. The product of two compact spaces is compact. The compact subsets of Euclidean space. Sequential compactness. [3]

1 M ETRICSPACES

Contents 2 1 Metric spaces 1.1 Introduction 5 1.2 Open balls and open sets 7 1.3 Limits and continuity 9 1.4 Completeness 9 2 Topological spaces 2.1 Introduction 13 2.2 Interiors and closures 2.3 Base of open subsets for a topology 14 2.4 Subspaces, quotients and products 16 3 Connectedness 3.1 Various results 19 3.2 Path-connectedness 20 3.3 Products of connected spaces 21 4 Compactness 4.1 Various results 25 4.2 Sequential compactness

§1M ETRIC SPACES 1.1 Introduction

Consider the Euclidean space Rn equipped with the standard Euclidean inner product: Given n Pn x; y R with coordinates xi ; yi respectively, we define .x; y/ i 1 xi yi , sometimes denoted by the2 dot product x y. ´ D
1 From this we have the Euclidean norm on Rn, x .x; x/ =2, representing the length of 1 k k ´ P 2
=2 the vector x. We have a distance function d2.x; y/ x y i .xi yi / , very often written as d simply. This is an example of a metric.´ k k D

Definition 1.1. A metric space .X; d/ consists of a set X and a function, called the metric, d X X R such that, for all P;Q;R X: W  ! 2 (i) d.P; Q/ 0, with equality iff P Q; > D (ii) d.P; Q/ d.Q; P /; D (iii) d.P; Q/ d.Q; R/ d.P; R/. C > Condition (iii) is called the triangle inequality. With the Euclidean metric, for any (possibly degenerate) triangle with vertices P;Q;R, the sum of the lengths of two sides of the triangle is at least the length of the third side.

2 1 . 1 I NTRODUCTION

n Proposition 1.2. The Euclidean distance function d2 on R is a metric in the sense of Definition 1.1.

Proof. (i) and (ii) are immediate. For (iii), use Cauchy-Schwarz inequality which says

n !2 n ! n ! X X 2 X 2 xi yi 6 xi yi ; i 1 i 1 i 1 D D D or in inner product notation, .x; y/2 x 2 y 2 for x; y Rn. (See Lemma 1.3) 6 k k k k 2 R y x y C Q x P

We take P to be the origin, Q with position vector x with respect to P , and R with position vector y with respect to Q. Then R has position vector x y with respect to P . Now C x y 2 x 2 2.x; y/ y 2 k C k D k k C C k k x 2 2 x y y 2 . x y /2: 6 k k C k k
k k C k k D k k C k k This implies d.P; R/ x y x y d.P; Q/ d.Q; R/. D k C k 6 k k C k k D C

Lemma 1.3. (Cauchy-Schwarz). .x; y/2 x 2 y 2 for any x; y Rn. 6 k k k k 2 Proof. For x 0, the quadratic polynomial in the real variable  ¤ x y 2 2 x 2 2.x; y/ y 2 k C k D k k C C k k is non-negative for all . Considering the discriminant, we have 4.x; y/2 4 x 2 y 2. 6 k k k k

Remarks.

1. In the Euclidean case, equality in the triangle inequality Q lies on the straight line segment PR. (See Example Sheet 1, Question 2) , 2. The argument for Cauchy-Schwarz above generalises to integrals. For example, if f; g are continuous functions on Œ0; 1, consider

Z 1 ÂZ 1 Ã2 Z 1 Z 1 2 2 2 .f g/ > 0 fg 6 f g : 0 C ) 0 0 0

3 1 . 1 I NTRODUCTION

More examples of metric spaces n Pn (i) Let X R , and d1.x; y/ i 1 xi yi or d .x; y/ maxi xi yi . These are both´ metrics. ´ D j j 1 ´ j j (ii) Let X be any set, and for x; y X, define the discrete metric to be 2 ( 1 if x y; ddisc.x; y/ ¤ ´ 0 if x y: D (iii) Let X C Œ0; 1 f Œ0; 1 R where f is continuous . We can define ´ ´ f W ! g metrics d1, d2, d on X by 1 Z 1 d1.f; g/ f g ; ´ 0 j j 1 ÂZ 1 à 2 2 d2.f; g/ .f g/ ; ´ 0 d .f; g/ sup f .x/ g.x/ : 1 ´ x Œ0;1 j j 2 For d2, the triangle inequality follows from Cauchy-Schwarz for integrals, i.e. R
2 R 2
R 2
fg 6 f g . See Remark 2 after Lemma 1.3, and use the same argument as in Proposition 1.2.

(iv) British rail metric. Consider Rn with the Euclidean metric d, and let O denote the origin. Define a new metric  on Rn by ( d.P; O/ d.O; Q/ if P Q; .P; Q/ C ¤ ´ 0 if P Q; D i.e. all journeys from P to Q P go via O. (All rail journeys go via London.) ¤ Some metrics in fact satisfy a stronger triangle inequality. A metric space .X; d/ is called ultra-metric if d satisfies condition (iii)0: d.P; R/ max d.P; Q/; d.Q; R/ 6 f g for all P;Q;R X. 2 Example. Let X Z and p be a prime. The p-adic metric is defined by ´ ( 0 if m n; dp.m; n/ D ´ 1=pr if m n where r max s N ps .m n/ : ¤ D f 2 W j g

r1 We claim that d is an ultra metric. Indeed, suppose that dp.m; n/ 1=p and r2 r1 r2 D dp.n; q/ 1=p for distinct m; n; q Z. Then p .m n/ and p .n q/ together minD r1;r2 2 j j imply p .m q/. So, for some r min r1; r2 , f g j > f g r min r1;r2 dp.m; q/ 1=p 1=p f g D 6 max 1=pr1 ; 1=pr2 D f g max dp.m; n/; dp.n; q/ : D f g

4 1 . 2 O PENBALLSANDOPENSETS

This extends to a p-adic metric on Q: For any rational x y, we can write x y r ¤ r D p m=n, where r Z, m; n are coprime to p, and define dp.x; y/ 1=p similarly. Then 2 D we also have .Q; dp/ as an ultra-metric space.

n 1 Example. The sequence an 1 p p , where p is prime, is convergent in ´ 1C C


C n .Q; dp/ with limit a .1 p/ . This is because dp.an; a/ 1=p for all n, and so ´ D dp.an; a/ 0 as n . ! ! 1 Lipschitz equivalence

Definition 1.4. Two metrics 1; 2 on a set X are Lipschitz equivalent if 0 < 1 2 R such that 11 2 21. 9 6 2 6 6

n Remark. For metrics d1; d2; d on R , one can show that d1 > d2 > d > d2=pn > 1 1 d1=n, so they are all Lipschitz equivalent. (See Example Sheet 1, Question 9)

Proposition 1.5. d1; d on C Œ0; 1 are not Lipschitz equivalent. 1

Proof. For n > 1, let fn C Œ0; 1 be as follows: 2

pn

fn

0 1 2 1 n n

d1.fn; 0/ Area of the triangle 1=pn 0 as n , while d .fn; 0/ pn as n . D D ! ! 1 1 D ! 1 ! 1

q 2 Exercise. Show that d2.fn; 0/ for all n, and so d2 is not Lipschitz equivalent to D 3 either d1 or d on C Œ0; 1. 1

1.2 Open balls and open sets

Let .X; d/ be a metric space, and let P X, ı > 0. We define the open ball by Bd .P; ı/ 2 ´ Q X d.P; Q/ < ı . This is often written as B.P; ı/ or Bı .P / simply. f 2 W g Examples.

1. In .R; d1/, we get open intervals of the form .P ı; P ı/. 2 2 C 2. In .R ; d2/, we get open discs of radius ı; With .R ; d / we get squares; With 2 1 .R ; d1/, we get tilted squares. (See Figures 1 (i), (ii) and (iii) below.) 3. In .C Œ0; 1; d /, see the example in Figure 1 (iv) below. 1 4. In .X; d / where X is any set, B P; 1
P for all P X. disc 2 D f g 2

5 1 . 2 O PENBALLSANDOPENSETS

P P P

ı ı ı (i) (ii) (iii)

ı f g

0 1 (iv)

2 2 2 Figure 1: B.P; ı/ in (i) .R ; d2/, (ii) .R ; d / and (iii) .R ; d1/. In (iv), g B.f; ı/, where f; g C Œ0; 1 and we1 use the metric d . 2 2 1

Definition 1.6. A subset U X of a metric space .X; d/ is an open subset if P U , open ball B.P;Â ı/ U for some ı > 0. 8 2 9 Â Note that an open subset is just a union of (usually infinitely many) open balls. Here is an 2 example of an open subset U of the space .R ; d2/:

ı P U

From Definition 1.6, we also define that a subset F X is closed if X F is open. Â n Example. The closed ball B.P; ı/ Q X d.P; Q/ 6 ı is a closed subset of X. Indeed, we show its complement is open.´ f 2 W g

ı0 ı Q P

If Q B.P; ı/, then d.P; Q/ > ı. We take some 0 < ı < d.P; Q/ ı. Suppose … 0 R B.Q; ı0/. Then using the triangle inequality, d.P; R/ > d.P; Q/ d.R; Q/ > d.P;2 Q/ ı > ı. So B.Q; ı / X B.P; ı/, as required. 0 0 Â n

6 1 . 3 L IMITSANDCONTINUITY

Lemma 1.7. Let .X; d/ be a metric space. Then:

(i) X and are open subsets of .X; d/; ; S (ii) If, i I , Ui are open subsets of .X; d/, then so too is i I Ui ; 8 2 2 (iii) If U1;U2 X are open, then so too is U1 U2. Â \ Proof. (i) and (ii) are immediate.

For (iii): Let P U1 U2, then open balls B.P; ı1/ U1 and B.P; ı2/ U2. Take 2 \ 9   ı min ı1; ı2 . Then B.P; ı/ U1 U2. D f g  \

Definition 1.8. Let P be a point in .X; d/. An open neighbourhood (nbhd) of P is an open subset N P , for example, the open balls centred at P . 3

Example. Open neighbourhoods around P R2 with the British rail metric : 2 If P O, and 0 < ı < d.P; O/, we have B.P; ı/ P . If P O, then B.P; ı/ ¤ D f g D are the open Euclidean discs of radius ı. So U .R2; / open means either O U , then U can be arbitrary, or O U , then for some ıÂ > 0 the Euclidean disc B .O;… ı/ U . 2 Eucl Â

1.3 Limits and continuity

Suppose x1; x2;::: is a sequence of points in the metric space .X; d/. We define xn x X, ! 2 and say xn converges to limit x, to mean d.xn; x/ 0 as n . Equivalently, we have ! ! 1  > 0, N such that xn B.x; / n > N . See an example about convergence in .Q; dp/ on page8 5. 9 2 8

Example. Consider fn X C Œ0; 1 in the proof of Proposition 1.5 on page 5. fn 0 2 ´ ! in .X; d1/, but not so in .X; d2/ or .X; d /. 1

Proposition. xn x in .X; d/ open neighbourhoods U x, N such ! , 8 3 9 that xn U for all n N . 2 > Proof. ( ): Take U B.x; / for any given  > 0. ( D ( ): Given an open set U x,  > 0 such that B.x; / U . So N such that xn B.x;) / U n N . 3 9 Â 9 2 Â 8 >

This means that the convergence of xn may be rephrased solely in terms of open subsets in .X; d/.

Caveat. Consider X C Œ0; 1, and the space .X; d1/. Let gn X be as follows: ´ 2

7 1 . 3 L IMITSANDCONTINUITY

1

gn

0 1 1 n

Then gn.0/ 1 n, but gn 0 as n . Is this not counter-intuitive? D 8 ! ! 1

Definition 1.9.(-δ definition of continuity). A function f .X; 1/ W ! .Y; 2/ is

(i) continuous at x X if  > 0, ı > 0 such that 1.x ; x/ < ı 2 8 9 0 ) 2.f .x0/; f .x// < ;

(ii) uniformly continuous on X if  > 0, ı > 0 such that 1.x1; x2/ 8 9 < ı 2.f .x1/; f .x2// < . )

Note that item (i) of the above definition may be rephrased as: f .X; 1/ .Y; 2/ is continuous at x X iff  > 0, ı > 0 such that f .B.x; ı// B.f .x/;W /, or! equivalently, B.x; ı/ f 1.B.f2 .x/; //8 x 9 X f .x / B.f .x/; / . Â Â D f 0 2 W 0 2 g

Lemma 1.10. If f .X; 1/ .Y; 2/ is continuous and xn x in .X; 1/, W ! ! then f .xn/ f .x/ in .Y; 2/. !

Proof.  > 0, ı > 0 such that 1.x ; x/ < ı 2.f .x /; f .x// < . As xn x, N such 8 9 0 ) 0 ! 9 that n N 1.xn; x/ < ı. So n N 2.f .xn/; f .x// < . Therefore f .xn/ f .x/. > ) > ) !

Example. Consider the identity map id .C Œ0; 1; d / .C Œ0; 1; d1/. Since d .f; g/ W 1 ! 1 <  supx Œ0;1 f .x/ g.x/ <  d1.f; g/ < , we see that id is continuous. , 2 j j ) We can use the functions fn C Œ0; 1 in the proof of Proposition 1.5 on page 5 to 2 show that the identity in the other direction, id .C Œ0; 1; d1/ .C Œ0; 1; d /, is not W ! 1 continuous, again by noting that d1.fn; 0/ 0 but d .fn; 0/ as n . ! 1 ! 1 ! 1 We wish to express the continuity of a map purely in terms of open sets.

Proposition 1.11. A map f .X; 1/ .Y; 2/ is continuous W ! , (i) open subsets U Y , f 1.U / are open in X; or equivalently 8 Â (ii) closed subsets F Y , f 1.F / are closed in X. 8 Â

1 Proof. (i) ( ): For given  > 0 and x X, take U B.f .x/; /. Then f .U / is open ( 1 2 D ) ı > 0 such that B.x; ı/ f .U /. This means 1.x ; x/ < ı 2.f .x /; f .x// < . 9 Â 0 ) 0 1 ( ): For an open subset U Y and any x f .U /, we can choose an open ball ) 2 2 1 B.f .x/; / U . Since f is continuous at x, ı > 0 such that B.x; ı/ f .B.f .x/; // 1 Â 9 1 Â Â f .U /. This is true for all such x, therefore f .U / is open in X.

8 1 . 4 C OMPLETENESS

1 (ii) ( ): F is closed in Y Y F is open in Y . By (i), this implies f .Y F/, which is equal to)X f 1.F /, is open, in X n f 1.F / is closed in X. n n , 1 ( ): Suppose U is open in Y . Then Y U is closed in Y so by the hypothesis, f .Y U/ X (f 1.U / is closed in X f 1.U / isn open in X. By (i), f is continuous. n D n , 1.4 Completeness

A metric space .X; d/ is called complete if, for all sequences of the form x1; x2;::: in X satisfying “  > 0, N such that d.xm; xn/ <  m; n N ”, we have xn x for some x X. 8 9 8 > ! 2 .R; d1/ is complete. This is known as Cauchy’s principle of convergence. However, .Q; d1/ and ..0; 1/ R; dEucl/ are both not complete. Â

Example. Let X C Œ0; 1. We show that .X; d1/ is not complete. ´ For n > 1, take fn X as follows: 2

1

fn

0 1 1 1
1 2 2 C n

1 Then d1.fm; fn/ 6 N for m; n > N , so fn form a Cauchy sequence. Suppose R 1 R 1 fn f X as n , i.e. 0 fn f 0. By the triangle inequality, 0 fn ! R 1=2 ! 1 jR 1 j ! R 1=2 R 1 j f > 0 . f 1 fn 1 / 1=2. f fn / 0 f 1 1=2 f , giving 1j j j j j C j j j j ! j j C j j R =2 f 1 R 1 f 0. 0 j j D 1=2 j j D Assuming f is continuous, this means

( 1 1 if x < 2 ; f .x/ 1 D 0 if x > 2 ;

which is discontinuous at x 1 . Contradiction. D 2

§2T OPOLOGICAL SPACES 2.1 Introduction Consider the open subsets of a metric space with properties as described in Lemma 1.7 on page 7. These properties may be abstracted out for a definition of a topological space:

Definition 2.1. A topological space .X; / consists of a set X and a set (the “topology”)  of subsets of X (hence  P.X/, the power set of X), where by definition we call the elements of  the “open” subsets, satisfying:

9 2 . 1 I NTRODUCTION

(i) X; ; ; 2 S (ii) If Ui  i I , then i I Ui ; 2 8 2 2 2 (iii) If U1;U2 , then U1 U2 . 2 \ 2 By induction, we also get the closure property (iii) for finite intersections. A subset Y X in a topological space .X; / is called closed if X Y is open. Therefore we can describe a topology on a set X by specifying the closed sets in X nwhich satisfy

(i) X; are closed; ; T (ii) If Fi are closed i I , then so too is i I Fi ; 8 2 2 (iii) If F1;F2 are closed, then so too is F1 F2. [ This description of a topology is sometimes more natural: Consider the non-metric topologies in examples (ii) and (iii) below. Lemma 1.7 on page 7 implies that every metric space .X; d/ gives rise to a topology. This is called the metric topology.

Two metrics 1; 2 on X are said to be topologically equivalent if their associated topologies are the same.

Exercise. Show that 1; 2 are Lipschitz equivalent 1; 2 are topologically equivalent. ) Example. The discrete metric on a set X gives rise to the discrete topology in which every subset in X is open, i.e.  P.X/. D Examples of non-metric topologies (i) Let X be a set with at least two elements, and  X; , the indiscrete topology. ´ f ;g (ii) Let X be any infinite set, and  Y X X Y is finite , the co-finite ´ f;g [ f  W n g topology. If X R or C then this is known as the Zariski topology, where open D sets are “complements of zeros of polynomials”. The Zariski topologies on Rn or Cn are very important in algebraic geometry.

(iii) Let X be any uncountable set e.g. R or C, and  Y X X Y is countable , the co-countable topology. ´ f;g [ f  W n g (iv) For the set X a; b , there are exactly 4 distinct topologies:  ˚ ; a ; b ; a; D f g ´ ; f g f g f b « i.e. the discrete / metric topology,  ˚ ; a; b « i.e. the indiscrete topology, g ˚ ; a ; a; b «, or  ˚ ; b ; a;´ b «.; f g ´ ; f g f g ´ ; f g f g

Example. The half-open interval topology  on R consists of arbitrary unions of half- open intervals Œa; b/ where a < b, a; b R. Clearly R;  and  is closed under unions. 2 ; 2 Suppose U1;U2 . We show that P U1 U2, Œa; b/ such that P Œa; b/ U1 U2 2 8 2 \ 9 2 Â \ and hence U1 U2 . \ 2 Since P U1 and P U2, we have P Œa1; b1/ U1 and P Œa2; b2/ U2 2 2 2 Â 2 Â for some a1 < b1, a2 < b2. Let a max a1; a2 and b min b1; b2 . Then D f g D f g P Œa; b/ U1 U2, as required. 2 Â \

10 2 . 1 I NTRODUCTION

More definitions

Definition 2.2. Let .X; 1/ and .Y; 2/ be topological spaces. Let P .X; 1/. 2 (i) An open neighbourhood (nbhd) of P is an open set U X with P U . (cf. Definition 1.8 on page 7.) Â 2 (ii) A sequence of points xn converges to limit x if, for any open neigh- bourhood U x, N such that xn U n > N . (cf. Section 1.3 on page 7.) 3 9 2 8

(iii) A map f .X; 1/ .Y; 2/ is continuous if, for any open set U Y , f W1.U / is open! in X. (cf. Proposition 1.11(i) on page 8.) Â The proof of Proposition 1.11(ii) on page 8 shows that f is continuous closed set F Y , 1 , 8 Â f .F / is closed in X.

Example. The identity map id .R; Eucl/ .R; co-finite topology/ is continuous, W ! because closed sets in the co-finite topology, namely, the finite sets and R, are closed in the Euclidean topology.

The identity map id .R; Eucl/ .R; co-countable topology/ is not continuous, W ! because Q R is closed in the co-countable topology, but not so in the Euclidean topology. Â

Definition 2.3. A map f .X; 1/ .Y; 2/ is a homeomorphism if: W ! (i) f is bijective; 1 (ii) Both f and f are continuous.

In this case, the open subsets of X correspond precisely to the open subsets of Y under the bijection f . This can be used to define an equivalence relation between topological spaces.

A property on topological spaces is called a topological property if .X; 1/ has the property, and .X; 1/ is homeomorphic to .Y; 2/ implies that .Y; 2/ has the same property.
Example. Consider the topological spaces .R; Eucl/ and . 1; 1/; Eucl . Let f R W ! . 1; 1/, defined by f .x/ x=.1 x /. f is bijective with inverse g . 1; 1/ R given by g.y/ y=.1 D y /. BothC j jf and g are continuous, and soW f and g!are homeomorphisms.D j j
Note that .R; dEucl/ is complete, whilst . 1; 1/; dEucl is not. Therefore this example shows that ‘completeness’ is a property of the metric on a metric space, and not just a topological property. However, it also shows that using the homeomorphism, we can construct a complete metric on . 1; 1/ coming from dEucl on R which is topologically equivalent to dEucl on . 1; 1/. Definition 2.4. A topological space .X; / is called Hausdorff if P;Q X with P Q, disjoint open sets U P and V Q, i.e. we can8 separate2 points by¤ open9 sets. This is a topological3 property. 3

11 2 . 1 I NTRODUCTION

U V P Q

Example. R with the indiscrete, co-finite or co-countable topology is not Hausdorff, because any two non-empty open sets intersect non-trivially (see the Examples (i), (ii) and (iii) on page 10). But any metric space is clearly Hausdorff. So these topologies are non-metric.

Example. The half-open interval topology on R is Hausdorff. If a; b R with a < b, then take Œa; b/ a, Œb; b 1/ b will do. Example Sheet 2, Question 182 shows that this topology is non-metric.3 C 3

Definition 2.5. In a topological space X, for any A X, we say that x0 X is an accumulation point or limit point of A if any open neighbourhood U2 of x0 satisfies U A . \ ¤ ; Lemma 2.6. A X is closed A contains all of its accumulation points, i.e.  , if x0 X is an accumulation point of A then x0 A. 2 2

Proof. ( ): Suppose A is closed and x0 X A. Then take U X A, an open ) 2 n ´ n neighbourhood of x0. Now U A , so x0 is not an accumulation point. \ D; ( ): Suppose A is not closed, then X A is not open. x0 X A such that no open ( n 9 2 n neighbourhood U of x0 is contained in X A, i.e. any open neighbourhood U of x0 satisfies n U A . So x0 is an accumulation point of A but not in A. \ ¤ ;

Remark. Suppose we have a convergent sequence xn x X with xn A n. Then for ! 2 2 8 any open neighbourhood U of x, N such that xn U n > N . So x is an accumulation point of A. Therefore if A is closed,9 we must have2x 8A. 2 A topological space .X; / has a countable base of open neighbourhoods (or is first countable) if, P X, open neighbourhoods N1 N2 N3 of P with the property that, open 8 2 9 Ã Ã Ã


8 neighbourhood U of P , m such that Ni U i m. 9 Â 8 > Example. Any metric space is first countable: Given P X, take the open balls B P; 1
2 n for n N. 2 Lemma 2.7. If .X; / is first countable, and A X has the property that “ Â 8 convergent sequences xn x X with xn A n, the limit x is in A”, then A is closed. ! 2 2 8

Proof. By Lemma 2.6, it suffices to prove that any accumulation point x of A is the limit of some sequence xn with xn A n. Let N1 N2 N3 be a base of open neighbourhoods 2 8 Ã Ã Ã


for x. Then i, since x is an accumulation point, xi A Ni , giving the sequence .xi /. 8 9 2 \ 8 open neighbourhood U of x, by first countability m such that Ni U i m. So xi U 9 Â 8 > 2 i m, i.e. xn x. 8 > ! Remark. Example Sheet 1, Question 19 (revised version on the DPMMS website) shows that we do need the first countability condition here.

12 2 . 2 I NTERIORSANDCLOSURES

2.2 Interiors and closures Given any A X, we define the interior Int .A/ or A0 of A to be the union of all open subsets contained in A.ÂInt .A/ is the largest open subset contained in A: If U is open and U A, then U Int .A/.   Define the closure Cl .A/ or A of A to be the intersection of all closed sets containing A. Cl .A/ is the smallest closed set containing A: If F is closed and A F , then Cl .A/ F .   Example. Consider R with the Euclidean topology. Int .Q/ (because any non- D; empty open subset contains irrationals) and Cl .Q/ R. Also, Int .Œ0; 1/ .0; 1/ and Cl ..0; 1// Œ0; 1. D D D For any set, we have Int .A/ A Cl .A/. The boundary or frontier of A is defined to be @A Cl .A/ Int .A/.   ´ n A set A X is called dense if Cl .A/ X.  D If A B are subsets of X, then Int .A/ Int .B/ and Cl .A/ Cl .B/.    Proposition 2.8. Let A X. Then:  (i) Int .Cl .Int .Cl .A//// Int .Cl .A//; D (ii) Cl .Int .Cl .Int .A//// Cl .Int .A//. D Proof. (i): Since Int .Cl .A// is open and Int .Cl .A// Cl .Int .Cl .A///, taking interiors of both sides we have Int .Cl .A// Int .Cl .Int .Cl .A////. Since Cl .A/ is closed and Int .Cl .A// Cl .A/, taking closures then interiors of both sides we have Int .Cl .Int .Cl .A//// Int .Cl .A//Â.  (ii): Similar argument. See Example Sheet 1, Question 11.

This shows that if we start from an arbitrary set A X and take successive interiors and clo- sures, we may obtain at most 7 distinct sets, namely, AÂ, Int .A/, Cl .A/, Cl .Int .A//, Int .Cl .A//, Int .Cl .Int .A/// and Cl .Int .Cl .A///. There is such an A where all 7 of these are distinct: See Example Sheet 1, Question 11.

2.3 Base of open subsets for a topology

Given a topological space .X; /, we say that a collection B Ui i I of “basic” open sets ´ f g 2 form a base or basis for the topology if any open set is the union of elements of B.

Question: When does an arbitrary collection B Ui i I of subsets of a set X form the base ´ f g 2 for some topology on X? Answer: If, i; j , Ui Uj is the union of some Uk’s in B. If so, we 8 \ can define a topology by specifying that an open set is just a union of some Ui ’s in B (and also include X, if necessary). A topological space .X; / is called second countable if it has a countable base of open sets. .X; / is second countable .X; / is first countable, because P X and open neighbourhood ) 8 2 U of P , U is the union of some basic open sets and so Ui U with P Ui i N. 9 Â 2 8 2

13 2 . 4 S UBSPACES , QUOTIENTSANDPRODUCTS

2.4 Subspaces, quotients and products

Let .X; / be a topological space, and Y X. The subspace topology on Y is  Y U Y U  . Consider the inclusion map i Y, X. It is continuous (becausej ´ fi 1.U\ / W U 2Y ),g and the subspace topology is the smallestW !topology on which i is continuous. D \ Proposition 2.9.

(i) Let B be a base for the topology . Then B Y U Y U B is a base for the subspace topology; j ´ f \ W 2 g

(ii) Let .X; / be a metric space and 1 be the restriction of the metric to Y . Then the subspace topology on Y induced from the -metric topology on X is the same as the 1-metric topology on Y . S S Proof. (i): Any open set U X is U U˛ where U˛ B, so U Y .U˛ Y/. Â D ˛ 2 \ D ˛ \ (ii): A base for the metric topology on X is given by open balls B.x; ı/ for a fixed x X. 2 y B.x; ı/ Y , ı > 0 such that B.y; ı / B.x; ı/, then B .y; ı / B.y; ı / Y 8 2 \ 9 0 0 Â 1 0 D 0 \ Â B.x; ı/ Y . Therefore B.x; ı/ Y , a base for the subspace topology, is the union of open \ \ 1-balls.

Example. .0; 1 R with the Euclidean topology has a base for the subspace topology consisting of open intervals .a; b/ and half-open intervals .a; 1, where 0 < a < b < 1.

Let .X; / be a topological space and be an equivalence relation on X. Denote the quotient set by Y X= and the quotient map by q X Y where x Œx. The quotient topology ´  1 W ! 7! on Y is then given by U Y q .U /  , i.e. the subsets U of the quotient set Y , for which the union of the equivalencef  classesW in X2correspondingg to points of U is an open subset of X. Remark. q is continuous, and the quotient topology is the largest topology on Y for which this is so. If f X Z is any continuous map between topological spaces such that x y f .x/W !f .y/, then there is a unique factorisation and f , defined  ) D 1 1 1 by f .Œx/ f .x/, is continuous. (q .f .U // f .U / is open in X f is continuous.)D D )

q X X=  f Šf Z 9

Examples.

(i) Define on R by x y iff x y Z. The map  R= T z C z 1 , where Œx e2ix, is well-defined 2 and a homeomorphism.W  ! (See´ f Example2 W j Sheetj D 1,g Question 15).7! 2 (ii) Define the 2-D torus T to be R = where .x1; y1/ .x2; y2/ iff x1 x2 Z and   2 y1 y2 Z. The topology in fact comes from a metric on T (See Example Sheet 1, Question 2 18) and hence well-behaved.

14 2 . 4 S UBSPACES , QUOTIENTSANDPRODUCTS

In general, one can get rather nasty (non-Hausdorff, for instance) topologies from an arbitrary equivalence relation. If A X we can define on X by x y iff x y or x; y A. The quotient space is sometimes written as X=A, in which A is scrunched downD to a point.2 Usually closed A is taken.

Example. Let D to be the closed unit disc in C, with boundary C , the unit circle. Then D=C is homeomorphic to S 2, the 2-sphere. (See Example Sheet 2, Question 13.)

Proposition 2.10. Suppose .X; / is Hausdorff and A X is closed, and x X A, open sets U;V with U V , A U , Âx V , then X=A is Hausdorff.8 2 n 9 \ D; Â 2

Proof. Given any two points x y in X=A, either one of the following holds: ¤ (i) Neither x or y correspond to A: then Šx; y X corresponding to x; y. Now 9 2 Ux A and Vx x such that Ux Vx . Uy A and Vy y such 9 Ã 3 \ D; 9 Ã 3 that Uy Vy . Since X is Hausdorff, wlog we assume Vx Vy . The \ D; \ D; corresponding open sets q.Vx/; q.Vy/ in X=A separate x and y. (ii) x q.x/ where x X A and y corresponds to A. Then open sets U A and V D x such that U2 Vn . The corresponding open sets9 q.U /; q.V /Ã X=A separate3 x and y. \ D; 2

Given topological spaces .X; / and .Y; /, we define the product topology   on X Y by W X Y is open iff .x; y/ W , open sets U X and V Y such that x U , y V Âand U V W . A base8 of open2 sets9 for .X Y;  / consists of sets U V , where2 2      U is open in X and V is open in Y . Note that .U1 V1/ .U2 V2/ .U1 U2/ .V1 V2/, so this does define a base for the topology.  \  D \  \

n Similarly if .Xi ; i /i 1 are topological spaces, the product topology on X1 Xn has a D 


 base of open sets of the form U1 Un, where Ui is open in Xi . 


 Example. Consider R with the Euclidean topology. The product topology for R R R2  D is just the metric topology, where open rectangles I1 I2, with I1 and I2 being open  intervals in R, form its base and also a base for the Euclidean topology on R2.

Y W ) V .x; y/ ( U V  ( ) U X

15 C ONNECTEDNESS

Lemma 2.11. Given topological spaces .X1; 1/ and .X2; 2/, the projection maps i .X1 X2; 1 2/ .Xi ; i / are continuous. Given a topological W   ! space .Y; / and continuous maps fi Y Xi , there is a unique factorisation, and f (in the diagram) is continuous.W !

f1 X1 Šf 1 Y 9 X1 X2  2 f2 X2

1 1 Proof. For the first part, use  .U / U X2 and  .V / X1 V . For the second part, 1 D  2 D  we have f .y/ .f1.y/; f2.y//. basic open set U V X1 X2, where U is open in X1 and D 1 8 1 1  Â  V is open in X2, f .U V/ f1 .U / f2 .V / is open in Y . Since any any open set W in  D \1 1 X1 X2 is the union of basic open sets, f .W / is the union of their inverses, and so f .W / is open in Y . Therefore f is continuous.

§3C ONNECTEDNESS 3.1 Various results

Definition 3.1. A topological space X is disconnected if non-empty open subsets U;V such that U V and U V X. We say9 U;V disconnect X. Otherwise X is connected\ .D; [ D

From this definition, we have X is connected iff open sets U;V where U V and U V X, either U (whence V X) or V 8 (whence U X). Connectedness\ D; is a topological[ D property. D; D D; D If Y X, then Y is disconnected in the subspace topology iff open sets U;V X such that U Y  , V Y , U V Y and Y U V . We9 say that U;V disconnect Y . \ ¤ ; \ ¤ ; \ \ D;  [ Proposition 3.2. Let X be a topological space. Then the following are equiva- lent:

(i) X is connected; (ii) The only subsets of X which are both open and closed are ;X; ; (iii) Every continuous function f X Z is constant. W ! Proof. (i) (ii): Trivial. For U X, consider U .X U/ X. , Â [ n D Not (iii) Not (i): Suppose non-constant f X Z which is continuous, then m; n ) 1 9 W1 ! 9 2 f .X/ with m < n. f . k k 6 m / and f . k k > m / are non-empty, open sets disconnecting X. f W g f W g

16 3 . 1 VARIOUSRESULTS

Not (i) Not (iii): non-empty disjoint open sets U;V such that U V X. Let f X Z where f .x/) 0 if x 9U , f .x/ 1 if x V . This is continuous but[ non-constant.D (ItW is locally! constant, however.)D 2 D 2

Proposition 3.3. Continuous images of a connected space are connected.

Proof. If f X Y is a continuous surjective map between topological spaces, and if U;V disconnect Y , thenW !f 1.U /; f 1.V / disconnect X. So X is connected Y is connected. ) Connectedness in R A subset I R is called an interval if given x; z I where x z, we have y I y  2 6 2 8 satisfying x y z. There are the cases inf I a R and a I , or a I , or inf I . 6 6 D 2 2 … D 1 Similarly we have the cases sup I b R and b I , or b I , or sup I . So any interval I takes the form Œa; b, Œa; b/, .a; bD, .a;2 b/, Œa; /2, .a; /, …. ; b, . D; b/ 1or . ; /. 1 1 1 1 1 1 Theorem 3.4. A subset of R is connected it is an interval. , Proof. ( ): If X R is not an interval, x; y; z where x < y < z, and x; z X but y X. Then . ); y/ and .y; / disconnect X. 9 2 … 1 1 ( ): Let I be an interval and U;V be open subsets of R disconnecting I . u U I and v (V I where, wlog, u < v. Since I is an interval, Œu; v I . Let s sup .Œu;9 2 v \U/. 2 \  D \ If s U , then s v so s < v and, since U is open, ı > 0 such that .s ı; s ı/ U . So s Œu;2 v U with¤ s > s. This contradicts s being an9 upper bound. C  9 0 2 \ 0 If s V , then ı > 0 such that .s ı; s ı/ V . In particular, .s ı; s ı/ U , so Œu; v 2U Œu; s9 ı. This contradicts s beingC  the least upper bound. C \ D; \ Â

Corollary 3.5. (Intermediate value theorem). Let a < b and f Œa; b R be continuous. If y Œf .a/; f .b/ (or Œf .b/; f .a/) then x Œa;W b such! that f .x/ y. 2 9 2 D Proof. Suppose not, then f 1. ; y/
and f 1.y; /
disconnect Œa; b. 1 1 Connected subsets and subspaces

Proposition 3.6. Given connected subspaces Y˛ ˛ A of a topological space f gS2 X, where Y˛ Yˇ ˛; ˇ A, the union Y Y˛ is connected. \ ¤ ; 8 2 D ˛ Proof. Using Proposition 3.2(iii), it suffices to prove that any continuous function f Y Z W ! is constant. Using the same proposition, f˛ f is constant, equal to n˛, say, on Y˛ ˛ A. ´ jY˛ 8 2 Since ˇ ˛, z Y˛ Yˇ , we have n˛ f .z/ nˇ . Therefore f is constant on Y . 8 ¤ 9 2 \ D D A connected component of a topological space X is a maximal connected subset Y X, i.e. if Z X is connected and Z Y then Z Y .   à D Each point x X is contained in a unique connected component of X, namely, S Z X Z is connected and2x Z . The fact that this is connected is a result of Proposition 3.6.f  W 2 g 17 3 . 1 VARIOUSRESULTS

˚ 1 « Example. Let X 0 n n 1; 2; : : : R, with the subspace topology. The ´ f g [ W 1D ˚ 1 « Â connected component containing n is n , both open and closed in X. The connected component containing 0 is 0 which is closed but not open in X. f g Proposition 3.7. If Y is a connected subset of a topological space X, then the closure Y is connected.

Proof. Suppose f Y Z is continuous. Proposition 3.2(iii) on page 16 f Y is constant, W ! 1 ) j equal to m, say. Given x Y , let n f .x/, then f .n/ is an open neighbourhood of x in Y , i.e. of the form U Y where2U is open´ in X. Since U Y , (considering Y X Int .X Y /) it contains a point\ z Y at which f .z/ m, so n \ m¤, i.e. ; f is constant onDY . n n 2 D D This shows that connected components of a topological space must be closed (because they are maximal connected subsets), but are not necessarily open, as the above example shows. We call a topological space X totally disconnected if its connected components are just the single points, or equivalently, its only connected subsets are single points. Any discrete topological space is totally disconnected, as is the above example.

Lemma. If X is a topological space, and x; y X where x y, X may be disconnected by U;V X where x 8U and2y V , then¤X is totally disconnected. Â 2 2

Proof. For any Y X with points x; y where x y, take U;V as given. Then U;V disconnect Y . Â ¤

The set of irrationals R Q is totally disconnected: Use the above lemma and the fact that between any two distinct irrationalsn there is a rational.

Cantor set

1 2
 1   2  Consider starting with I0 Œ0; 1 and removing 3 ; 3 to get I1 0; 3 3 ; 1 . Now ´  1   2  ´ [ remove the middle third from both 0; and ; 1 to get I2, and so on. The Cantor set is T 3 3 C In. ´ n>0 1 2 1 4 5 2 7 8 0 9 9 3 9 9 3 9 9 1

I0 Œ 

I1 Œ  Œ 

I2 Œ  Œ  Œ  Œ 

We can understand this in terms of ternary (i.e. base 3) expansions. I1 consists of numbers 1 with ternary expansion 0:a1a2a3 ::: where a1 0 or 2. (Wlog we impose 0:022 : : : .) I2 D 3 D consists of numbers where a1 0 or 2, and a2 0 or 2, and so on for I3 etc. So C consists of D D numbers with a ternary expansion where each ai 0 or 2. D Suppose now we are given two distinct points x; y in C . For some n, the ternary expansions n will differ first in the n-th place. C In, and In consists of 2 disjoint closed intervals, one of which contains x and another contains y.

18 3 . 2 PATH - CONNECTEDNESS

So we can disconnect In by open U;V Œ0; 1 where x U In and y V In. This implies U;V disconnect C where x U ÂC and y U 2C . By\ the previous2 lemma,\ C is totally disconnected. Also, both C and2 its\ complement2 are\ uncountable: Consider the ternary expansions.

3.2 Path-connectedness Let X be a topological space and x; y X.A continuous path from x to y is a continuous function  Œa; b X such that .a/ 2 x, .b/ y. (You may take a 0, b 1 if you prefer.) WeW say X is!path-connected if Dx; y X, Da path from x to y. ThisD is a topologicalD property. 8 2 9

Proposition 3.3∗. Continuous images of a path-connected space are path- connected. (cf. Proposition 3.3 on page 17.)

1 Proof. Suppose f X Y is a continuous surjection. Given y1; y2 Y , pick xi f .yi / W ! 2 2 for i 1; 2, and a path Œa; b X from x1 to x2. Then f Œa; b Y is a path from y1 D W ! B W ! to y2.

Proposition 3.8. Path-connected Connected. ) Proof. Let X be a path-connected topological space, and suppose U;V X disconnect X. Choose some u U , v V . Then a continuous function  Œa; b X Âwith .a/ u and .b/ v. Now 2 1.U /2,  1.V / 9Œa; b disconnect Œa; b, contradiction.W ! D D Â

But connected ; path-connected:

2 Example. In R , let I .x; 0/ 0 < x 6 1 and J .0; y/ 0 < y 6 1 . For n 1; 2; : : : , let L ´˚ f1 ; y
0W y 1«. Setg X ´I f J SW L
, withg the n n 6 6 n>1 n subspaceD topology. ´ W D [ [

1 J


0 1 1 1 1 4 3 2 I
X is not path-connected: For any continuous path .t/ 1.t/; 2.t/ from .1; 0/ to D .0; 1/, s such that 1.s/ 0 and 1.t/ > 0 t < s, so we must also have 2.s/ 0. This means9 passes throughD .0; 0/ X. 8 D … X is connected: Suppose f X Z is a continuous function. f is constant on J and on Y I S L
. However,W ! the points 1 ; 1
Y have limit 0; 1
J as n>1 n n 2 2 n , hence´ continuity[ of f the two constants agree,2 i.e. f is constant on X2. ! 1 )

19 3 . 3 P RODUCTSOFCONNECTEDSPACES

Given a topological space X, we can define an equivalence relation on X by x y iff a path from x to y in X. Reflexivity of is trivial. Symmetry of : If  Œa; b Xis a path9 from x to y, then .t/ . t/ gives a path Œ b; a Xfrom yW to x.! Transitivity of : If  Œa; b X andD Œc; d X are pathsW from x!to y and y to z respectively, then  Œa; b W d c! X whereW ! W C ! ( .t/ if t Œa; b; .t/ 2 ´ .t c b/ if t Œb; b d c; C 2 C is a path from x to z. The equivalence classes under are called the path-connected components of X. 

Theorem 3.9. Let X be an open subset of the Euclidean space Rn, then X is connected X is path-connected. , Proof. ( ): Proved as Proposition 3.8. ( ): Let x X and U be the equivalence class of x under defined( as above. ) 2  U is open in X: Let y U , whence x y. Since X is open, ı > 0 such that B.y; ı/ X. Then z B.y; ı/, we have2 y z (take the straight line segment),9 so transitivity of x z, i.e. B.y;8 ı/2 U .   )   Similarly, X U is open in X. Now X is connected and U , so we have X U U X x ny y X. ¤ ; n D;, D ,  8 2 3.3 Products of connected spaces Let X;Y be topological spaces. Consider X Y with the product topology.  Proposition 3.10. If X;Y are path-connected, then so too is X Y . 

Proof. Let .x1; y2/ and .x2; y2/ be in X Y . paths 1 Œ0; 1 X and 2 Œ0; 1 Y  9 W ! W !
from x1 to x2 and y1 to y2, respectively. Define Œ0; 1 X Y by .t/ 1.t/; 2.t/ . Any base for the topology on X Y consists of openW sets U! V where U is open´ in X and V is open in Y . Now 1.U V/  1.U / 1.V / is open. So is continuous and gives a path  D 1 \ 2 from .x1; y1/ to .x2; y2/.

Proposition 3.11. If X;Y are connected, then so too is X Y .  First we make some general comments about the product topology X Y . y Y , X y  8 2  f g with the subspace topology is homeomorphic to X via the projection map 1 X y X. Similarly, x X, x Y is homeomorphic to Y . So X, Y are connected W X f gy !and x Y are8 connected.2 f g  )  f g f g  Proof of 3.11. Let f X Y Z be continuous. By connectedness, f is constant on each W  ! of x Y and X y . Take two points .x1; y1/ and .x2; y2/ in X Y , then f .x1; y1/ f g   f g  D f .x1; y2/ f .x2; y2/ (See the following diagram), hence f constant on X Y , and X Y is connected.D  

20 C OMPACTNESS

Y .x2; y2/ y2

y1 .x1; y1/ 0 x1 x2 X

A similar argument proves Proposition 3.10: paths from .x1; y2/ to .x1; y2/ and from .x1; y2/ 9 to .x2; y2/ in X Y . 

§4C OMPACTNESS 4.1 Various results

Definition 4.1. Let X be a topological space and Y X. An open cover of  S Y is a collection U of open subsets such that Y U . f W 2 g  2

Such an open cover of Y provides a base of open sets U Y for Y with the f \ W 2 g subspace topology, and conversely. A subcover of an open cover U U of Y is a D f W 2 g subcollection V U which is still an open cover of Y . Â

Example. The intervals In . n; n/ where n 1; 2; : : : form an open cover of R, and D D In2 is a proper sub-cover. The intervals Jn .n 1; n 1/ where n Z form an open D C 2 cover of R with no proper subcover.

Definition 4.2. A topological space X is compact if every open cover has a finite subcover.

By the above example, R is not compact. Any finite topological space is compact, as is any set with the indiscrete or co-finite topology. By the way we defined it, compactness is a topological property.

Lemma 4.3. Let X be a topological space. Then Y X with the subspace  topology is compact every open cover U of Y has a finite subcover. , f g S Proof. ( ): Let U be an open cover of Y , then Y U Y , and the ) f W 2 g D 2 fSn\ g U Y ’s are open in Y . Since Y is compact, 1; : : : ; n such that Y i 1 U i Y \ 9 2 D D \ ) U i 1; : : : ; n covers Y . f i W D g S ( ): Suppose Y V where the V ’s are open in Y . Write V U Y where the ( D 2 D \ Sn U ’s are open in X and form an open cover of Y . So 1; : : : ; n such that Y i 1 U i , Sn 9 2  D hence Y i 1 V i . D D The open interval .0; 1/ R is not compact: Consider the open cover consisting of intervals 1 ; 1 1
for n 3; 4; : : :Â. n n D

21 4 . 1 VARIOUSRESULTS

Theorem 4.4. (Heine-Borel). The closed interval Œa; b R is compact.  S Proof. Let Œa; b U , where the U ’s are open in R. Let K x Œa; b Œa; x is  2 ´ f 2 W covered by finitely many U ’s . K since a K. K is bounded above by b. Let r sup K, g ¤ ; 2 ´ then r Œa; b, so r U for some 1. Since U is open, ı > 0 such that Œr ı; r ı U . 2 2 1 1 9 C  1

U 1 ./ a r ı r r ı b C

By definition of sup K, c Œr ı; r such that Œa; c is covered by finitely many of the U ’s. 9 2 Including U 1 , we have Œa; r ı Œa; b is covered by finitely many of the U ’s. This contradicts r being an upper bound unlessCr \b, in which case the above argument says that Œa; b is covered D ¼ by finitely many of the U ’s. So Œa; b is compact.

Proposition 4.5. Continuous images of a compact set are compact.

Proof. Suppose f X Y is a continuous map between topological spaces, and K X is W !S S 1  compact. Suppose f .K/ U , where U is open in Y . Then K f .U /, and  21  2 since f is continuous, each f .U / is open in X. Since K is compact, 1; : : : ; n such Sn 1 Sn 9 2 that K i 1 f .U i /. So f .K/ i 1 U i .  D  D Proposition 4.6. Closed subsets of a compact topological space are compact.

Proof. Let X be a compact topological space, and K X be closed. If K then S Â D; this is trivial, so assume not. Suppose K U , where the U ’s are open in X. Then S
 2 X .X K/ U , where X K is also open. Since X is compact, 1; : : : ; n D n [ 2 Sn
n Sn 9 2 such that X .X K/ i 1 U i . So K i 1 U i . D n [ D Â D Proposition 4.7. Compact subsets of a Hausdorff space are closed.

Proof. Suppose X is a Hausdorff space, and K X is compact. If K X then this is trivial, so assume not. We show that X K is open. Â D n Let x X K. Since X is Hausdorff, y K, disjoint open sets Uy x and Vy y. 2 n 8 2 9 3 3 Now Vy y K is an open cover of K. Since K is compact, y1; : : : ; yn K such that fSn W 2 g Tn 9 2 K i 1 Vyi . Then U i 1 Uyi is an open neighbourhood of x with U K , so U Â X DK, as required. ´ D \ D; Â n Corollary 4.8. A subset X R is compact X is closed and bounded. Â , Proof. ( ): Since R is Hausdorff, Proposition 4.7 X is closed. Suppose X is not bounded, then the intervals) . n; n/ where n 1; 2; : : : form an) open cover of X with no finite subcover. Contradiction. D

( ): M > 0 such that X Œ M; M . Since R X is open, we have .R X/ Œ M; M  is open( in9 Œ M; M , i.e. X is closed in Œ M; M . Byn Heine-Borel (Theoremn 4.4), \Œ M; M  is compact. By Proposition 4.6, X is compact.

22 4 . 1 VARIOUSRESULTS

Together with Theorem 3.4 on page 17, this shows that the only connected compact subsets of R are the closed intervals. See also the remark after Theorem 4.11 on page 24.

Example. Consider the Cantor set C T I as defined on page 18. Each I is the n>0 n n n D disjoint union of 2 closed intervals, so In is closed. So C is closed and bounded C is compact. )

Corollary 4.9. Let f X Y be a continuous bijection from a compact space X to a Hausdorff spaceW !Y . Then f is a homeomorphism.

1 Proof. Write g f . Let F X be closed. X is compact, so Proposition 4.6 F is ´ Â 1 ) compact. f is continuous, so Proposition 4.5 g .F / f .F / is compact. Y is Hausdorff, so Proposition 4.7 g 1.F / is closed in Y . So)g is continuous.D )

This result is particularly useful in identifying quotient spaces. E.g. If we define on R by  x y iff x y Z, and T z C z 1 (the unit circle with the subspace topology  2 ´ f 2 W j j D g coming from C), then it is clear that the map f R T, given by x e2ix, is continuous. It W ! 7! induces a bijection f R= T, which, by definition of quotient topology, is also continuous. (See the remark aboutW quotient ! topologies on page 14.)

The restricted quotient map Œ0; 1 R= is a continuous surjection. Œ0; 1 is compact, so !  Proposition 4.5 R= is compact. T is Hausdorff, so Corollary 4.9 f is a homeomorphism. )  ) Similarly, we can show that the 2-D torus R2=Z2, defined on page 14, is homeomorphic both to the product space S 1 S 1 (where S 1 is the unit circle) and to the embedded torus X R3 consisting of points .2 cos / cos ; .2 cos / sin ; sin 
, where ;  Œ0; 2/.  C C 2

(i) (ii)

Figure 2: (i) The 2-D torus R2=Z2; (ii) The embedded torus X R3.  In Analysis, you learnt that continuous real-valued functions on Œa; b R are bounded and attain their bounds. What is being used here is the compactness of Œa; b. This statement is still true for continuous real-valued functions on the Cantor set, for instance.

Proposition 4.10. Continuous real-valued functions on a compact space X is bounded and attain their bounds.

Proof. Let X be a compact space, and f X R be continuous. Proposition 4.5 f .X/ is compact. Corollary 4.8 f .X/ is closedW and! bounded. Now f .X/ is closed ) f .X/ contains all its accumulation) points. sup f .X/ ; inf f .X/ (exist because f .X/ is bounded), are accumulation points for f .X/, so sup ff .X/g; inf ff .X/g f .X/, hence the result. f g f g 2

23 4 . 1 VARIOUSRESULTS

Theorem 4.11. The product of two compact spaces is compact.

S Proof. Let X;Y be compact spaces and X Y U , where the U ’s are open in X Y .  D 2  Each U is the union of open sets of the form V W , where V is open in X and W is open in S  Y . So X Y ı
.Vı Wı /, where each Vı is open in X, each Wı is open in Y , and each  D 2  Vı Wı U for some .  Â Let x X. Then x Y S V W . In particular, Y S W . Since Y ı
x Vı ı ı ı
x Vı ı 2 f g  Â 2 W 2  DSm 2 W 2 is compact, ı1; : : : ; ım
, where x Vıi for each i, such that Y i 1 Wıi . 9 2 2 D D Y Vı Wı 

x X

Tm Sm Now let Vx i 1 Vıi , an open neighbourhood of x satisfying Vx Y i 1 Vıi ´ D  Â D  Wıi . Considering all x X, we see the Vx’s form an open cover of X. Since X is compact, 2 Sn Sn x1; : : : ; xn X such that X j 1 Vxj . So X Y j 1 Vxj Y . 9 2 D D  D D  Each Vx Y has a finite cover by the Vı Wı ’s, so the same is true for X Y . Using j    Vı Wı U , we see that X Y has a finite cover by the U ’s, hence the result.  Â 

Remark. Given topological spaces X;Y;Z, the product X Y Z is homeomorphic to X .Y Z/ (the open sets correspond). By induction, the above theorem implies the product of finitely many compact spaces is compact. As a consequence, and with the use of Corollary 4.8 on page 22, Œ M; M n is a compact subset of Rn. Now a set X Rn is bounded means M > 0 such that X Œ M; M n.  9  So the proof of the same corollary may be extended to show X Rn is compact X is closed and bounded.  ,

Proposition 4.12. Let X be a compact metric space, Y be any metric space, and f X Y be a continuous map. Then f is uniformly continuous, i.e. W !  > 0, ı > 0 such that x; y X, dX .x; y/ < ı dY .f .x/; f .y// < . 8 9 8 2 )

Proof. Since f is continuous, x X, ıx > 0 such that dX .x; x0/ < 2ıx dY .f .x/;f .x0//  8 2 9 ) < 2 . Let Ux x0 X dX .x; x0/ < ıx . Considering all x X, we see the Ux’s form an ´ f 2 W g 2 Sn open cover of X. Since X is compact, x1; : : : ; xn X such that X i 1 Uxi . 9 2 D D Let ı min ıx . Suppose dX .y; z/ < ı for some y; z X. Since the Ux form an open ´ f i g 2 i cover of X, i 1; : : : ; n such that d.y; xi / < ıx . Since dX .y; z/ < ı ıx , we have 9 2 f g i 6 i dX .z; xi / < 2ıxi by the triangle inequality. Therefore dY .f .y/; f .z// 6 dY .f .y/; f .xi //   C dY .f .xi /; f .z// < . 2 C 2 D

24 4 . 2 S EQUENTIALCOMPACTNESS

4.2 Sequential compactness

Definition 4.13. A topological space X is sequentially compact if every sequence in X has a convergent subsequence.

For a general topological space, compactness and sequential compactness do not imply each other. However: Proposition 4.14. Compact metric spaces are sequentially compact.

Proof. Let .X; d/ be a metric space and .xn/ be a sequence in X with no convergent subse- quence. Then .xn/ has infinitely many distinct members. x X, ı > 0 such that d.x; xn/ < ı 8 2 9 1 for finitely many n only. (If not, then x X such that m N, d.x; xn/ < for infinitely 9 2 8 2 m many n, and so a subsequence of .xn/ tending to x.) 9 We let Ux y X d.x; y/ < ı . Each Ux contains finitely many xn’s only, so ´ f 2 W g Ux x X is an open cover of X with no finite subcover. X is not compact. f W 2 g This result implies the Bolzano-Weierstrass theorem: Any closed bounded subset of Rn is sequentially compact.

n Example. Let X R be sequentially compact. Then (1) X is bounded: Otherwise .xn/ Â 9 such that d.x; xn/ > n, with no convergent subsequence; (2) X is closed: otherwise .xn/ 9 such that xn x X. ! … x X X 2 n x X n x2 x1

In fact, we have the following more general result: Theorem 4.15. Let .X; d/ be a sequentially compact metric space. Then: Sn (i)  > 0, x1; : : : ; xn X such that X i 1 B.xi ; /; 8 9 2 D D (ii) open cover U of X,  > 0 such that x X, B.x; / U for 8 9 8 2 Â some U U; 2 (iii) .X; d/ is compact.

Proof. (i): Suppose not. Then by induction we can construct a sequence .xn/ in X such that d.xm; xn/  m n. Its subsequences are not Cauchy, hence divergent. Contradiction. > 8 ¤ 1
(ii): Suppose not. Then an open cover U of X such that n, xn X such that B xn; 9 8 9 2 n › U , U U. But .xn/ has a subsequence .xn.r// tending to x X. So let x U0 for some 8 2 2
2 2 U0 U. Since U0 is open, m > 0 such that B x; U0. 2 9 m  1
Now N such that xn.r/ B x; r N . Additionally if n.r/ > m, and y 9 2 m 8 > 2  1 Á 2  1 Á B xn.r/; , then d.x; y/ d.x; xn.r// d.xn.r/; y/ < . So for such n.r/, B xn.r/; n.r/ 6 C m n.r/ 2
B x; U0. Contradiction. (See the diagram.) Â m Â

25 4 . 2 S EQUENTIALCOMPACTNESS

 1 Á B xn.r/; n.r/ 1
xn.r/ B x; m x 2
B x; m

(iii): Let U be an open cover of X. Choose an  > 0 as provided by (ii). For this , using Sn (i) x1; : : : ; xn X such that X i 1 B.xi ; /. For each i, B.xi ; / Ui for some Ui U 9 2 Sn D D Â 2 because of (ii). So X i 1 Ui . X is compact. D D

  

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