Metric and Topological Spaces

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Metric and Topological Spaces

T. W. K¨orner August 17, 2015

Small printThe syllabus for the course is defined by the Faculty Board Schedules (which are minimal for lecturing and maximal for examining). What is presented here contains some results which it would not, in my opinion, be fair to set as book-work although they could well appear as problems. In addition, I have included a small amount of material which appears in other 1B courses. I shouldvery muchappreciate being told of any corrections or possible improvements and might even part with a small reward to the first finder of particular errors. These notes are written in L^ATEX 2ε and should be available in tex, ps, pdf and dvi format from my home page

http://www.dpmms.cam.ac.uk/˜twk/

I can send some notes on the exercises in Sections 16 and 17 to supervisors by e-mail.

1 Preface

Within the last sixty years, the material in this course has been taught at Cambridge in the fourth (postgraduate), third, second and first years or left to students to pick up for themselves. Under present arrangements, students may take the course either at the end of their first year (before they have met metric spaces in analysis) or at the end of their second year (after they have met metric spaces).

Because of this, the first third of the course presents a rapid overview of metric spaces (either as revision or a first glimpse) to set the scene for the main topic of topological spaces.

The first part of these notes states and discusses the main results of the course. Usually, each statement is followed by directions to a proof in the final part of these notes. Whilst I do not expect the reader to find all the proofs by herself, I do ask that shetries to give a proof herself before looking one up. Some of the more difficult theorems have been provided with hints as well as proofs.

In my opinion, the two sections on compactness are the deepest part of the course and the reader who has mastered the proofs of the results therein is well on the way to mastering the whole course.

May I repeat that, as I said in the small print, I welcome corrections and comments.

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The reader should be acquainted with the convention that, if we have a function f : X → Y, then f is associated with two further set-valued functions

f :P(X)→ P(Y) and f⁻¹ :P(Y)→ P(X) (here P(Z) denotes the collection of subsets of Z) given by

f(A) ={f(a) : a ∈A} and f⁻¹(B) ={x∈X : f(x)∈B}.

We shall mainly be interested in f⁻¹ since this is better behaved as a set- valued function than f.

Exercise 1.1. We use the notation just introduced.

(i) LetX =Y ={1, 2, 3, 4} andf(1) = 1, f(2) = 1, f(3) = 4, f(4) = 3.

Identify

f⁻¹({1}), f⁻¹({2}) and f⁻¹({3, 4}).

(ii) IfUθ ⊆Y for all θ ∈Θ, show that f⁻¹ [

θ∈Θ

U_θ

!

= [

θ∈Θ

f⁻¹(U_θ) and f⁻¹ \

θ∈Θ

U_θ

!

= \

θ∈Θ

f⁻¹(U_θ).

Show also that f⁻¹(Y) =X, f⁻¹(∅) =∅ and that, if U ⊆Y, f⁻¹(Y \U) =X\f⁻¹(U).

(iii) If Vθ ⊆X for all θ ∈Θ, show that

f [

θ∈Θ

Vθ

!

= [

θ∈Θ

f(Vθ) and observe that f(∅) =∅.

(iv) By finding appropriate X, Y, f and V, V₁, V₂ ⊆ X, show that we may have

f(V1∩V2)6=f(V1)∩f(V2), f(X)6=Y and f(X\V)6=Y \f(V).

Solution. The reader should not have much difficulty with this, but if necessary, she can consult page 64.

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2 What is a metric?

If I wish to travel from Cambridge to Edinburgh, then I may be interested in one or more of the following numbers.

(1) The distance, in kilometres, from Cambridge to Edinburgh ‘as the crow flies’.

(2) The distance, in kilometres, from Cambridge to Edinburgh by road.

(3) The time, in minutes, of the shortest journey from Cambridge to Edinburgh by rail.

(4) The cost, in pounds, of the cheapest journey from Cambridge to Edinburgh by rail.

Each of these numbers is of interest to someone and none of them is easily obtained from another. However, they do have certain properties in common which we try to isolate in the following definition.

Definition 2.1. LetX be a set¹andd:X² →Ra function with the following properties:

(i) d(x, y)≥0 for all x, y ∈X.

(ii) d(x, y) = 0 if and only if x=y.

(iii)d(x, y) =d(y, x) for all x, y ∈X.

(iv) d(x, y) +d(y, z) ≥ d(x, z) for all x, y, z ∈ X. (This is called the triangle inequality after the result in Euclidean geometry that the sum of the lengths of two sides of a triangle is at least as great as the length of the third side.)

Then we say that d is a metric on X and that(X, d) is a metric space.

You should imagine the author muttering under his breath

‘(i) Distances are always positive.

(ii) Two points are zero distance apart if and only if they are the same point.

(iii) The distance from A toB is the same as the distance from B toA.

(iv) The distance from A to B via C is at least as great as the distance from A toB directly.’

Exercise 2.2. If d:X² →R is a function with the following properties:

1We thus allow X =∅. This is purely a question of taste. If we did not allow this possibility, then, every time we defined a metric space (X, d), we would need to prove that X was non-empty. If we do allow this possibility, and we prefer to reason about non-empty spaces, then we can begin our proof with the words ‘If X is empty, then the result is vacuously true, so we may assume thatX is non-empty.’ (Of course, the result may be false for X = ∅, in which case the statement of the theorem must include the conditionX 6=∅.)

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(ii) d(x, y) = 0 if and only if x=y, (iii)d(x, y) =d(y, x) for all x, y ∈X,

(iv) d(x, y) +d(y, z)≥d(x, z) for all x, y, z ∈X, show that d is a metric on X.

[Thus condition (i) of the definition is redundant.]

Solution. See page 66 for a solution.

Exercise 2.3. LetX be the set of towns on the British railway system. Con- sider the d corresponding to the examples (1) to (4) and discuss informally whether conditions (i) to (iv) apply.

[An open ended question like this will be more useful if tackled in a spirit of good will.]

Exercise 2.4. Let X = {a, b, c} with a, b and c distinct. Write down functions dj :X² →R satisfying condition (i) of Definition 2.1 such that:

(1) d₁ satisfies conditions (ii) and (iii) but not (iv).

(2) d2 satisfies conditions (iii) and (iv) and d2(x, y) = 0 implies x = y, but it is not true that x=y implies d2(x, y) = 0.

(3) d₃ satisfies conditions (iii) and (iv) and x = y implies d₃(x, y) = 0.

but it is not true that d3(x, y) = 0 implies x=y.

(4) d4 satisfies conditions (ii) and (iv) but not (iii).

You should verify your statements.

Solution. See page 67.

Other axiom grubbing exercises are given as Exercise 16.1 and 17.1.

Exercise 2.5. Let X be a set and ρ:X² →R a function with the following properties.

(i) ρ(x, y)≥0 for all x, y ∈X.

(ii) ρ(x, y) = 0 if and only if x=y.

(iv) ρ(x, y) +ρ(y, z)≥ρ(x, z) for all x, y, z ∈X.

Show that, if we set d(x, y) =ρ(x, y) +ρ(y, x), then (X, d) is a metric space.

3 Examples of metric spaces

We now look at some examples. The material from Definition 3.1 to Theo- rem 3.10 inclusive is covered in detail in Analysis II. You have met (or you will meet) the concept of a normed vector space both in algebra and analysis courses.

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Definition 3.1. Let V be a vector space overF (with F=R or F=C) and N :V →R a map such that, writingN(u) =kuk, the following results hold.

(i) kuk ≥0 for all u∈V. (ii) Ifkuk= 0, then u=0.

(iii) If λ∈F and u ∈V, then kλuk=|λ|kuk.

(iv) [Triangle law.] If u, v∈V, then kuk+kvk ≥ ku+vk.

Then we call k k a norm and say that (V,k k) is a normed vector space.

Exercise 3.2. By putting λ= 0 in Definition 3.1 (iii), show thatk0k= 0.

Any normed vector space can be made into a metric space in a natural way.

Lemma 3.3. If (V,k k) is a normed vector space, then the condition d(u,v) =ku−vk

defines a metric d on V.

Proof. The easy proof is given on page 67.

The concept of an inner product occurs both in algebra and in many physics courses.

Definition 3.4. Let V be a vector space over R and M : V × V → R a map such that, writing M(u,v) = hu,vi, the following results hold for u, v, w∈V, λ∈R.

(i) hu,ui ≥0.

(ii) Ifhu,ui= 0, then u=0.

(iii)hu,vi=hv,ui.

(iv) hu+w,vi=hu,vi+hw,vi.

(v) hλu,vi=λhu,vi.

Then we callh, ian inner product and say that(V,h, i)is an inner product space.

Lemma 3.5. Let (V,h , i) be an inner product space. If we write kuk = hu,ui^1/2 (taking the positive root), then the following results hold.

(i) (The Cauchy–Schwarz inequality) If u,v∈V, then kukkvk ≥ |hu,vi|.

(ii) (V,k k) is a normed vector space.

Proof. The standard proofs are given on page 68.

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Lemma 3.6. If we work on Rⁿ made into a vector space in the usual way, then

hx,yi=

n

X

j=1

x_jy_j

is an inner product.

Proof. Direct verification.

We call the norm

kxk₂ =

n

X

j=1

x²_j

!1/2

,

derived from this inner product the Euclidean norm (or sometimes just ‘the usual norm’). Although several very important norms are derived from inner products most are not.

Lemma 3.7. (The parallelogram law) Using the hypotheses and notation of Lemma 3.5, we have

ku+vk²+ku−vk²= 2kuk²+ 2kvk². Proof. Direct computation .

We need one more result before we can unveil a collection of interesting norms.

Lemma 3.8. Suppose that a < b, that f : [a, b] → R is continuous and f(t)≥0 for all t∈[a, b]. Then, if

Z b a

f(t)dt= 0, it follows that f(t) = 0 for all t ∈[a, b].

Proof. See page 69.

Exercise 3.9. Show that the result of Lemma 3.8 is false if we replace ‘f continuous’ by ‘f Riemann integrable’.

Solution. See page 69 if necessary.

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Theorem 3.10. Suppose that a < b and we consider the space C([a, b]) of continuous functions f : [a, b] → R made into a vector space in the usual way.

(i) The equation

hf, gi= Z b

a

f(t)g(t)dt defines an inner product on C([a, b]). We write

kfk₂ = Z b

a

f(t)²dt ^1/2

for the derived norm.

(ii) The equation

kfk1 = Z b

a

|f(t)|dt

defines a norm on C([a, b]). This norm does not obey the parallelogram law.

(iii) The equation

kfk_∞ = sup

t∈[a,b]

|f(t)|.

defines a norm on C([a, b]). This norm does not obey the parallelogram law.

Proof. The routine proofs are given on page 69.

However, not all metrics can be derived from norms. Here is a metric that turns out to be more important and less peculiar than it looks at first sight.

Definition 3.11. If X is a set and we define d:X² →R by d(x, y) =

(0 if x=y, 1 if x6=y, then d is called the discrete metric on X.

Lemma 3.12. The discrete metric onX is indeed a metric.

Exercise 3.13. (We deal with the matter somewhat better in Exercise 5.7) (i) If V is a vector space over Randdis a metric derived from a norm in the manner described above, show that, if u∈V we have d(0,2u) = 2d(0,u).

(ii) IfV is non-trivial (i.e. not zero-dimensional) vector space overRand d is the discrete metric on V, show that d cannot be derived from a norm on V.

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You should test any putative theorems on metric spaces on bothRⁿ with the Euclidean metric and Rⁿ with the discrete metric.

Exercise 3.14. [The counting metric.] If E is a finite set and E is the collection of subsets of E, we write cardC for the number of elements in C and

d(A, B) = cardA△B.

Show thatd is a metric onE. The reader may be inclined to dismiss this metric as uninteresting but it plays an important role (as the Hamming metric) in the Part II course Codes and Cryptography.

Here are two metrics which are included simply to show that metrics do not have to be as simple as the ones above. I shall use them as examples once or twice, but they do not form part of standard mathematical knowledge and you do not have to learn their definition.

Definition 3.15. (i) If we define d:R²×R² →R by d(u,v) =

(kuk₂+kvk₂, if u6=v,

0 if u=v,

then d is called the British Rail express metric. (To get from A to B travel via London.)

(ii) If we define d:R²×R² →R by d(u,v) =

(ku−vk₂ ifu and v are linearly dependent, kuk₂+kvk₂ otherwise,

then d is called the British Rail stopping metric. (To get from A to B travel via London unless A and B are on the same London route.)

(Recall that u and v are linearly dependent if u = λv for some real λ and/or v=0.)

Exercise 3.16. Show that the British Rail express metric and the British Rail stopping metric are indeed metrics.

Solution. On page 72 we show that the British Rail stopping metric is indeed a metric. The British Rail express metric can be dealt with similarly.

However, fascinating as exotic metrics may be, the reader should reflect on the number of different useful metric spaces that exist. We have Rⁿ with the usual Euclidean norm, C([a, b]) with the three norms described in

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Theorem 3.10, the counting (Hamming) metric, the p-adic metric (used in number theory) described in Exercise 16.23, the metric space described in Exercise 16.21 (a model for coin tossing) and many others. The notion of a metric provides a common thread, suggesting appropriate theorems and proofs.

4 Continuity and open sets for metric spaces

Some definitions and results transfer essentially unchanged from classical analysis on R to metric spaces. Recall the classical definition of continuity.

Definition 4.1. [Old definition.] A function f :R→R is called continuous if, given t∈R and ǫ >0, we can find a δ(t, ǫ)>0such that

|f(t)−f(s)|< ǫ whenever |t−s|< δ(t, ǫ).

It is not hard to extend this definition to our new, wider context.

Definition 4.2. [New definition.] Let (X, d) and (Y, ρ) be metric spaces.

A function f : X → Y is called continuous if, given t ∈ X and ǫ > 0, we can find a δ(t, ǫ)>0 such that

ρ(f(t), f(s))< ǫ whenever d(t, s)< δ(t, ǫ).

It may help you grasp this definition if you read ‘ρ(f(t), f(s))’ as ‘the distance from f(t) to f(s) in Y’ and ‘d(t, s)’ as ‘the distance from t to s in X’.

Lemma 4.3. [The composition law.] If (X, d) and (Y, ρ) and (Z, σ) are metric spaces and g : X → Y, f : Y → Z are continuous, then so is the composition f g.

Proof. This is identical to the one we met in classical analysis. If needed, details are given on page 73.

Exercise 4.4. Let R and R² have their usual (Euclidean) metric.

(i) Suppose that f :R→Rand g :R→Rare continuous. Show that the map (f, g) : R² →R² is continuous.

(ii) Show that the mapM :R² →Rgiven byM(x, y) =xy is continuous.

(iii) Use the composition law to show that the map m:R² →R given by m(x, y) =f(x)g(y)is continuous.

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Exercise 4.4 may look perverse at first sight, but, in fact, we usually show functions to be continuous by considering them as compositions of simpler functions rather than using the definition directly. Think about

x7→log

2 + sin 1 1 +x²

.

If you are interested, we continue the chain of thought in Exercise 16.2. If you are not interested or are mildly confused by all this, just ignore this paragraph.

Just as there are ‘well behaved’ and ‘badly behaved’ functions between spaces, so there are ‘well behaved’ and ‘badly behaved’ subsets of spaces.

In classical analysis and analysis on metric spaces, the notion of continuous function is sufficiently wide to give us a large collection of interesting functions and sufficiently narrow to ensure reasonable behaviour². In intro- ductory analysis we work on R with the Euclidean metric and only consider subsets in the form of intervals. Once we move to R² with the Euclidean metric, it becomes clear that there is no appropriate analogue to intervals.

(We want appropriate rectangles to be well behaved, but we also want to talk about discs and triangles and blobs.)

Cantor identified two particular classes of ‘well behaved’ sets. We start with open sets.

Definition 4.5. Let (X, d) be a metric space. We say that a subset E is open in X if, whenever e ∈ E, we can find a δ > 0 (depending on e) such that

x∈E whenever d(x, e)< δ.

Suppose we work in R² with the Euclidean metric. If E is an open set then any point e inE is the centre of a disc of strictly positive radius all of whose points lie inE. If we are sufficiently short sighted, every point that we can see from elies in E. This property turns out to be a key to many proofs in classical analysis (remember that in the proof of Rolle’s theorem it was vital that the maximum did not lie at an end point) and complex analysis (where we examine functions analytic on an open set).

Here are a couple of simple examples of an open set and a simple example of a set which is not open.

Example 4.6. (i) Let (X, d) be a metric space. If r >0, then B(x, r) = {y : d(x, y)< r}

2Sentences like this are not mathematical statements, but many mathematicians find them useful.

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is open.

(ii) If we work in Rⁿ with the Euclidean metric, then the one point set {x} is not open.

(iii) If (X, d) is a discrete metric space, then {x}=B(x,1/2) and all subsets of X are open.

Proof. See page 74.

We call B(x, r) the open ball with centre x and radius r. The following result is very important for the course, but is also very easy to check.

Theorem 4.7. If (X, d) is a metric space, then the following statements are true.

(i) The empty set ∅ and the space X are open.

(ii) If Uα is open for all α ∈A, then S

α∈AUα is open. (In other words, the union of open sets is open.)

(iii) If Uj is open for all 1≤j ≤n, then Tn

j=1Uj is open.

Proof. See page 75.

It is important to realise that we place no restriction on the size of A in (ii). In particular, A could be uncountable. However, conclusion (iii) cannot be extended.

Example 4.8. Let us work in Rⁿ with the usual metric. Then B(x,1/j) is open, but T∞

j=1B(x,1/j) ={x} is not.

Proof. See Example 4.6.

There is a remarkable connection between the notion of open sets and continuity.

Theorem 4.9. Let(X, d)and(Y, ρ)be metric spaces. A functionf :X →Y is continuous if and only if f⁻¹(U) is open in X whenever U is open in Y. Proof. See page 76.

Note that the theorem does not work ‘in the opposite direction’.

Example 4.10. Let X =R and d be the discrete metric. Let Y =R and ρ be the usual (Euclidean) metric.

(i) If we define f : X → Y by f(x) = x, then f is continuous but there exist open sets U in X such that f(U) is not open.

(ii) If we define g : Y → X by g(y) = y, then g is not continuous but g(V) is open in X whenever V is open in Y.

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Proof. Very easy, but see page 76 if you need.

The message of this example is reinforced by the more complicated Ex- ercise 16.3.

Observe that Theorem 4.9 gives a very neat proof of the composition law.

Theorem 4.3. If (X, d), (Y, ρ), (Z, σ) are metric spaces and g : X → Y, f :Y →Z are continuous, then so is the composition f g.

New proof. If U is open in Z, then, by continuity, f⁻¹(U) is open in Y and so, by continuity, (f g)⁻¹(U) = g⁻¹ f⁻¹(U)

is open in X. Thus f g is continuous.

This confirms our feeling that the ideas of this section are on the right track.

We finish with an exercise, which may be omitted at first reading, but which should be done at some time since it gives examples of what open sets can look like.

Exercise 4.11. ConsiderR². For each of the British rail express and British rail stopping metrics:

(i) Describe the open balls. (Consider both large and small radii.)

(ii) Describe the open sets as well as you can. (There is a nice description for the British rail express metric.) Give reasons for your answers.

5 Closed sets for metric spaces

The second class of well behaved sets identified by Cantor were the closed sets. In order to define closed sets in metric spaces, we need a notion of limit.

Fortunately, the classical definition generalises without difficulty.

Definition 5.1. Consider a sequence x_n in a metric space (X, d). If x∈X and, given ǫ >0, we can find an integer N ≥1 (depending on ǫ) such that

d(xn, x)< ǫ for all n≥N ,

then we say that x_n → x as n → ∞ and that x is the limit of the sequence xn.

Lemma 5.2. Consider a metric space (X, d). If a sequence xn has a limit, then that limit is unique.

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Proof. The simple proof is given on page 78. Just as in the next exercise, it suffices to follow the ‘first course in analysis’ proof with minimal changes.

Exercise 5.3. Consider two metric spaces (X, d) and (Y, ρ). Show that a function f : X → Y is continuous if and only if, whenever xn ∈ X and xn →x as n → ∞, we have f(xn)→f(x)

Solution. See page 78, if necessary.

Exercise 5.4. In this exercise we consider the identity map between a space and itself when we equip the space with differentmetrics. We look at the three norms (and their associated metrics) defined on C([0,1]) in Theorem 3.10.

Define jα,β : C([0,1]),k kα

→ C([0,1]),k kβ

by jα,β(f) = f.

(i) Show that j∞,1 and j∞,2 are continuous but j1,∞ and j2,∞ are not.

(ii) By using the Cauchy–Schwarz inequality |hf, gi| ≤ kfk₂kgk₂ with g = 1, or otherwise, show that j2,1 is continuous. Show that j1,2 is not.

[Hint: Consider functions of the form fR,K(x) =Kmax{0,1−Rx}.]

Solution. See page 79, if necessary.

Definition 5.5. Let (X, d) be a metric space. A set F in X is said to be closed if, whenever xn ∈F and xn→x as n→ ∞, it follows that x∈F.

The following exercises are easy, but instructive.

Exercise 5.6. (i) If(X, d)is any metric space, thenX and ∅are both open and closed.

(ii) If we consider R with the usual metric and take b > a, then [a, b] is closed but not open, (a, b) is open but not closed and [a, b) is neither open nor closed.

Exercise 5.7. (i) If (X, d) is a metric space with discrete metric d, then all subsets of X are both open and closed.

(ii) If V is a vector space over R and ρ is a metric derived from a norm, show that the one point sets {x} are not open in this metric.

(iii) Deduce that the discrete metric d on the vector space V cannot be derived from a norm on V.

It is easy to see why closed sets will be useful in those parts of analysis which involve taking limits. The reader will recall theorems in elementary analysis (for example the boundedness of continuous functions) which were true for closed intervals, but not for other types of intervals.

Life is made much easier by the very close link between the notions of closed and open sets given by our next theorem.

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Theorem 5.8. Let (X, d) be a metric space. A set F in X is closed if and only if its complement is open.

Proof. There is a proof on page 80.

We can now deduce properties of closed sets from properties of open sets by complementation. In particular, we have the following complementary versions of Theorems 4.7 and 4.9

Theorem 5.9. If (X, d) is a metric space, then the following statements are true.

(i) The empty set ∅ and the space X are closed.

(ii) IfFα is closed for all α∈A, then T

α∈AFα is closed. (In other words the intersection of closed sets is closed.)

(iii) If Fj is closed for all 1≤j ≤n, then Sn

j=1Fj is closed.

Proof. See page 80.

Theorem 5.10. Let(X, d) and(Y, ρ) be metric spaces. A function f :X → Y is continuous if and only if f⁻¹(F) is closed in X whenever F is closed in Y.

Proof. See page 81.

6 Topological spaces

We now investigate general objects which have the structure described by Theorem 4.7.

Definition 6.1. Let X be a set and τ a collection of subsets of X with the following properties.

(i) The empty set ∅∈τ and the space X ∈τ.

(ii) IfU_α ∈τ for all α ∈A, then S

α∈AU_α ∈τ.

(iii) If Uj ∈τ for all 1≤j ≤n, then Tn

j=1Uj ∈τ.

Then we say that τ is a topology on X and that (X, τ) is a topological space.

Theorem 6.2. If (X, d) is a metric space, then the collection of open sets forms a topology.

Proof. This is Theorem 4.7.

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If (X, d) is a metric space we call the collection of open sets the topology induced by the metric.

If (X, τ) is a topological space we extend the notion of open set by calling the members of τ open sets. The discussion above ensures what computer scientists call ‘downward compatibility’.

Just as group theory deals with a collection of objects together with an operation of ‘multiplication’ which follows certain rules, so we might say that topology deals with a collection τ of objects (subsets of X) under the two operations of ‘union’ and ‘intersection’ following certain rules. A remarkable application of this philosophy is provided by Exercise 17.7. However, many mathematicians simply use topology as a language which emphasises certain aspects of Rⁿ and other metric spaces.

Exercise 6.3. If (X, d)is a metric space with the discrete metric, show that the induced topology consists of all the subsets of X.

We call the topology consisting of all subsets of X the discrete topology on X.

Exercise 6.4. If X is a set and τ ={∅, X}, then τ is a topology.

We call{∅, X} the indiscrete topology on X.

Exercise 6.5. (i) If F is a finite set and (F, d) is a metric space, show that the induced topology is the discrete topology.

(ii) If F is a finite set with more than one point, show that the indiscrete topology is not induced by any metric.

You should test any putative theorems on topological spaces on the discrete topology and the indiscrete topology,Rⁿwith the topology derived from the Euclidean metric and [0,1] with the topology derived from the Euclidean metric.

The following exercise is tedious but instructive (the tediousness is the instruction).

Exercise 6.6. Write P(Y) for the collection of subsets ofY. IfX has three elements, how many elements does P P(X)

have?

How many topologies are there on X?

The idea of downward compatibility suggests ‘turning Theorem 4.9 in a definition’.

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Definition 6.7. Let (X, τ) and (Y, σ) be topological spaces. A function f : X →Y is said to be continuous if and only iff⁻¹(U)is open inX whenever U is open in Y.

Theorem 4.9 tells us that, if (X, d) and (Y, ρ) are metric spaces, the notion of a continuous function f : X → Y is the same whether we consider the metrics or the topologies derived from them.

The proof of Theorem 4.3 given on page 13 carries over unchanged to give the following generalisation.

Theorem 6.8. If(X, τ), (Y, σ), (Z, µ)are topological spaces and g :X →Y, f :Y →Z are continuous, then so is the composition f g.

Downward compatibility suggests the definition of a closed set for a topological space based on Theorem 5.8.

Definition 6.9. Let (X, τ) be a topological space. A set F in X is said to be closed if its complement is open.

Theorem 5.8 tells us that if (X, d) is a metric space the notion of a closed set is the same whether we consider the metric or the topology derived from it.

Just as in the metric case, we can deduce properties of closed sets from properties of open sets by complementation. In particular, the same proofs as we gave in the metric case give the following extensions of Theorems 5.9 and 5.10

Theorem 6.10. If(X, τ)is a topological space, then the following statements are true.

(i) The empty set ∅ and the space X are closed.

(ii) IfFα is closed for allα ∈A, thenT

α∈AFα is closed. (In other words, the intersection of closed sets is closed.)

(iii) If F_j is closed for all 1≤j ≤n, then Sn

j=1F_j is closed.

Theorem 6.11. Let (X, τ) and (Y, σ) be topological spaces. A function f : X → Y is continuous if and only if f⁻¹(F) is closed in X whenever F is closed in Y.

7 Interior and closure

The next section is short, not because the ideas are unimportant, but because they are so useful that the reader will meet them over and over again in other courses.

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Definition 7.1. Let (X, τ) be a topological space and A a subset of X. We write

IntA=[

{U ∈τ : U ⊆A} and ClA=\

{F closed : F ⊇A}

and call ClA the closure of A and IntA the interior of A.

Simple complementation, which I leave to the reader, shows how closely the two notions of closure and interior are related. (Recall that A^c =X\A, the complement of A.)

Lemma 7.2. With the notation of Definition 7.1

(ClA^c)^c = IntA and (IntA^c)^c = ClA.

There are other useful ways of viewing IntA and ClA.

Lemma 7.3. Let (X, τ) be a topological space and A a subset of X.

(i) IntA={x∈A : ∃ U ∈τ with x∈U ⊆A}.

(ii) IntA is the unique V ∈ τ such that V ⊆ A and, if W ∈ τ and V ⊆ W ⊆ A, then V = W. (Informally, IntA is the largest open set contained in A.)

Exercise 7.4. ConsiderR with its usual topology (i.e. the one derived from the Euclidean norm). We look at the open interval I = (0,1). Show that ifF is closed and F ⊆(0,1), there is a closedG withF ⊆G⊆(0,1)andG6=F. (Thus there is no largest closed set contained in (0,1).)

Solution. See page 83 if necessary.

Simple complementation, which I leave to the reader, gives the corresponding results for closure.

Lemma 7.5. Let (X, τ) be a topological space and A a subset of X.

(i) ClA={x∈X :∀U ∈τ with x∈U, we have A∩U 6=∅}.

(ii) ClA is the unique closed set G such that G ⊇ A and, if F is closed with G ⊇ F ⊇ A, then F = G. (Informally, ClA is the smallest closed set containing A.)

Exercise 7.6. Prove Lemma 7.5 directly without using Lemma 7.3.

Sometimes, when touring an ancient college, you may be shown a 14th century wall which still plays an important part in holding up the building.

The next lemma goes back to Cantor and the very beginnings of topology.

(It would then have been a definition rather than a lemma.)

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Lemma 7.7. Let (X, d) be a metric space and A a subset of X. Then ClA consists of all those x such that we can find x_n ∈ A with d(x, x_n) → 0. (In old fashioned terminology, the closure of A is its set of closure points³.) Proof. The easy proof is given on page 83.

The idea of closure is strongly linked to the idea of a dense subset.

Definition 7.8. Let (X, τ) be a topological space and F a closed subset of X. We say that A ⊆X is a dense subset of F if ClA=F.

In some sense A is a ‘skeleton’ of F and we may hope to prove results about F by first proving them on the dense subset A and then extending the result by ‘density’. Sometimes this idea works (see, for example, part (ii) of Exercise 7.9) and sometimes it does not (see, for example, part (iii) of Exercise 7.9). When it does work, this is very powerful technique.

Exercise 7.9. (i) Let (X, τ)be a topological space and (Y, d)a metric space.

If f, g:X →Y are continuous show that the set {x∈X : f(x) = g(x)}

is closed.

(ii) Let (X, τ) be a topological space and (Y, d) a metric space⁴. If f, g : X →Y are continuous and f(x) = g(x) for all x∈ A, where A is dense in X, show that f(x) =g(x) for all x∈X.

(iii) Consider the unit interval [0,1] with the Euclidean metric and A= [0,1]∩Q with the inherited metric. Exhibit, with proof, a continuous map f :A →R (where R has the standard metric) such that there does not exist a continuous map f˜: [0,1]→R with f˜(x) =f(x) for all x∈A.

Solution. There is a solution on Page 83.

8 More on topological structures

Two groups are the same for the purposes of group theory if they are (group) isomorphic. Two vector spaces are the same for the purposes of linear algebra if they are (vector space) isomorphic. When are two topological spaces (X, τ)

3I strongly advise caution in employing terms like ‘limit point’, ‘accumulation point’,

‘adherent point’ and ‘closure point’ since both the literature and your lecturer are confused about what they mean. If an author uses one of these terms, check what definition they are using. If you wish to use these terms, define them explicitly.

4Exercise 9.7 gives an improvement of parts (i) and (ii).

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and (Y, σ) the same for the purposes of topology? In other words, when does there exist a bijection between X and Y for which open sets correspond to open sets, and the grammar of topology (things like union and inclusion) is preserved? A little reflection shows that the next definition provides the answer we want.

Definition 8.1. We say that two topological spaces (X, τ) and (Y, σ) are homeomorphic if there exists a bijection θ :X →Y such that θ and θ⁻¹ are continuous. We call θ a homeomorphism.

The following exercise acts as useful revision of concepts learnt last year.

Exercise 8.2. Show that homeomorphism is an equivalence relation on topological spaces.

Homeomorphism implies equivalence for the purposes of topology.

Exercise 8.3. Suppose that(X, d)and (Y, ρ)are metric spaces and f :X → Y is a homeomorphism. Show that

d(xn, x)→0⇔ρ(f(xn), f(x))→0.

Thus the limit structure of a metric space is a topological property.

To give an interesting example of a property which is not preserved by homeomorphism, we introduce a couple of related ideas which are fundamen- tal to analysis on metric spaces, but which will only be referred to occasionally in this course.

Definition 8.4. (i) If (X, d)is a metric space, we say that a sequence xn in X is Cauchy if, given ǫ >0, we can find an N₀(ǫ) with

d(xn, xm)< ǫ whenever n, m≥N0(ǫ).

(ii) We say that a metric space (X, d) is complete if every Cauchy sequence converges.

Example 8.5. LetX =R and letd be the usual metric onR. LetY = (0,1) (the open interval with end points 0 and 1) and let ρ be the usual metric on (0,1). Then (X, d) and (Y, ρ) are homeomorphic as topological spaces, but (X, d) is complete and (Y, ρ) is not.

Proof. See page 84.

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We say that ‘completeness is not a topological property’. Exercise 17.32 shows that there exist metric spaces which are not homeomorphic to any complete metric space.

In group theory, we usually prove that two groups are isomorphic by constructing an explicit isomorphism and that two groups are not isomorphic by finding a group property exhibited by one but not by the other. Similarly, in topology, we usually prove that two topological spaces are homeomorphic by constructing an explicit homeomorphism and that two topological spaces are not homeomorphic by finding a topological property exhibited by one but not by the other. Later in this course we will meet some topological properties like being Hausdorff and compactness and you will be able to tackle Exercise 16.17.

We also want to be able to construct new topological spaces from old. To do this we we make use of a simple, but useful, lemma.

Lemma 8.6. Let X be a space and let H be a collection of subsets of X.

Then there exists a unique topology τH such that (i) τ_H ⊇ H, and

(ii) if τ is a topology with τ ⊇ H, then τ ⊇τH.

Proof. The proof, which follows the standard pattern for such things, is given on page 85.

We callτ_H the smallest (or coarsest) topology containing H.

Lemma 8.7. Suppose thatAis non-empty, the spaces(Xα, τα)are topological spaces and we have maps fα : X → Xα [α ∈ A]. Then there is a smallest topology τ on X for which the maps f_α are continuous.

Proof. A topology σ on X makes all the fα continuous if and only if it contains

H={f_α⁻¹(U) : U ∈τα, α∈A}.

Now apply Lemma 8.6.

Recall that, if Y ⊆ X, then the inclusion map j : Y → X is defined by j(y) =y for all y ∈Y.

Definition 8.8. If (X, τ) is a topological space and Y ⊆ X, then the subspace topologyτY on Y induced byτ is the smallest topology onY for which the inclusion map is continuous.

Lemma 8.9. If (X, τ) is a topological space and Y ⊆X, then the subspace topology τY on Y is the collection of sets Y ∩U with U ∈τ.

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Proof. The very easy proof is given on page 85.

Exercise 8.10. (i) If (X, τ)is a topological space and Y ⊆X is open, show that the subspace topologyτY onY is the collection of setsU ∈ τ withU ⊆Y. (ii) Consider R with the usual topology τ (that is, the topology derived from the Euclidean metric). If Y = [0,1], show that [0,1/2) ∈ τ_Y but [0,1/2)∈/ τ.

Exercise 8.11. Let (X, d) be a metric space, Y a subset of X and dY the metricdrestricted to Y (formally, d_Y :Y² →Ris given byd_Y(x, y) =d(x, y) for x, y ∈ Y). Then if we give X the topology induced by d, the subspace topology on Y is identical with the topology induced by dY.

[This is an exercise in stating the obvious.]

Next recall that ifXandY are sets the projection mapsπX :X×Y →X and πY :X×Y →Y are given by

π_X(x, y) = x, π_Y(x, y) = y.

Definition 8.12. If(X, τ)and(Y, σ)are topological spaces, then the product topologyµonX×Y is the smallest topology onX×Y for which the projection maps π_X and π_Y are continuous.

Lemma 8.13. Let (X, τ) and (Y, σ) be topological spaces and λ the product topology on X×Y. Then O ∈λ if and only if, given (x, y)∈O, we can find U ∈τ and V ∈σ such that

(x, y)∈U ×V ⊆O.

Proof. See page 86.

Exercise 14.9 gives a very slightly different treatment of the matter.

Exercise 8.14. Suppose that (X, τ)and (Y, σ) are topological spaces and we give X×Y the product topology µ. Now fix x ∈ X and give E = {x} ×Y the subspace topology µE. Show that the map k : (Y, σ) → (E, µE) given by k(y) = (x, y) is a homeomorphism.

Solution. The proof is a direct application of Lemma 8.13. See page 87 if necessary.

The next remark is useful for proving results like those in Exercise 8.16.

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Lemma 8.15. Let τ1 and τ2 be two topologies on the same space X.

(i) We have τ₁ ⊆τ₂ if and only if, given x∈ U ∈τ₁, we can find V ∈τ₂ such that x∈V ⊆U.

(ii)We have τ1 =τ2 if and only if, given x∈U ∈τ1, we can find V ∈τ2

such that x ∈ V ⊆ U and, given x ∈ U ∈ τ₂, we can find V ∈ τ₁ such that x∈V ⊆U.

Proof. The easy proof is given on Page 87

Exercise 8.16. Let (X₁, d₁) and (X₂, d₂) be metric spaces. Let τ be the product topology on X1 ×X2 where Xj is given the topology induced by dj

[j = 1,2].

Define ρ_k : (X₁×X₂)² →R by

ρ1((x, y),(u, v)) =d1(x, u),

ρ2((x, y),(u, v)) =d1(x, u) +d2(y, v), ρ3((x, y),(u, v)) = max(d1(x, u), d2(y, v)), ρ4((x, y),(u, v)) = (d1(x, u)²+d2(y, v)²)^1/2.

Establish that ρ1 is not a metric and that ρ2, ρ3 and ρ4 are. Show that each of the ρ_j with2≤j ≤4 induces the product topology τ on X₁×X₂.

It is easy to extend our definitions and results to any finite product of topological spaces⁵. In fact, it is not difficult to extend our definition to the product of an infinite collection of topological spaces, but I feel that it is important for the reader to concentrate on first thoroughly understanding the finite product case and I have relegated the infinite case to an exercise (Exercise 16.7).

We conclude this section by looking briefly at the quotient topology. This will not play a major part in our course and the reader should not worry too much about it.

If ∼ is an equivalence relation on a set X, then we know from previous courses that it gives rise to equivalence classes

[x] ={y ∈X : y ∼x}.

There is a natural map q from X to the space X/∼ of equivalence classes given by q(x) = [x]. When we defined the subspace and product topologies, we used natural maps from the new spaces to the old spaces. Here, we have a natural map from the old space to the new, so our definition has to take a different form.

5Once you are confident with the material you may wish to look at Exercise 17.11, but this exercise is confusing for the beginner and trivial to the expert.

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Exercise 8.17. Let X = {1,2,3} and θ = {∅,{1},{2}, X}. Check that there does not exist a topology τ₁ ⊆θ such that, if τ ⊆θ is a topology, then τ ⊆τ1. (Thus there does not exist a largest topology contained in θ.)

However, since intersection and union behave well under inverse map- pings, it is easy to check the following statement.

Lemma 8.18. Let (X, τ) be a topological space and Y a set. If f :X →Y is a map and we write

σ={U ⊆Y : f⁻¹(U)∈τ}, then σ is a topology on Y such that

(i) f : (X, τ)→(Y, σ) is continuous and

(ii) if θ is a topology on Y with f : (X, τ) → (Y, θ) continuous, then θ ⊆σ.

Lemma 8.18 allows us to make the following definition.

Definition 8.19. Let (X, τ) be a topological space and ∼ an equivalence relation onX. Write q for the map fromX to the quotient space X/∼ given by q(x) = [x]. Then we call the largest topology σ on X/∼ for which q is continuous that is to say

σ ={U ⊆X/∼: q⁻¹(U)∈τ}

the quotient topology

The following is just a restatement of the definition.

Lemma 8.20. Under the assumptions and with the notation of Definition 8.19, the quotient topology consists of the sets U such that

[

[x]∈U

[x]∈τ.

Later we shall give an example (Exercise 11.7) of a nice quotient topology.

Exercise 16.24, which requires ideas from later in the course, is an example of really nasty quotient topology.

In general, the quotient topology can be extremely unpleasant (basically because equivalence relations form a very wide class) and although nice equivalence relations sometimes give very useful quotient topologies, you should always think before using one. Exercise 16.9 gives some further information.

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9 Hausdorff spaces

When we work in a metric space, we make repeated use of the fact that, if d(x, y) = 0, then x=y. The metric is ‘powerful enough to separate points’.

The indiscrete topology, on the other hand, clearly cannot separate points.

When Hausdorff first crystallised the modern idea of a topological space, he included an extra condition to ensure ‘separation of points’. It was later discovered that topologies without this extra condition could be useful, so it is now considered separately.

Definition 9.1. A topological space (X, τ) is called Hausdorff if, whenever x, y ∈ X and x 6= y, we can find U, V ∈ τ such that x ∈ U, y ∈ V and U ∩V =∅.

In the English educational system, it is traditional to draw U and V as little huts containing x and y and to say that x and y are ‘housed off from each other’.

The next exercise requires a one line answer, but you should write that line down.

Exercise 9.2. Show that, if(X, d)is a metric space, then the derived topology is Hausdorff.

Although we defer the discussion of neighbourhoods in general to towards the end of the course, it is natural to introduce the following locution here.

Definition 9.3. If (X, τ)is a topological space and x∈U ∈τ, we call U an open neighbourhood of x.

Exercise 9.4. If (X, τ) is a topological space, then a subset A of X is open if and only if every point of A has an open neighbourhood U ⊆A.

Lemma 9.5. If (X, τ) is a Hausdorff space, then the one point sets {x} are closed.

The following exercise shows that the converse to Lemma 9.5 is false and that, if we are to acquire any intuition about topological spaces, we will need to study a wide range of examples.

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Exercise 9.6. Let X be infinite (we could take X =Z or X =R). We say that a subset E of X lies in τ if either E =∅ or X\E is finite. Show that τ is a topology and that every one point set {x} is closed, but that (X, τ) is not Hausdorff.

What happens if X is finite?

Exercise 9.7. Prove Exercise 7.9 (i) and (ii) with ‘(Y, d) a metric space’

replaced by ‘(Y, σ) a Hausdorff topological space’.

It is easy to give examples of topologies which are not derived from metrics. It is somewhat harder to give examples of Hausdorff topologies which are not derived from metrics. An important example is given in Exercise 16.10.

The next two lemmas are very useful.

Lemma 9.8. If (X, τ) is a Hausdorff topological space and Y ⊆X, then Y with the subspace topology is also Hausdorff.

Lemma 9.9. If(X, τ)and(Y, σ)are Hausdorff topological spaces, thenX×Y with the product topology is also Hausdorff.

Proof. The proof is easy (but there is one place where you can make a silly mistake). See page 89.

Exercise 16.9 shows that, even when the original topology is Hausdorff, the resulting quotient topology need not be.

10 Compactness

Halmos says somewhere that if an idea is used once it is a trick, if used twice it is a method, if used three times a theorem but if used four times it becomes an axiom.

Several important theorems in analysis hold for closed bounded intervals.

Heine used a particular idea to prove one of these. Borel isolated the idea as a theorem (the Heine–Borel theorem), essentially Theorem 10.6 below.

Many treatments of analysis (for example, Hardy’s Pure Mathematics) use the Heine–Borel theorem as a basic tool. The notion of compactness repre- sents the last stage in the Halmos progression.

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Definition 10.1. A topological space (X, τ) is called compact if, whenever we have a collection U_α of open sets [α∈ A]with S

α∈AU_α =X, we can find a finite subcollection Uα(1), Uα(2), . . . , Uα(n) with α(j)∈A [1≤ j ≤ n] such that Sn

j=1Uα(j) =X.

Definition 10.2. If (X, τ) is a topological space, then a subset Y is called compact if the subspace topology on Y is compact.

The reader should have no difficulty in combining these two definitions to come up with the following restatement,

Lemma 10.3. If(X, τ)is a topological space, then a subset Y is compact if, whenever we have a collection Uα of open sets [α ∈A] with S

αUα ⊇Y, we can find a finite subcollection U_α(1), U_α(2), . . . , U_α(n) with α(j)∈A [1≤j ≤ n] such that Sn

j=1Uα(j) ⊇Y.

In other words, ‘a set is compact if any cover by open sets has a finite subcover’.

The reader is warned that compactness is a subtle property which requires time and energy to master⁶. (At the simplest level, a substantial minority of examinees fail to get the definition correct.) Up to this point most of the proofs in this course have been simple deductions from definitions. Several of our theorems on compactness go much deeper and have quite intricate proofs.

Here are some simple examples of compactness and non-compactness.

Exercise 10.4. (i) Show that, ifX is finite, every topology onX is compact.

(ii) Show that the discrete topology on a set X is compact if and only if X is finite.

(iii) Show that the indiscrete topology is always compact.

(iv) Show that the topology described in Exercise 9.6 is compact.

(v) Let X be uncountable (we could take X = R). We say that a subset A of X lies in τ if either A = ∅ or X \A is countable. Show that τ is a topology but that (X, τ) is not compact.

Solution. There is a partial solution for parts (iv) and (v) on page 89.

The next lemma will serve as a simple exercise on compactness but is also important in its own right.

6My generation only reached compactness after a long exposure to the classical Heine–

Borel theorem.

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Lemma 10.5. Suppose that (X, d)is a compact metric space (that is to say, the topology induced by the metric is compact).

(i) Given any δ > 0, we can find a finite set of points E such that X = S

e∈EB(e, δ).

(ii) X has a countable dense subset.

Proof. See 90.

Observe that R with the usual metric has a countable dense subset but is not compact.

We now come to our first major theorem.

Theorem 10.6. [The Heine–Borel Theorem.] Let R be given its usual topology (that is to say the topology derived from the usual Euclidean metric).

Then the closed bounded interval [a, b] is compact.

Proof. I give a hint on page 62 and a proof on 91. An alternative proof, which is much less instructive, is given on page 34.

Lemma 10.3 gives the following equivalent statement.

Theorem 10.7. Let[a, b]be given its usual topology (that is to say the topology derived from the usual Euclidean metric). Then the derived topology is compact.

We now have a couple of very useful results.

Theorem 10.8. A closed subset of a compact set is compact. [More precisely, if E is compact and F closed in a given topology, then, if F ⊆ E, it follows that F is compact.]

Proof. This is easy if you look at it the right way. See page 91.

Theorem 10.9. If (X, τ) is Hausdorff, then every compact set is closed.

Proof. This is harder, though it becomes easier if you realise that you must use the fact that τ is Hausdorff (see Example 10.11 below). There is a hint on page 63 and a proof on page 92.

Exercise 10.10. Why does Theorem 10.9 give an immediate proof of Lemma 9.5?

Example 10.11. Give an example of a topological space (X, τ) and a compact set in X which is not closed.

Proof. There is a topological space with two points which will do. See page 92.

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Combining the Heine–Borel theorem with Theorems 10.8 and 10.9 and a little thought, we get a complete characterisation of the compact subsets of R (with the standard topology).

Theorem 10.12. Consider (R, τ) with the standard (Euclidean) topology.

A set E is compact if and only if it is closed and bounded (that is to say, there exists a M such that |x| ≤M for all x∈E).

In Example 4.10 we saw that the continuous image of an open set need not be open. It also easy to see that the continuous image of a closed set need not be closed.

Exercise 10.13. Let R have the usual metric. Give an example of a continuous injective function f :R→R such thatf(R) is not closed.

Hint. Look at the solution of Example 8.5 if you need a hint.

However, the continuous image of a compact set is always compact.

Theorem 10.14. Let (X, τ) and(Y, σ)be topological spaces and f :X →Y a continuous function. IfKis a compact subset ofX, thenf(K)is a compact subset of Y.

Proof. This is easy if you look at it the right way. See page 93.

This result has many delightful consequences. Recall, for example, that the quotient topology X/∼ is defined in such a way that the quotient map q :X →X/∼ is continuous. Since q(X) =X/∼, Theorem 10.14 gives us a positive property of the quotient topology.

Theorem 10.15. Let (X, τ) be a compact topological space and∼ an equivalence relation on X. Then the quotient topology on X/∼ is compact.

The next result follows at once from our characterisation of compact sets for the real line with the usual topology.

Theorem 10.16. Let Rhave the usual metric. If K is a closed and bounded subset of R and f :K →R is continuous, then f(K) is closed and bounded.

This gives a striking extension of one of the crowning glories of a first course in analysis.

Theorem 10.17. Let Rhave the usual metric. If K is a closed and bounded subset of R and f :K →R is continuous, then f is bounded and attains its bounds.

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Proof. The straightforward proof is given on page 93.

Theorem 10.17 is complemented by the following observation.

Exercise 10.18. Let R have the usual metric.

(i) If K is subset of R with the property that, whenever f : K → R is continuous, f is bounded, show that that K is closed and bounded.

(ii) If K is subset of R with the property that, whenever f : K → R is continuous and bounded, then f attains its bounds show that K is closed and bounded.

Proof. See page 94.

Exercise 17.3 gives a stronger result, but will be easier to tackle when the reader has done the section on sequential compactness (Section 12). Theo- rem 10.17 has the following straightforward generalisation whose proof is left to the reader.

Theorem 10.19. If K is a compact space and f : K → R is continuous then f is bounded and attains its bounds.

We also have the following useful result.

Theorem 10.20. Let(X, τ) be a compact and (Y, σ) a Hausdorff topological space. If f :X →Y is a continuous bijection, then it is a homeomorphism.

Proof. There is a hint on page 63 and a proof on page 94.

Theorem 10.20 is illuminated by the following almost trivial remark.

Lemma 10.21. Letτ1 andτ2 be topologies on the same spaceX. The identity map

ι: (X, τ1)→(X, τ2)

from X with topologyτ1 toX with topologyτ2 given byι(x) =xis continuous if and only if τ₁ ⊇τ₂.

Theorem 10.22. Let τ1 and τ2 be topologies on the same space X.

(i) If τ1 ⊇τ2 and τ1 is compact, then so is τ2. (ii) Ifτ₁ ⊇τ₂ and τ₂ is Hausdorff, then so is τ₁.

(iii) If τ1 ⊇τ2, τ1 is compact and τ2 is Hausdorff, then τ1 =τ2. Proof. The routine proof is given on page 95.

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The reader may care to recall that ‘Little Bear’s porridge was neither too hot nor too cold but just right’.

With the hint given by the previous theorem it should be fairly easy to do do the next exercise.

Exercise 10.23. (i) Give an example of a Hausdorff space (X, τ) and a compact Hausdorff space (Y, σ) together with a continuous bijectionf :X → Y which is not a homeomorphism.

(ii) Give an example of a compact Hausdorff space (X, τ) and a compact space (Y, σ) together with a continuous bijection f : X → Y which is not a homeomorphism.

We shall give a (not terribly convincing) example of the use of Theo- rem 10.20 in our proof of Exercise 11.7.

The reader may have gained the impression that compact Hausdorff spaces form an ideal backdrop for continuous functions to the reals. Later work shows that the impression is absolutely correct, but it must be remem- bered that many important spaces (including the real line with the usual topology) are not compact.

11 Products of compact spaces

The course contains one further major theorem on compactness.

Theorem 11.1. The product of two compact spaces is compact. (More formally, if (X, τ)and(Y, σ)are compact topological spaces andλis the product topology, then (X×Y, λ) is compact.)

Proof. There is a very substantial hint on page 63 and a proof on page 95.

Tychonov showed that the general product of compact spaces is compact (see the note to Exercise 16.7) so Theorem 11.1 is often referred to as Tychonov’s theorem.

The same proof, or the remark that the subspace topology of a product topology is the product topology of the subspace topologies (see Exer- cise 16.11), gives the closely related result.

Theorem 11.2. Let (X, τ) and (Y, σ) be topological spaces and let λ be the product topology. If K is a compact subset of X and L is a compact subset of Y, then K×L is a compact in λ.

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We know (see Exercise 8.16) that the topology on R² derived from the Euclidean metric is the same as the product topology when we give R the topology derived from the Euclidean metric. Theorem 10.7 thus has the following corollary.

Theorem 11.3. [a, b]×[c, d] with its usual (Euclidean) topology is compact.

The arguments of the previous section carry over to give results like the following⁷.

Theorem 11.4. ConsiderR² with the standard (Euclidean) topology. A set E is compact if and only if it is closed and bounded (that is to say, there exists a M such thatkxk ≤M for all x∈E).

Theorem 11.5. Let R² have the usual metric. If K is a closed and bounded subset of R² and f :K →R is continuous, thenf is bounded and attains its bounds.

Exercise 11.6. Let R² have the usual metric. If K is a subset of R² with the property that, whenever f : K → R is continuous, then f is bounded, show that K is closed and bounded. Let R² have the usual metric. If K is a subset of R² with the property that, whenever f :K →R is continuous, and bounded, then f attains its bounds, show that K is closed and bounded.

The generalisation to Rⁿ is left to the reader.

The next exercise brings together many of the themes of this course. The reader should observe that we know what we want the circle to look like.

This exercisechecks that defining the circle via quotient maps gives us what we want.

Exercise 11.7. Consider the complex plane with its usual metric. Let

∂D ={z ∈C : |z|= 1}

and give ∂D the subspace topology τ. Give R its usual topology and define an equivalence relation ∼by x∼yif x−y∈Z. We writeR/∼=T and give

7Stated more poetically by Conway.

IfE’s closed and bounded, says Heine–Borel, And also Euclidean, then we can tell

That, if it we smother With a large open cover,

There’s a finite refinement as well.

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T the quotient topology. The object of this exercise is to show that ∂D and T are homeomorphic.

(i) Verify that ∼ is indeed an equivalence relation.

(ii) Show that, if we definef :R→∂D by f(x) = exp(2πix), then f(U) is open whenever U is open.

(iii) If q : R → T is the quotient map q(x) = [x] show that q(x) = q(y) if and only if f(x) =f(y). Deduce that q f⁻¹({exp(2πix)})

= [x] and that the equation F(exp(2πix)) = [x] gives a well defined bijection F :∂D→T.

(iv) Show that F⁻¹(V) = f q⁻¹(V)

and deduce that F is continuous.

(v) Show that T is Hausdorff and explain why ∂D is compact. Deduce that F is a homeomorphism.

12 Compactness in metric spaces

When we work in R (or, indeed, in Rⁿ) with the usual metric, we often use the theorem of Bolzano–Weierstrass that every sequence in a bounded closed set has a subsequence with a limit in that set. It is also easy to see that closed bounded sets are the only subsets ofRⁿ which have the property that every sequence in the set has a subsequence with a limit in that set. This suggests a series of possible theorems some of which turn out to be false.

Example 12.1. Give an example of a metric space (X, d) which is bounded (in the sense that there exists an M with d(x, y) ≤M for all x, y ∈X) but for which there exist sequences with no convergent subsequence.

Solution. We can find such a space within our standard family of examples.

See page 98.

Fortunately we do have a very neat and useful true theorem.

Definition 12.2. A metric space (X, d) is said to be sequentially compact if every sequence in X has a convergent subsequence.

Theorem 12.3. A metric space is sequentially compact if and only if it is compact.

We prove the if and only if parts separately. The proof of the if part is quite simple when you see how.

Theorem 12.4. If the metric space(X, d) is compact, then it is sequentially compact.

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Proof. There is a hint on page 63 and a proof on page 98 Here is a simple but important consequence.

Theorem 12.5. If the metric space(X, d) is compact, then d is complete.

Proof. The easy proof is given on page 99. It uses a remark of independent interest given below as Lemma 12.6.

Lemma 12.6. Let (X, d) be a metric space. If a subsequence of a Cauchy sequences converges, then the series converges.

Observe that R with the usual Euclidean metric is complete but not compact.

The only if part of Theorem 12.3 is more difficult to prove (but also, in my opinion, less important). We start by proving a result of independent interest.

Lemma 12.7. Suppose that (X, d) is a sequentially compact metric space and that the collection Uα with α ∈ A is an open cover of X. Then there exists a δ > 0 such that, given any x ∈ X, there exists an α(x) ∈ A such that the open ball B(x, δ)⊆Uα(x).

We now prove the required result.

Theorem 12.8. If the metric space(X, d)is sequentially compact, it is compact.

This gives an alternative, but less instructive, proof of the theorem of Heine–Borel.

Alternative proof of Theorem 10.6. By the Bolzano–Weierstrass theorem, [a, b]

is sequentially compact. Since we are in a metric space, it follows that [a, b]

is compact.

If you prove a theorem on metric spaces using sequential compactness it is good practice to try and prove it directly by compactness. (See, for example, Exercise 16.16.)

The reader will hardly need to be warned that this section dealt only with metric spaces. Naive generalisations to general topological spaces are likely to be meaningless or false.

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13 Connectedness

This section deals with a problem which the reader will meet (or has met) in her first complex variable course. Here is a similar problem that occurs on the real line. Suppose that U is an open subset of R(in the usual topology) and f :U →Ris a differentiable function withf^′(u) = 0 for allu∈U. We would like to conclude that f is constant, but the example U = (−2,−1)∪(1,2), f(u) = 1 if u > 0,f(u) =−1 if u <0 shows that the general result is false.

What extra condition should we put on U to make the result true?

After some experimentation, mathematicians have come up with the following idea.

Definition 13.1. A topological space (Y, σ) is said to be disconnected if we can find non-empty open sets U and V such that U∪V =Y andU∩V =∅.

A space which is not disconnected is called connected.

Definition 13.2. If E is a subset of a topological space (X, τ), then E is called connected (respectively disconnected) if the subspace topology on E is connected (respectively disconnected).

The definition of a subspace topology gives the following alternative characterisation which the reader may prefer.

Lemma 13.3. If E is a subset of a topological space (X, τ), then E is disconnected if and only if we can find open sets U and V such thatU∪V ⊇E, U ∩V ∩E =∅, U ∩E 6=∅ and V ∩E 6=∅

We now look at another characterisation of connectedness which is very useful but requires a little preliminary work

Lemma 13.4. Let (X, τ) be a topological space and A a set. Let ∆ be the discrete topology onA. The following statements about a function f :X →A are equivalent.

(i) If x∈ X we can find a U ∈τ with x ∈U such that f is constant on U.

(ii) Ifx∈A, f⁻¹({x})∈τ

(iii) The map f : (X, τ)→(A,∆) is continuous.

Proof. Immediate.

If the conditions of Lemma 13.4 apply, we say that f islocally constant.

Theorem 13.5. If A contains at least two points, then a topological space (X, τ) is connected if and only if every locally constant function f :X →A is constant.

Metric and Topological Spaces