Metric and Topological spaces

difference between metric space and topological spaces, metric spaces topologically equivalent introduction to metric and topological spaces by w. a. sutherland
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Published Date:26-07-2017
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Metric and Topological Spaces T. W. Ko¨rner August 17, 2015 Small print The 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 should very much appreciate being told of any corrections or possible improvements and might even part with a small reward to A the first finder of particular errors. These notes are written in LT X2 and should be E ε 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. Contents 1 Preface 2 2 What is a metric? 4 3 Examples of metric spaces 5 4 Continuity and open sets for metric spaces 10 5 Closed sets for metric spaces 13 6 Topological spaces 15 7 Interior and closure 17 8 More on topological structures 19 9 Hausdorff spaces 25 10 Compactness 26 111 Products of compact spaces 31 12 Compactness in metric spaces 33 13 Connectedness 35 14 The language of neighbourhoods 38 15 Final remarks and books 41 16 Exercises 43 17 More exercises 51 18 Some hints 62 19 Some proofs 64 20 Executive summary 108 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 she tries to give a proofherself beforelooking 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. 2The 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 −1 f :P(X)→P(Y) and f :P(Y)→P(X) (hereP(Z) denotes the collection of subsets of Z) given by −1 f(A)=f(a) : a∈A and f (B) =x∈X : f(x)∈B. −1 We shall mainly be interested in f 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 −1 −1 −1 f (1), f (2) and f (3, 4). (ii) If U ⊆Y for all θ∈ Θ, show that θ \ \ −1 −1 −1 −1 f U = f (U ) and f U = f (U ). θ θ θ θ θ∈Θ θ∈Θ θ∈Θ θ∈Θ −1 −1 Show also that f (Y) =X, f (∅) =∅ and that, if U ⊆Y, −1 −1 f (Y \U) =X\f (U). (iii) If V ⊆X for all θ∈ Θ, show that θ f V = f(V ) θ θ θ∈Θ θ∈Θ and observe that f(∅) =∅. (iv) By finding appropriate X, Y, f and V, V , V ⊆ X, show that we 1 2 may have f(V ∩V )= 6 f(V )∩f(V ), f(X)6=Y and f(X\V) = 6 Y \f(V). 1 2 1 2 Solution. The reader should not have much difficulty with this, but if neces- sary, she can consult page 64. 32 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. Eachofthesenumbers isofinterest tosomeone andnoneofthemiseasily obtained fromanother. However, they dohave certain properties in common which we try to isolate in the following definition. 1 2 Definition2.1. LetX beaset andd :X →Rafunctionwiththefollowing 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 to B is the same as the distance from B to A. (iv) The distance from A to B via C is at least as great as the distance from A to B directly.’ 2 Exercise 2.2. If d :X →R is a function with the following properties: 1 We 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 that X is non-empty.’ (Of course, the result may be false for X = ∅, in which case the statement of the theorem must include the condition X 6=∅.) 4(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. Exercise2.3. 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 2 functions d :X →R satisfying condition (i) of Definition 2.1 such that: j (1) d satisfies conditions (ii) and (iii) but not (iv). 1 (2) d satisfies conditions (iii) and (iv) and d (x,y) = 0 implies x = y, 2 2 but it is not true that x =y implies d (x,y)= 0. 2 (3) d satisfies conditions (iii) and (iv) and x = y implies d (x,y) = 0. 3 3 but it is not true that d (x,y)= 0 implies x =y. 3 (4) d satisfies conditions (ii) and (iv) but not (iii). 4 You should verify your statements. Solution. See page 67. Other axiom grubbing exercises are given as Exercise 16.1 and 17.1. 2 Exercise 2.5. Let X be a set and ρ :X →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. 5Definition 3.1. Let V be a vector space overF (withF =R orF=C) and N :V →R a map such that, writing N(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=λkuk. (iv) Triangle law. If u, v∈V, then kuk+kvk≥ku+vk. Then we callk 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 that k0k =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, i an innerproduct and say that (V,h, i) is an innerproduct space. Lemma 3.5. Let (V,h , i) be an inner product space. If we write kuk = 1/2 hu,ui (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. 6n Lemma 3.6. If we work on R made into a vector space in the usual way, then n X hx,yi = x y j j j=1 is an inner product. Proof. Direct verification. We call the norm 1/2 n X 2 kxk = x , 2 j j=1 derived from this inner product the Euclidean norm (or sometimes just ‘the usual norm’). Although several very important norms arederived frominner products most are not. Lemma 3.7. (The parallelogram law) Using the hypotheses and notation of Lemma 3.5, we have 2 2 2 2 ku+vk +ku−vk = 2kuk +2kvk . 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 f(t)dt=0, a 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. 7Theorem 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 Z b hf,gi = f(t)g(t)dt a defines an inner product on C(a,b). We write   Z 1/2 b 2 kfk = f(t) dt 2 a for the derived norm. (ii) The equation Z b kfk = f(t)dt 1 a defines a norm on C(a,b). This norm does not obey the parallelogram law. (iii) The equation kfk = sup f(t). ∞ t∈a,b 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. 2 Definition 3.11. If X is a set and we define d :X →R by ( 0 if x =y, d(x,y)= 1 if x6=y, then d is called the discrete metric on X. Lemma 3.12. The discrete metric on X is indeed a metric. Proof. The easy proof is given on page 72. Exercise 3.13. (We deal with the matter somewhat better in Exercise 5.7) (i) IfV is a vector space overR andd is 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 overR and d is the discrete metric on V, show that d cannot be derived from a norm on V. 8n You should test any putative theorems on metric spaces on bothR with n the Euclidean metric andR 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 met- ric 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 nothave tobeassimple astheonesabove. Ishallusethemasexamples once ortwice, butthey donotformpartofstandardmathematicalknowledge and you do not have to learn their definition. 2 2 Definition 3.15. (i) If we define d :R ×R →R by ( kuk +kvk , if u = 6 v, 2 2 d(u,v)= 0 if u =v, then d is called the British Rail express metric. (To get from A to B travel via London.) 2 2 (ii) If we define d :R ×R →R by ( ku−vk if u and v are linearly dependent, 2 d(u,v)= kuk +kvk otherwise, 2 2 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. Onpage72weshowthattheBritishRailstoppingmetricisindeed a metric. The British Rail express metric can be dealt with similarly. However, fascinating as exotic metrics may be, the reader should reflect n on the number of different useful metric spaces that exist. We have R with the usual Euclidean norm, C(a,b) with the three norms described in 9Theorem 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 Continuityand open sets for metric spaces Some definitions and results transfer essentially unchanged from classical analysis onR to metric spaces. Recall the classical definition of continuity. Definition 4.1. Old definition. A function f :R→R is called continu- ous if, given t∈R and ǫ 0, we can find a δ(t,ǫ) 0 such that f(t)−f(s)ǫ whenevert−sδ(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 fg. Proof. This is identical to the one we met in classical analysis. If needed, details are given on page 73. 2 Exercise 4.4. LetR andR have their usual (Euclidean) metric. (i) Suppose thatf :R→R andg :R→R are continuous. Show that the 2 2 map (f,g):R →R is continuous. 2 (ii) Show that the mapM :R →R given byM(x,y)=xy is continuous. 2 (iii) Use the composition law to show that the map m :R →R given by m(x,y) =f(x)g(y) is continuous. Solution. See page 73. 10Exercise 4.4may lookperverse atfirst 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   1 x7→ log 2+sin . 2 1+x 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 continu- ous function is sufficiently wide to give us a large collection of interesting 2 functions and sufficiently narrow to ensure reasonable behaviour . In intro- ductory analysis we work onR with the Euclidean metric and only consider 2 subsets in the form of intervals. Once we move to R 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. 2 Suppose we work in R with the Euclidean metric. If E is an open set then any point e in E is the centre of a disc of strictly positive radius all of whose pointslie inE. Ifwe aresufficiently short sighted, every pointthatwe can see frome lies inE. 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). Hereareacoupleofsimple examples ofanopensetandasimple 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 2 Sentences like this are not mathematical statements, but many mathematicians find them useful. 11is open. n (ii) If we work in R 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. S (ii) If U is open for all α∈A, then U is open. (In other words, α α α∈A the union of open sets is open.) T n (iii) If U is open for all 1≤j≤n, then U is open. j j j=1 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. n Example 4.8. Let us work in R with the usual metric. Then B(x,1/j) is T ∞ open, but B(x,1/j)=x is not. j=1 Proof. See Example 4.6. There is a remarkable connection between the notion of open sets and continuity. Theorem4.9. Let (X,d)and(Y,ρ) be metric spaces. A functionf :X →Y −1 is continuous if and only if f (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. 12Proof. 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 thatTheorem 4.9gives avery neat proofofthe 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 fg. −1 New proof. If U is open in Z, then, by continuity, f (U) is open in Y  −1 −1 −1 and so, by continuity, (fg) (U) = g f (U) is open in X. Thus fg 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. 2 Exercise4.11. ConsiderR . 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. Solution. See page 77. 5 Closed sets for metric spaces The second class of well behaved sets identified by Cantor were the closed sets. Inordertodefine closedsets inmetricspaces, weneed anotionoflimit. Fortunately, the classical definition generalises without difficulty. Definition 5.1. Consider a sequence x in a metric space (X,d). If x∈X n and, given ǫ0, we can find an integer N ≥ 1 (depending on ǫ) such that d(x ,x)ǫ for all n≥N, n then we say that x → x as n→∞ and that x is the limit of the sequence n x . n Lemma 5.2. Consider a metric space (X,d). If a sequence x has a limit, n then that limit is unique. 13Proof. The simple proof is given on page 78. Just as in the next exercise, it suffices tofollowthe‘firstcourseinanalysis’proofwithminimalchanges. 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 x ∈ X and n x →x as n→∞, we have f(x )→f(x) n n 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 spacewith different metrics. We lookat the three norms (and their associated metrics) defined on C(0,1) in Theorem 3.10.   Define j : C(0,1),k k → C(0,1),k k by j (f)=f. α,β α β α,β (i) Show that j and j are continuous but j and j are not. ∞,1 ∞,2 1,∞ 2,∞ (ii) By using the Cauchy–Schwarz inequality hf,gi ≤ kfk kgk with 2 2 g = 1, or otherwise, show that j is continuous. Show that j is not. 2,1 1,2 Hint: Consider functions of the form f (x) =Kmax0,1−Rx. R,K 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 x ∈F and x →x as n→∞, it follows that x∈F. n n 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 overR and ρ is a metric derived from a norm, show that the one point setsx 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. 14Theorem 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. T (ii) IfF is closed for allα∈A, then F is closed. (In other words α α α∈A the intersection of closed sets is closed.) S n (iii) If F is closed for all 1≤j≤n, then F is closed. j j j=1 Proof. See page 80. Theorem 5.10. Let (X,d) and (Y,ρ) be metric spaces. A function f :X → −1 Y is continuous if and only if f (F) is closed inX wheneverF 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. S (ii) If U ∈τ for all α∈A, then U ∈τ. α α α∈A T n (iii) If U ∈τ for all 1≤j≤n, then U ∈τ. j j j=1 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. 15If (X,d) is a metric space we call the collection of open sets the topology induced by the metric. If(X,τ) isatopologicalspace we extend the notionofopen 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 operationof‘multiplication’ which followscertainrules, sowe mightsaythat 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 n aspects ofR 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) IfF 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 dis- n cretetopologyandtheindiscretetopology,R withthetopologyderivedfrom the Euclidean metric and 0,1with the topologyderived fromthe Euclidean metric. The following exercise is tedious but instructive (the tediousness is the instruction). Exercise 6.6. WriteP(Y) for the collection of subsets of Y. If X has three  elements, how many elements doesP P(X) have? How many topologies are there on X? Solution. See page 81. The idea of downward compatibility suggests ‘turning Theorem 4.9 in a definition’. 16Definition 6.7. Let (X,τ) and (Y,σ) be topological spaces. A function f : −1 X →Y is said to be continuous if and only iff (U) is open inX whenever U is open in Y. Theorem4.9tellsusthat,if(X,d)and(Y,ρ)aremetricspaces,thenotion of a continuous function f : X → Y is the same whether we consider the metrics or the topologies derived from them. TheproofofTheorem 4.3given onpage13carriesover unchangedtogive the following generalisation. Theorem6.8. If(X,τ), (Y,σ), (Z,µ ) are topologicalspaces andg :X →Y, f :Y →Z are continuous, then so is the composition fg. Downward compatibility suggests thedefinition ofaclosed set foratopo- logical 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 Theorem6.10. If(X,τ) is atopologicalspace, then thefollowingstatements are true. (i) The empty set∅ and the space X are closed. T (ii) IfF is closed for allα∈A, then F is closed. (In other words, α α α∈A the intersection of closed sets is closed.) S n (iii) If F is closed for all 1≤j≤n, then F is closed. j j j=1 Theorem 6.11. Let (X,τ) and (Y,σ) be topological spaces. A function f : −1 X → Y is continuous if and only if f (F) is closed in X whenever F is closed in Y. 7 Interior and closure Thenextsectionisshort,notbecausetheideasareunimportant,butbecause theyaresousefulthatthereaderwillmeetthemoverandoveragaininother courses. 17Definition 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 c the two notions of closure and interior are related. (Recall thatA =X\A, the complement of A.) Lemma 7.2. With the notation of Definition 7.1 c c c c (ClA ) =IntA and (IntA ) = 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.) Proof. The easy proof is given on page 82. Exercise 7.4. ConsiderR with its usual topology (i.e. the one derived from the Euclidean norm). We look at the open intervalI =(0,1). Show that ifF is closed andF ⊆ (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 corre- sponding 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.) 18Lemma 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 ∈ A with d(x,x )→ 0. (In n n 3 old fashioned terminology, the closure of A is its set of closure points .) 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. 4 (ii) Let (X,τ) be a topological space and (Y,d) a metric space . 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 (whereR 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 arethesame forthe purposes ofgrouptheory iftheyare(group) isomorphic. Twovectorspacesarethesameforthepurposesoflinearalgebra iftheyare(vectorspace)isomorphic. Whenaretwotopologicalspaces(X,τ) 3 I strongly advise caution in employing terms like ‘limit point’, ‘accumulation point’, ‘adherentpoint’and‘closurepoint’sinceboththeliteratureandyourlecturerareconfused 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. 4 Exercise 9.7 gives an improvement of parts (i) and (ii). 19and (Y,σ)the same forthe 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 −1 homeomorphic if there exists a bijection θ :X →Y such that θ and θ are continuous. We call θ a homeomorphism. The following exercise acts as useful revision of concepts learnt last year. Exercise8.2. Show that homeomorphism is an equivalence relation on topo- logical spaces. Homeomorphism implies equivalence for the purposes of topology. Exercise 8.3. Suppose that (X,d) and (Y,ρ) are metric spaces andf :X → Y is a homeomorphism. Show that d(x ,x)→ 0⇔ρ(f(x ),f(x))→ 0. n n 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 introduceacoupleofrelatedideas which arefundamen- taltoanalysisonmetricspaces,butwhichwillonlybereferredtooccasionally in this course. Definition 8.4. (i) If (X,d) is a metric space, we say that a sequence x in n X is Cauchy if, given ǫ 0, we can find an N (ǫ) with 0 d(x ,x )ǫ whenever n, m≥N (ǫ). n m 0 (ii) We say that a metric space (X,d) is complete if every Cauchy se- quence 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. 20

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