Topology I Lecture Notes

Transcription

1 Topology I Lecture Notes A80029 Version 2.3a, July 5, 2007 by Mahmoud Filali Edited by V-M Sarenius according to lectures in fall 1998 and Further editing in spring 2007 by Iikka Mylly. Topologically Speaking Topologically speaking, circles are the same as squares. Topologically speaking, people are the same as bears. Your donut is the same as the coffee cup you dunk it in each day. "Upside-down s the same as downside-up," a topologist might say. Topologically speaking, a quarter is the same as a dime. Topologists can count, but rarely finite amounts. You can bet they haven t yet to tell a 6 from a 9. A topologist can make a t-shirt with a piece of paper and a three-hole punch. The topological secret is the homeomorphic scrunch! Topolocical spaces are the places where topologists live. They like to drive compact manifolds, for when they bump into each other they give. You can visit your favorite topologist in a land called RP2, and if the hand that you favor is right you just might become a new left-handed you! Topologically speaking, peaches are the same as pears. Topologically speaking, people are the same everywhere. Topologically faces all look just like the one on you! Topologically races all share one gender, shape and hue. Topological spaces will expand your point of view. Cause topologically speaking, I m just the same as you. It s true - I m just the same as you. We re homeomorphic! I m just the same as you! c Monty Harper

2 Contents 1 Preliminaries Basic set theory Functions 3 2 Metric Spaces Distance Interior and closure Sequences Continuous functions 13 3 Topological spaces Topology Continuous functions Subspaces 21 4 Operations on topological spaces Sum of spaces Product spaces 25 5 Convergence Sequences Nets 31 6 Separation axioms 35 7 The T 4 -spaces 39 8 Compact spaces 44 9 Connected spaces 48 A Exercises 56 A.1 Exercises for Chapter 1 56 A.2 Exercises for Chapter 2 57 A.3 Exercises for Chapter 3 60 A.4 Exercises for Chapter 4 64 A.5 Exercises for Chapter 5 64 A.6 Exercises for Chapter 6 66 A.7 Exercises for Chapter 7 67 A.8 Exercises for Chapter 8 68 A.9 Exercises for Chapter 9 69

3 Chapter 1 Preliminaries This chapter discusses the very basics of set theory and functions. It is intended as a review of facts the reader is already assumed to be somewhat familiar with, and should not be seen as a rigorous introduction to the subject. However, all the rest of the material in these notes should be intelligible to anyone able to understand the basic results presented here. 1.1 Basic set theory The discussion in this section will be based on the so called intuitive or naïve set theory. Although a more exact axiomatic approach is available, given proper caution the approach taken here will work fine for us. Definition 1.1. A set, family or collection is a composition of objects, called the elements or points of the set. If an object a is an element of a given set A, we write a A. Otherwise, we write a / A. A straightforward way of describing a set is to list all of its elements, for example the set of symbols in the binary numeral system can be written as {0,1}, whereas the set of all odd integers could be written as {±1, ±3, ±5,...}. It is easy to realize that this method will be impossible to apply in many cases, even with sets with finitely many elements. Try listing the set of all the cells in your body, for instance! To avoid such troubles, we give the following definition. Definition 1.2. If S is a statement which applies to some of the elements of a given set A, then {a A : S(a)} is the set of all elements of A for which S(a) is true. One might wonder what was to happen if S from Definition 1.2 was a universally false sentence, like x x, for one. Our next definition answers that question. Definition 1.3. Let A be a given set. Then the set {x A : x x}, which clearly has no elements, is called the empty set, and denoted by. Example 1.1. Here we have some examples of how Definition 1.2 can be applied.

4 1.1 Basic set theory 2 1) {n N : 0 < n < 4} = {1,2,3}. 2) {n Z : n is odd} = {±1, ±3, ±5,...}. 3) {z C : z = 1} = T (the complex unit circle). 4) {x R : 0 x 1} = [0,1]. 5) {x R : 0 < x < 1} =]0,1[. Unfortunately, our naïve definition of sets leads to unexpected difficulties, of which the next famous result is the prime example. Russel s Paradox First note that a set can easily have elements which are themselves sets, like in {0, {1,2},5}. So, a set might well contain itself as one of its elements. In such a situation we say that our set is exceptional. Otherwise we say that it is ordinary. Let now A be the set of all ordinary sets. We are running into a logical contradiction; this set is both ordinary and exceptional for if A is exceptional then it must be one of its own elements and so it is ordinary. If it is ordinary then it is contained in A and so it is exceptional. Remark. Because of Russel s paradox, we agree from now on that no set shall contain itself as an element. Definition 1.4. A set A is said to be a subset of a set B if each element in A also belongs to B. We denote this by A B. The set B is said to be a superset of A, denoted by B A. Two sets A and B are equal if A B and B A. Remark. It follows from the definition above, that the empty set is unique. Definition 1.5. The union of two sets A and B is the set: A B = {a : a A or a B}. The intersection of two sets A and B is the set: A B = {a : a A and a B}. The concept of union and intersection can be extended to any collection of sets. Let {A i : i I} = {A i } i I be the collection of sets A i indexed by elements of a given set I. The union of this collection is the set A i = {a : a A i for some i I}, i I and the intersection is the set A i = {a : a A i for all i I}. i I The difference of two sets A and B is the set: A \ B = A B = {a : a A and a / B}. Let A and X be sets, with A defined as A = {a X : S(a)}. Then the complement of the set A in X, denoted by A, is the set of elements x X which are not elements of A, i.e., for which S(x) is false. Note that A = {x X : x X and a / A} = X \ A.

5 1.2 Functions 3 The product of two sets A and B is the set of ordered pairs A B = {(a,b) : a A and b B}. For any finite collection A 1,...,A n of sets, the product of the sets is defined as the set of ordered n-tuplets A 1 A n = {(x 1,...,x n ) : x i A i,i = 1,...,n}. If for each i = 1,...,n, A i = A, the product can be abbreviated as A n. The power set of a set A is the set of all subsets of A. The power set is denoted by P(A) or 2 A. Theorem 1.1. Let X be a given set, let {A i } i I be a family of subsets of X, and let P(A) be the power set of A. Then 1) X \ ( ) A i = (X \ A i ), i I i I 2) X \ ( ) A i = (X \ A i ), i I i I 3) A ( ) A i = (A A i ), i I i I 4) A ( ) A i = (A A i ), 5) 6) i I i I P(A i ) = P ( ) A i, i I i I P(A i ) P ( ) A i. i I i I The first two identities are called the de Morgan laws. Proof. See Exercise Functions Definition 1.6. A function (or mapping) consists of three objects: two nonempty sets X and Y and a rule f which assigns to each element x X a single element y Y. We write y = f(x) and say that y is the image of x under f or the value of f at x. We can also say that f is a function from X to Y, and denote this by f : X Y. The set X is called the domain of the function f and the set Y the range of f. Definition 1.7. Let A X and B Y. Then the image of A under f is the set f(a) Y defined by f(a) = {f(x) Y : x A}. The inverse image of B under f is the subset f 1 (B) X with f 1 (B) = {x X : f(x) B}. Theorem 1.2. Let f : X Y be a function. Let {A i } i I and {B j } j J be families of subsets of X and Y respectively. Then 1) f ( ) A i = f(a i ), i I i I

6 1.2 Functions 4 2) f ( ) A i f(a i ), i I i I 3) f 1( ) B j = f 1 (B j ), j J j J 4) f 1( ) B j = f 1 (B j ), j J j J 5) A f 1( f(a) ) for any A X, 6) f ( f 1 (B) ) B for any B Y. Proof. See Exercise 6. Remark. The equality in 2) does not hold in general. For example, let f : R R be the constant function f(x) = 1. Let A 1 = [0,1] and A 2 = [2,3]. Then = f( ) = f(a 1 A 2 ) f(a 1 ) f(a 2 ) = {1} {1} = {1}. Definition 1.8. Let X and Y be non-empty sets, and let f : X Y be a function. We say that f is surjective, a surjection or onto when f(x) = Y. We say that f is injective, an injection or one-to-one when f(x 1 ) = f(x 2 ) implies x 1 = x 2. When f is both injective and surjective, we say that it is bijective or a bijection. Definition 1.9. A set X is said to be countable if it is either finite (in other words has a finite number of members) or there is a bijection between X and N. If X is not countable, it is said to be uncountable. Example ) The set Z of integers is countable. A bijection between Z and N can be given by { 2n if n 0, f : Z N,n 1 2n if n < 0. 2) The set Q of rationals is countable. Consider any rational p q. Both p and q have a unique prime number decomposition, i.e., can be written uniquely as p = p n1 1 pn2 2 pni i,q = q m1 1 qm2 2 q mj j for some prime numbers p 1,...,p i and q 1,...,q j. Now a bijection between Q and Z is established by letting g : Q Z, p q p2n1 1 p 2n2 2 p 2ni i }{{} even powers q 2m1 1 1 q 2m2 1 2 q 2mj 1 j. }{{} odd powers The composition g f is a bijection between Q and N. 3) The set R of reals is uncountable. We will use a version of Cantor s diagonal argument to show this. Consider the set B of all real numbers between 0 and 1 with a decimal expansion consisting only of the digits 0

7 1.2 Functions 5 and 1. Let h : N B be an injective mapping such that 1 0.d 11 d 12 d d 21 d 22 d d 31 d 32 d k 0.d k1 d k2 d k3.... where d ij {0,1}. Now define an element x in B by letting x = 0.c 1 c 2 c 3... where c n = 0 if d nn = 1 and c n = 1 if d nn = 0, for all n N. Consequently, there is no natural number that maps to x, and thus h cannot be surjective. This shows that there exists no bijection between B and N, and so B is uncountable. This clearly implies that the whole of R is uncountable as well. 4) The Cantor set. Let C be a set obtained from [0,1] by removing first the open middle third ] 1 3, 2 3 [, so we are left with C 1 = [0, 1 3 ] [2 3,1]. Next, remove the open middle thirds ] 1 9, 2 9 [ and ]7 9, 8 9 [ from [0, 1 3 ] and [2 3,1], respectively. We are left with C 2 = [0, 1 9 ] [2 9, 1 3 ] [2 3, 7 9 ] [8 9,1]. With this process, we obtain a sequence of closed sets C n. The Cantor set is C = C n. The n=1 Cantor set is uncountable (see Exercise 30).

8 Chapter 2 Metric Spaces The concept of a metric space was first introduced by Maurice Fréchet in his doctoral dissertation, published in The fundamental idea of Fréchet s work was to generalize the distance properties of the real line, and subsequently extend the concept of convergence to a more abstract setting. The most important result in this chapter, and arguably one of the key theorems in analysis on metric spaces in general, is Baire s theorem. It was given by René Baire in the beginning of 20th century, and has since helped produce significant results such as the Principle of Uniform Boundedness and Open Mapping theorem. 2.1 Distance Definition 2.1. Let X be a non-empty set. A metric (or distance) on X is a function d : X X R satisfying, for x, y, z X, the following conditions. M1. d(x,y) 0. M2. d(x,y) = 0 if and only if x = y. M3. d(x, y) = d(y, x). (symmetry) M4. d(x, y) d(x, z) + d(z, y). (triangular, triangle or - inequality) The image d(x,y) is called the distance from x to y, and the pair (X,d) is called a metric space. Remark. Often several different metrics can be defined on a given non-empty set. Definition 2.2. Let (X,d) be a metric space. Let A and B be subsets of X and let x X. Then d(x, A) = inf{d(x, y) : y A}, (distance from point to set) d(a,b) = inf{d(x,y) : x A and y B}, (distance between two sets) d(a) = sup{d(x,y) : x,y A}. (diameter of A) Definition 2.3. Let (X,d) be a metric space, let x X and r > 0. The set B(x,r) = {y X : d(x,y) < r} is the open ball of centre x and radius r.

9 2.1 Distance 7 Example ) The real line R with d(x, y) = x y is a metric space: M1. x y 0 for all x,y R. M2. x y = 0 x y = 0 x = y for all x,y R. M3. x y = (x y) = y x for all x,y R. M4. x y = x z + z y x z + z y for all x,y,z R. Now the open ball of centre x and radius r is B(x,r) = {y R : d(x,y) < r} = {y R : x y < r} = {y R : r < x y < r} = {y R : x r < y < x + r} =]x r,x + r[. 2) More generally, consider the space R n and let x = (x 1,x 2,...,x n ) and y = (y 1,y 2,...,y n ). Define d 1, d 2 and d as follows: d 1 (x,y) = n x k y k, k=1 ( n d 2 (x,y) = (x k y k ) 2) 1 2, k=1 d (x,y) = max{ x k y k : k = 1,2,3,...,n}. The functions d 1, d 2 and d are all metrics on R n (see Exercise 16). 3) Let X and define d : X X R by { 0 x = y, d(x,y) = 1 x y. Then (X,d) is a metric space, and d is called the discrete metric. The open balls are given by for r 1, and for r > 1. B(x,r) = {y X : d(x,y) < r 1} = {x} B(x,r) = {y X : d(x,y) < r} = X Definition 2.4. Let (X,d) be a metric space and Y be a non-empty subset of X. Regard d as a function on Y Y. Then (Y,d) is also a metric space. It is called a subspace of (X,d). Example 2.2. The spaces ([0,1], ), (Q, ), (N, ) are subspaces of (R, ). The unit circle (T, ) and (R, ) are subspaces of (C, ). Remark. From now on, if the metric in question is either evident from the context of the statement or is arbitrary, we may refer to a metric space (X,d) briefly as metric space X.

10 2.1 Distance 8 Definition 2.5. Let (X,d) be a metric space. A subset A of X is said to be open, if for every x A there exists r > 0 such that B(x,r) A. A subset of X is said to be closed if its complement is open. Theorem 2.1. Let (X,d) be a metric space. Then each open ball in (X,d) is an open set. Proof. Let B(x 0,r) be an open ball in (X,d). Now let x B(x 0,r) and fix ρ = r d(x 0,x). Since x B(x 0,r), we have d(x 0,x) < r and thus ρ > 0. To show that B(x,ρ) B(x 0,r), take y B(x,ρ). This means precisely that d(x,y) < ρ = r d(x 0,x), and by applying the triangle inequality, we obtain d(x 0,y) d(x 0,x) + d(x,y) < d(x 0,x) + r d(x 0,x) = r. This shows that for all x B(x 0,r), there exists ρ > 0 such that B(x,ρ) B(x 0,r); since x 0 X and r > 0 were arbitrary, we have our desired result. Theorem 2.2. Let (X, d) be a metric space. Then the following statements are true. Proof. 1) The empty set and the whole space X are open. 2) Any union of open sets is open. 3) Any finite intersection of open sets is open. 1) Clear. 2) Let {A i } i I be an arbitrary family of open sets in X. Let x A i. Then i I x A i0 for some i 0 I. Since A i0 is open, there exists r > 0 such that B(x,r) A i0. Thus B(x,r) A i, and so A i is open. i I i I 3) Let {A 1,A 2,...,A n } be a finite family of open sets in X and let x n A i. Then x A i for every i = 1,2,...,n. Since each A i is open, i=1 there exists r i > 0 for each i = 1,2,...,n such that B(x,r i ) A i. Let r = min{r 1,r 2,...,r n }. Then B(x,r) B(x,r i ) A i for each i = 1,2,...,n. Accordingly, B(x,r) n i=1 A i, and so n i=1 A i is open. Theorem 2.3. Let (X,d) be a metric space. Then 1) The empty set and the whole space X are closed. 2) Any intersection of closed sets is closed. 3) Any finite union of closed sets is closed. Proof. See Exercise 19. Remark. 1) Infinite intersection of open sets is not, in general, open. Let (R, ) be our metric space, and consider the family {] 1 n, 1 n [} n=1 of open sets. Then ] 1 n=1 n, 1 n [= {0}. The singleton {0} is not open, since it cannot contain an open interval.

11 2.2 Interior and closure 9 2) An infinite union of closed sets is not necessarily closed. Take for example {a} =]0,1[, which is not closed although each singleton {a} is a ]0,1[ closed. Definition 2.6. Let (X,d) be a metric space. Let x X and r 0. Then B (x,r) = {y X : d(x,y) r} is called the closed ball of centre x and radius r. Note that when r = 0, B (x,r) = {x}. Theorem 2.4. Let (X,d) be a metric space. Then each closed ball in (X,d) is a closed set. Proof. Let B (x,r) be any closed ball in (X,d). For B (x,r) to be closed we need to show that X \ B (x,y) is open. Let y X \ B (x,y), i.e., d(x,y) > r and so d(x,y) r > 0. Fix ρ = d(x,y) r. We check that B(y,ρ) X \ B (x,r). Let z B (y,ρ). Then d(y,z) < ρ, and so by triangle inequality d(x,z) d(x,y) d(y,z) >d(x,y) + r d(x,y) =r, showing that z / B (x,r). In other words, z X \ B (x,r) and thus X \ B (x,r) is open, so our result follows. 2.2 Interior and closure Definition 2.7. Let X be a metric space. Let A be a subset of X. A point x A is an interior point of A if it is the centre of some open ball contained in A. In other words, if there exists r > 0 such that B(x,r) A. We call the set of all such points the interior of A and denote it by Int(A) or A. Theorem 2.5. Let X be a metric space and A a subset of X. Then a) IntA is open. b) A is open if and only if A = IntA. Proof. See Exercise 25. Definition 2.8. Let X be a metric space and let A be a subset of X. A point x X is a limit point or accumulation point of A if each open ball with centre x contains at least one point of A different from x. In other words, for all r > 0, B(x,r) (A \ {x}). Definition 2.9. The closure of A, denoted by A or Cl(A) is the union of A and the set of all its limit points, i.e., A = A {limit points of A}.

12 2.3 Sequences 10 Definition Let X be a metric space and A X. When A = X, we say that A is dense in X. If there exists some A countable and dense in X, we say that X is a separable. Finally, if Int(A) =, we say that A is nowhere dense in X. Example ) Let N (R, ). Then N = N. However, Int(N) =, so N is nowhere dense in R. 2) Int([0,1]) =]0,1[. 3) The space ]0,1] has 0, which is not in the set, as a limit point. All other points of ]0,1] are also limit points, and so ]0,1] = [0,1]. 4) The subset {1, 1 2, 1 3,...} of (R, ) has 0 as the only limit point. Now {1, 1 2, 1 3,...} = {0,1, 1 2, 1 3,...} and Int({1, 1 2, 1 3,...}) =, so this set is nowhere dense in R. 5) Q = R, so Q is dense in R. The set Q is also countable and thus R is separable. Theorem 2.6. Let X be a metric space and A a subset of X. Then A is closed if and only if A = A. In other words, A contains all its limit points. Proof. Suppose first that A is closed. Clearly A A, so we only need to show that A A. Let x A. Then B(x,r) A for all r > 0, and so B(x,r) X \ A for all r > 0. Since X \A is open, the point x cannot be in X \A, and thus x A. Therefore A A, as required. Conversely, suppose that A contains all of its limit points, and proceed to show that X \A is open. If X \A =, the statement follows immediately. Assume that X \ A, and let x X \ A. Then x is not a limit point of A, so B(x,r) A = for some r > 0. Therefore B(x,r) X \ A for some r > 0, showing that X \ A is open, as required. Remark. In general B(x,r) B (x,r). For example, let d be the discrete metric on some set X which has more than two elements. Then B(x,1) B (x,1). 2.3 Sequences Definition Let (X,d) be a metric space and let (x n ) n=1 be a sequence of points in X. We say that (x n ) n=1 is convergent in X if there exists a point x in X, such that for every r > 0, there exists n 0 so that n n 0 implies d(x n,x) < r. Equivalently, (x n ) n=1 is convergent, if for all r > 0 there exists n 0 such that n n 0 implies x n B(x,r). If such a point x exists, we say that (x n ) n=1 converges to x, and denote this by x n x or lim x n = x. The point x is called the limit of (x n ) n=1. Note that the x limit of a sequence and a limit point of a set are different concepts. Theorem 2.7. Let X be a metric space and let (x n ) n=1 be a sequence in X. Then (x n ) n=1 has exactly one limit.

13 2.3 Sequences 11 Proof. Suppose that x n x and x n y. Then for every r > 0, there exists n 0 N such that d(x n,x) < r 2 whenever n n 0; and there exists n 1 N such that d(x n,y) < r 2 whenever n n 1. Let n 2 = max{n 0,n 1 }. Then 0 d(x,y) d(x,x n ) + d(x n,y) < r 2 + r 2, whenever n n 2. Thus, for all r > 0, we have 0 d(x,y) < r. This implies that d(x,y) = 0, and so x = y. Theorem 2.8. Let (X,d) be a metric space, A be a nonempty subset of X and x X. Then x A if and only if there is a sequence (x n ) n=1 in A such that x n x. Proof. Let x A. If x A, then the constant sequence with x n = x for all n = 1,2,... converges to x. Suppose that x is a limit point of A. Then for each n N, B(x, 1 n ) A, so we may take x n B(x, 1 n ) A. In this way, we form a sequence (x n ) n=1 in A which converges to x. To see this, let r > 0. Then there exists n 0 N such that 0 < 1 n r for every n n 0. Accordingly, d(x n,x) < 1 n r, whenever n n 0, as required. Conversely, suppose that (x n ) n=1 is a sequence in A converging to some point x in X. Then for every r > 0, there exists n 0 N such that x n B(x,r) A whenever n n 0. Thus for every r > 0, B(x,r) A, giving x A. Definition A sequence (x n ) n=1 is said to be a Cauchy sequence, if for every r > 0, there exists n 0 N such that n,m n 0 implies d(x n,x m ) < r. Theorem 2.9. Let X be a metric space and let (x n ) n=1 be a convergent sequence in X. Then (x n ) n=1 is a Cauchy sequence. Proof. Let (x n ) n=1 be a sequence in X converging to some point x X. Then, for every r > 0, there exists n r N such that d(x,x n ) < r 2, whenever n n 0. Now by the triangle inequality d(x n,x m ) d(x n,x) + d(x m,x) < r 2 + r 2 = r, whenever n,m n 0. Thus (x n ) n=1 is a Cauchy sequence. Example 2.4. In general, a Cauchy sequence need not converge. Let (]0,1], ) be our metric space and consider the sequence ( 1 n ) n=1. This is a Cauchy sequence because 1 n 1 m max{ 1 n, 1 m } 1 n 0 < r for all n n 0, where n 0 > 1 r. The sequence ( 1 n ) n=1, however, is not convergent in ]0,1], since its limit 0 is not in ]0,1]. Definition A metric space X is said to be complete if every Cauchy sequence in X is convergent in X. Example ) The space (]0,1], ) is not complete (see Example 2.4). 2) The space ([0,1], ) is complete. This is the completion of the space above. 3) The space (Q, ) is not complete.

14 2.3 Sequences 12 4) The space (R, ) is complete (see Exercise 32). 5) The space (R n,d 2 ) is complete (see Exercise 33). Theorem Let X be a complete metric space and let Y be a subspace of X. Then Y is complete if and only if Y is closed. Proof. Suppose that Y is complete, and proceed to show that Y = Y. Let y Y. Then by Theorems 2.7 and 2.8, there is a sequence (y n ) n=1 in Y converging to some unique y in X. By Theorem 2.9, (y n ) n=1 is a Cauchy sequence in Y, and since Y is complete, (y n ) n=1 converges in Y. So y = lim y n Y. Hence Y = Y, n and so Y is closed. Conversely, suppose that Y is closed and let (y n ) n=1 be a Cauchy sequence in Y. Then (y n ) n=1 is also a Cauchy sequence in X, and so (y n ) n=1 is convergent to some point y in X since X is complete. But Y is closed, so by Theorem 2.8, the limit y must be in Y. Thus Y is complete. Baire s theorem. Let X be a complete metric space and let {A n : n N} be a countable family of open subsets of X, each of which is dense in X (in other words A n = X for all n N). Then n=1 A n is also dense in X. Proof. We shall show that n=1 A n = X. Let x 1 be a fixed but arbitrary point in X, and let r 1 > 0. First we show that B(x 1,r 1 ) ( n=1 A n ). Since A 1 is dense in X, we see that B(x 1,r 1 ) A 1. Pick x 2 in B(x 1,r 1 ) A 1. Since B(x 1,r 1 ) A 1 is open, there exists r 2 > 0 such that B (x 2,r 2 ) B(x 2,2r 2 ) B(x 1,r 1 ) A 1. Since A 2 is also dense in X, then x 2 A 2 and so we may pick x 3 from B(x 2,r 2 ) A 2. Again, since B(x 2,r 2 ) A 2 is open, there exists r 3 > 0 such that B (x 3,r 3 ) B(x 3,2x 3 ) B(x 2,r 2 ) A 2. Inductively, we form a sequence (x n ) n=1 in X and a sequence (r n ) n=1 of positive real numbers such that B (x n+1,r n+1 ) B(x n,r n ) A n. Note that r n < r1 2. It follows that if n m, then x n n B(x m,r n ), and so d(x n,x m ) < r m r1 2. Thus, for all r > 0, there exists m m 0 such that n,m m 0 implies d(x n,x m ) < r m < r 1 2 m max{ 1 n, 1 m } < r. Hence (x n ) n=1 is a Cauchy sequence, and since X is complete, it converges to some point x 0 in X. Now since x n B (x m,r m ) for all n m and for all m N, we see that x 0 B (x m,r m ) for all m N. In particular, x 0 B (x 2,r 2 ) B(x 1,r 1 ). We also have x 0 B (x n+1,r n+1 ) B(x n,r n ) A n for all n = 1,2,... Thus x 0 A n for all n N, and accordingly x 0 B(x 1,r 1 ) ( n=1 A n ), as required.

15 2.4 Continuous functions 13 Corollary. Let X be a complete metric space and let {X n : n N} be a countable family of subsets of X such that n=1 X n = X. Then Int(X n ) for some n N. In other words, X cannot be covered by a countable union of nowhere dense subsets. Proof. Suppose that X = X n and Int(X n ) = for all n N. Then (by Exercise 25), X \ X n is dense in X for each n N. Since X n is closed, X \ X n is n=1 open for each n N. Baire s Theorem implies that X \ X n = X. However, n N we have = X \ X = X \ n=1 X n = n=1 (X \ X n ) n=1 (X \ X n ), implying that n N (X \ X n ) =. Clearly this is a contradiction, and so we must have Int(X n ) for some n N. Example 2.6. Baire s theorem can be used to show that there are plenty of continuous functions that are nowhere differentiable. In fact, in a topological sense there are more continuous real functions nowhere differentiable on some interval, say [0, 1], than those differentiable at even a single point. This is established by showing that the set of continuous functions on [0,1], that are differentiable at some point in [0,1] is of first category in the space of continuous real functions on [0,1], i.e., it is a countable union of nowhere dense subsets. Let b be an odd integer and 0 < a < 1 such that ab > π. Then the function defined on R by W (x) = a n cos(b n πx) n=0 is an example of a function that is continuous but nowhere differentiable. This was the first concrete example of such a function, presented by Weierstrass in One of the simplest examples of this type of functions, found later in 1930 by van der Waerden, can be written as V (x) = n=0 {10 n x} 10 n, where {a} is the difference between a and the nearest integer. 2.4 Continuous functions Definition Let (X,d 1 ) and (Y,d 2 ) be metric spaces and let f : X Y. We say that f is continuous at some point x 0 in X, when for every r > 0, there exists δ > 0 such that d 1 (x 0,x) < δ implies d 2 (f(x 0 ),f(x)) < r. We say that f is continuous on X if it is continuous at every point x X. Theorem Let X,Y,f and x 0 be as in Definition Then the following statements are equivalent.

16 2.4 Continuous functions 14 a) The function f is continuous at x 0. b) For every r > 0, there exists δ > 0 such that f(b 1 (x 0,δ)) B 2 (f(x 0 ),r). c) For every open set V in Y containing f(x 0 ), there exists an open set U in X containing x 0 such that f(u) V. d) Whenever (x n ) n=1 is a sequence in X with x n x 0, we have f(x n ) f(x 0 ) in Y. Proof. a) b) Let r > 0 and δ > 0 be such that if d 1 (x 0,x) < δ then d 2 (f(x 0 ),f(x)) < r. In other words, if x B 1 (x 0,δ), then f(x) B 2 (f(x 0 ),r). This says precisely that f(b 1 (x 0,δ)) B 2 (f(x 0 ),r). The converse is obvious. b) c) Let V be an open set in Y containing f(x 0 ). Then there exists r > 0 such that B 2 (f(x 0 ),r) V. By b) there exists δ > 0 such that f(b 1 (x 0,δ)) B 2 (f(x 0 ),r). We simply take U = B 1 (x 0,δ) to obtain f(u) V. c) b) Let r > 0. Then B 2 (f(x 0 ),r) is an open set in Y containing f(x 0 ). By c), there exists an open set U in X containing x 0 such that f(u) B 2 (f(x 0 ),r). Since U is open, there exists δ > 0 such that B 1 (x 0,δ) U. Thus, f(b 1 (x 0,δ)) f(u) B 2 (f(x 0 ),r). a) d) Let (x n ) n=1 be a convergent sequence in X with lim x n = x 0, and let r > 0. By a), there exists δ > 0 such that d 1 (x 0,x) < δ implies d 2 (f(x 0 ),f(x)) < r. Let n 0 N be such that d 1 (x n,x n ) < δ for all n n 0, thus implying that d 2 (f(x 0 ),f(x n )) < r whenever n n 0. Hence, f(x n ) f(x) in Y. d) a) We argue by contradiction, and suppose that f is not continuous at x 0. So there exists r > 0, such that for every δ > 0, there exists x X with d 1 (x 0,x) < δ but d 2 (f(x 0 ),f(x)) r. Accordingly, for each n N, we may pick x n X such that d 1 (x 0,x n ) < 1 n but d 2(f(x 0 ),f(x n )) r. In this way, we construct a sequence (x n ) n=1 in X for which x n x 0 in X, but f(x n ) f(x 0 ) in Y. Corollary. Let X and Y be metric spaces. A function f : X Y is continuous on X if and only if f(x n ) f(x) in Y whenever x n x in X. Proof. See Exercise 36.

17 Chapter 3 Topological spaces In this chapter we will abandon the concept of distance, and move on to even more abstract ways of characterizing openness, continuity and other such properties. This approach, although in a way a joint result of the work of dozens of the most brilliant mathematical minds of late 19th and early 20th century, is often said to have its origins in the book Basics of Set Theory by Felix Hausdorff, published in Topology Definition 3.1. A topological space is pair the (X, T), where X is a non-empty set and T is a family of subsets of X satisfying the following conditions. T1. The empty set and the whole space X belong to T. T2. If {A i } i I is a family of sets in T, then i I A i T. T3. If {A 1,A 2,...,A n } is a finite family of sets in T, then n i=1 A i T. The family T is called a topology on the set X and the elements of T are called open sets of X. Remark. As before with metric spaces, where there is little or no possibility of misinterpretation, we may denote a topological space (X,T) briefly by topological space X. Definition 3.2. Let X be a topological space and let A be a subset of X. We say that A is closed when X \ A is open. Definition 3.3. If x X and U T are such that x U, we say that U is a neighbourhood of x. Remark. A given set A in a topological space X is open if and only if for every x A there is a neighbourhood U x of x such that U x A. Proof. Suppose that A is open and let x A. We simply take U x = A. Conversely, suppose that for every x A there is a neighbourhood U x with x U x A. Then A = x A U x, and A is open by T2. Definition 3.4. A subfamily B T is called a base for X, if every A in T is a union of sets from B.

18 3.1 Topology 16 Remark. A family B is a base for X if and only if for all x X and for every neighbourhood U x of x, there exists B B such that x B U x. Proof. Let B be a base for X, let x X and U x be a neighbourhood of x. Then since U x is open, there is a family {B i : i I} B with U x = B i. So, there i I exists i I with x B i U x. Conversely, let A T be arbitrary. Then for every x A, there exists some B x B with x B x A. Thus A = B x, and so B is x A a base. Definition 3.5. A base at a point x X is a family B(x) of neighbourhoods of x, such that for every U containing x, there exists B B(x) with x B U. Definition 3.6. A family C is a subbase for X if the family of all finite intersections C 1 C 2 C n, where each C i is in C, is a base for X. Example ) A metric space is a topological space where the topology is defined by T = {all open sets of (X,d) in the sense of Definition 2.5}. A base for T can be given by B ={B(x,r) : x X,r R + }, B ={B(x,r) : x X,r Q + } or B ={B(x, 1 ) : x X,n N}. n 2) Let X. Then T = P(X) is a topology on X. This is called the discrete topology, and it is induced by the discrete metric (see Exercise 41). 3) The trivial (or indiscrete) topology on a set X is T = {,X}. 4) Are all topological spaces metric spaces? That is, does there exist a distance d such that T = T d. The following counterexample shows that this is not true. Let X be the Sierpinski space. In other words, let X = {a,b} and let T = {,X, {a}}. Then (X,T) is a topological space. This topology does not come from any metric. Proof. Sierpinski topology is clearly a topology. For the second claim suppose that there is d such that T d = T. Then there exists r > 0 such that B(a,r) {a}, since {a} T d and a {a}. This means that d(a,x) < r, implying that x {a}, i.e., x = a. Hence d(a,b) r, and B(b,r) = {b} is an open set in T d = T, which is a contradiction. If there exists such a distance d on X that T = T d, we say that the topological space X is metrizable. 5) Let X = R 2. Then {L R 2 : L is a line} is not a topology (and cannot be a base for any topology on X) because L 1 L 2 = {a}. The set {L R 2 : L is a line} is, however, a subbase for the discrete topology. Theorem 3.1. Let X be a topological space. Then the following statements are true. T1. The empty set and the whole space X are closed.

19 3.1 Topology 17 T2. Any intersection of closed sets is closed. T3. Any finite union of closed sets is closed. Proof. See Exercise 53. Definition 3.7. Let X be a topological space and let A X. Then the closure of A is the set A = Cl(A) = {F X : F is closed and A F }. Remark. Note that A is closed by Theorem 3.1, and that A is the smallest closed set containing A. Definition 3.8. As with metric spaces, we say that a point x in X is a limit point or accumulation point of a set A X, if each neighbourhood of x contains a point of A \ {x}. Remark. Let L(A) be the set of all limit points of A. Then A = A L(A) = {x X : for every neighbourhood U of x, U A }. Proof. Suppose that x / A L(A). Then there exists a neighbourhood U of x such that U A =. This means that A X \ U, where X \ U is closed since U is open. Therefore x / A. This shows that A A L(A). Conversely, suppose that x / A. Then x X \ A. Since A is closed, X \ A is open, and so X \ A is a neighbourhood of x. Now (X \ A) A =, and thus x / A L(A). This shows that A L(A) A. Theorem 3.2. Let X be a topological space and let A and B be subsets of X. Then 1) A A. 2) A B A B. 3) A = A. 4) A B = A B. 5) A is closed if and only if A = A. Proof. 1) Follows immediately from the definition of closure. 2) Let A B. Statement 1) says that A B B. It follows that A B, since A is the smallest closed set containing A. 3) From 1) A A. A = {F X : F closed and A F } A, since A is closed and is clearly one of the sets F. Thus A = A. 4) Since A A B, it follows from b) that A A B. Similarly B A B so A B A B. Now A B is closed and A B A B, so we must have A B A B. 5) Let A be closed. Then A = {F X : F closed and A F } A, and so by 1) A = A. Conversely, A = A makes A closed by definition.

20 3.1 Topology 18 Definition 3.9. Let X be a topological space and A X. The interior of A is the set Int(A) = A = {O X : O is open and O A}. Note that A is open and it is the largest open subset of A. Remark. As with metric spaces, we may also say that A = {x A : there exists a neighbourhood U of x such that x U A}. Proof. Let X be a topological space and A X. Then denote {x A : there exists a neighbourhood U of x such that x U A} = I(A). Let x A. It follows that there exists a set O open in X such that O A and x O. This says precisely that O is a neighbourhood of x and x O A, so x I(A). Conversely, let x I(A). Now there exists a neighbourhood U of x with x U A, and so x A. Theorem 3.3. Let X be a topological space and A X. Then X \ A = Int(X \ A). Proof. Let x X \ A, i.e., x / A. Then there exists a neighbourhood U of x such that U A =. Equivalently, U X \ A, and so x Int(X \ A). Theorem 3.4. Let X be a topological space and let A,B P(X). Then Proof. 1) Int(A) A. 2) A is open if and only if A = Int(A). 3) Int(Int(A)) = Int(A). 4) Int(A B) = Int(A) Int(B). 1. Evident from the definition. 2. A is open A Int(A) A = Int(A) (by 1)). The converse is clear. 3. Int(A) is open, so by 2), Int(Int(A)) = Int(A). 4. A B A Int(A B) Int(A) A B B Int(A B) Int(B) } Int(A B) Int(A) Int(B). Now Int(A) Int(B) is open and contained in A B. Thus we must have Int(A) Int(B) Int(A B), since Int(A B) is the largest open subset of A B. Remark. Try to do the proof using Theorems 3.2 and 3.3 (see Exercise 59). Definition Let X be a topological space and A X. The frontier or the boundary of A is the set δ(a) = Fr(A) = A (X \ A).

21 3.1 Topology 19 Note that Fr(A) is closed and that Example 3.2. Fr(X \ A) = X \ A X \ (X \ A) = X \ A A = Fr(A). a) Fr(]a,b[) = [a,b] (],a[ ]b, [) = {a,b}. b) Fr(X) = X X \ X = X =. c) Fr(Q) = Q R \ Q = R R = R. d) The frontier of R in R is the set Fr R (R) = R R \ R =. e) The frontier of R in R 2 is the set Fr R 2(R) = R R 2 \ R = R R 2 = R. As items d) and e) above show, the frontier of a set depends on the space where the set is considered. Theorem 3.5. Let X be a topological space and let A and B be subsets of X. Then 1) A = A Fr(A) = Int(A) Fr(A). 2) Int(A) = A \ Fr(A). 3) Int(A) Fr(A) Int(X \ A) = X. 4) Fr(A B) Fr(A) Fr(B). 5) Fr(A B) Fr(A) Fr(B). 6) Fr(A) = A \ A when A is open. 7) Fr(A) = if and only if A is closed and open (clopen). Proof. 1) First note that A Fr(A) = A (A (X \ A) = (A A) (A X \ A) = A X = A. Then for the other equality consider Int(A) Fr(A) = Int(A) (A X \ A) = (Int(A) A) (Int(A) X \ A) = A [Int(A) (X \ Int(A))] = A X = A. 2) Now A \ Fr(A) = A \ (A (X \ A)). By de Morgan s laws and Theorem 3.3, A \ (A (X \ A)) = (A \ A) A \ (X \ A) = (A X \ X \ A) = A Int(A) = Int(A). 3) Using 1) and Theorem 3.3, we obtain Int(A) Fr(A) Int(X \ A) = A Int(X \ A) = A (X \ A) = X. 4) With the help of Theorem 3.2 and by meticulous use of Definition 3.10,

22 3.2 Continuous functions 20 we can write Fr(A B) = A B X \ A B = (A B) ( (X \ A) (X \ B) ) (A B) (X \ A X \ B) = ( A (X \ A X \ B) ) ( B (X \ A X \ B) ) (A X \ A) (B X \ B) = Fr(A) Fr(B). 5) Again using the properties of closure, we have Fr(A B) = (A B) (X \ A B) (A B) (X \ A X \ B) = (A B X \ A) (A B X \ B) (A X \ A) (B X \ B) = Fr(A) Fr(B). 6) Let A be open, implying that X\A is closed. In other words X \ A = X\A, giving Fr(A) = A (X \ A) = A (X \ A) = A \ A. 7) Let A be open and closed. Then Fr(A) = A X \ A = A (X \ Int(A)) = A (X \ A) =. Conversely, let Fr(A) = A X \ A =. Then by Theorem 3.3, A A X \ (X \ A) = Int(A) A A, giving A = Int(A) = A, and so A is both open and closed. 3.2 Continuous functions Definition Let X and Y be topological spaces and f : X Y be a function (mapping). We say that f is continuous at a point x X if for every neighbourhood V of f(x) in Y, there exists a neighbourhood U of x in X such that f(u) V. We say that f is continuous on X when f is continuous at each point in X. Theorem 3.6. Let X and Y be topological spaces and f be a function from X to Y. Then the following statements are equivalent. a) The function f is continuous on X. b) For each open set V in Y, f 1 (V) is open in X. c) For each closed set F in Y, f 1 (F) is closed in X. d) f(a) f(a), for every A X, i.e., f(cl X (A)) Cl Y (f(a)). Proof. a) b) Let V be on open subset of Y and let x f 1 (V). Then f(x) V, and so there exists a neighbourhood U of x such that f(u) V, i.e., x U f 1 (V). Thus f 1 (V) is open.

23 3.3 Subspaces 21 b) c) Let F be closed in Y, implying that Y \ F is open in Y. By b), f 1 (Y \ F) is open in X, giving that X \f 1 (F) is open in X. Hence f 1 (F) is closed in X. c) d) Let A X. Then A f 1 (f(a)) f 1 (f(a)). Now by taking closure, we obtain A f 1 (f(a)) = c) f 1 (f(a)), showing that f(a) f(a). d) a) Let x X. We want to show that if d) holds, then f is continuous at x. Let V be an open set in Y with f(x) V, and let U = Int(f 1 (V)). Then U is open in X and f(u) = f(int(f 1 (V))) f(f 1 (V)) = V. So f(u) V. We only need to check that x U. Suppose that x / U. Then x X \ U = X \ Int(f 1 (V)) = X \ f 1 (V). Hence f(x) f(x \ f 1 (V)), which by d) is contained in f(x \ f 1 (V)) = f(x) \ f(f 1 (V)), meaning that f(x) Y \ V = Y \ V. This shows that f(x) / V, which is a contradiction. Thus x U. Definition Let X and Y be topological spaces. We say that f : X Y is a homeomorphism when i) f is bijective, ii) f is continuous and iii) f 1 : Y X is continuous. If there exists such a homeomorphism, we say that X and Y are homeomorphic. Definition Let X and Y be topological spaces. We say that f : X Y is an embedding of X into Y, when i) f is injective, ii) f is continuous and iii) f 1 : Y X is continuous. When such an embedding exists, we say that X is embedded into Y. We will denote this by X Y. Example ) Let a < b be in R. Then ]a,b[ and ]0,1[ are homeomorphic. The homeomorphism f :]a,b[ ]0,1[ is given by f(x) = x a b a. 2) The sets ]1,+ [ and ]0,1[ are homeomorphic, the homeomorphism being f(x) = 1 x. 3) The real line R and ] π 2, π 2 [ are homeomorphic, where the homeomorphism f : R ] π 2, π 2 [ is given by f(x) = arctan(x). 3.3 Subspaces Definition Let (X,T) be a topological space and let Y X. Then the family T = {A Y : A T }

24 3.3 Subspaces 22 is a topology on Y (see Exercise 58). It is called the relative or the induced topology on Y, and (Y,T ) a subspace of (X,T). Example ) The space R with its usual topology is a subspace of the Euclidean space R 2. 2) Usually Q is regarded as a subspace of R. 3) A subspace Z of a subspace Y of X is a subspace of X (see Exercise 71). Theorem 3.7. Let (X,T) be a topological space, (Y,T ) be a subspace of X and A Y. Then Proof. a) The set A is closed in Y if and only if A = F Y, where F is closed in X. b) Cl Y (A) = Cl X (A) Y. c) If B is a base for (X,T) then B = {B Y : B B} is a base for (Y,T ). a) Let A be closed in Y, and so Y \ A is open in Y. Then Y \ A = U Y for some open subset U of X. It follows that A =Y \ (Y \ A) = Y \ (U Y ) = (Y \ U) (Y \ Y ) =Y \ U = (X \ U) Y. So A = (X \ U) Y, where X \ U is closed in X. Conversely, suppose that A = F Y with F closed in X. Then Y \ A =Y \ (F Y ) = (Y \ F) (Y \ Y ) =Y \ F = (X \ F) Y, where X \ F is open in X, and so Y \ A is open in Y. Thus A is closed in Y. b) From the definition of closure, we obtain Cl Y (A) = {K Y : K closed in Y and A K} a) = {F Y : F closed in X and A F } ( ) = {F : F closed in X and A F } Y = Cl X (A) Y. c) We need to show that for any open set U in Y and for every x U, there exists B B such that x B U. By a), U = U Y for some open set U in X. Since B is a base for X, we pick B B with x B U. So x B Y U Y = U, as required. Remark. There is no mention about interior or frontier in Theorem 3.7. The reason is as follows: Let X = R 2, Y = R and A = R. Then Int Y (A) = Int R (R) = R, whereas Int X (A) = Int R 2(R) =. So Int Y (A) Int X (A) Y.

25 3.3 Subspaces 23 Similarly for frontiers, we have Fr Y (A) = Fr R (R) = Cl R (R) Cl R (R \ R) =, whereas Fr X (A) = Fr R 2(R) = Cl R 2(R) Cl R 2(R 2 \ R) = R R = R, so Fr Y (A) Fr X (A) Y. However, we still have Fr Y (A) Fr X (A) Y, Int Y (A) Int X (A) Y (see Exercise 58).

26 Chapter 4 Operations on topological spaces In Chapter 3 we saw how a subset of a given topological space naturally inherits a topology from the whole space, in a sense through the operation of set intersection. In this chapter we show that there are natural topologies corresponding to disjoint unions and cartesian products of sets as well. The latter, called product or Tychonoff topology, is vital in many important applications, and gives rise to one of the most universally applied theorems in general topology, the Tychonoff theorem (introduced in Chapter 8). 4.1 Sum of spaces Let {X s } s S be a family of disjoint topological spaces, that is, X s X t = for all s t. Let X = X s and T = {A X : A X s is open in X s s S}. Then s S T is a topology on X (see Exercise 76). Definition 4.1. The set X with topology T (from above) is called the sum of the spaces X s. We write (X,T) = s SX s. Theorem 4.1. A set F s SX s is closed if and only if F X s is closed in X s for all s S. Proof. Let F be a closed set in s SX s. This implies that the complement X \ F is open in X s. Now the set X \ F is open in s if and only if (X \ F) X s s S s SX is open in X s for all s S. This is again equivalent with [X \ ((X \ F) X s )] X s =(F X s ) [(X \ X s ) X s ] being closed in X s for all s S, as required. Remark. =(F X s ) = F X s a) Each set X s is open and closed in s SX s, since X s = X s X s is open in X s and X s = X s X s is closed in X s for all s S. b) Each X s is a subspace of s SX s, as is evident from the definition of the topology.

27 4.2 Product spaces 25 Theorem 4.2. Let X be a topological space and X = X s, where each X s is open in X and X s X t = when s t. Then X = s SX s, where each X s is equipped with the relative topology from X. Proof. Since X and s SX s have the same points, we only need to prove that they have the same open sets. Let U be open in X, implying that U X s is open in X s for all s S. This says exactly that U is open in s SX s. Conversely, let U be open in s SX s. This implies that U X s is open in X s for all s S. Now each X s is equipped with the subspace topology coming from X, and so we must have U X s = V X s for some V open in X. Since the sets X s are open in X for all s S as well, the intersection U X s is open in X. Ultimately s S (U X s) = U X s = U X = U s S s S is open in X, being a union of open subsets of X. Example The real line with its usual topology cannot be written as a sum of two topological spaces. 2. Let E be the Sorgenfrey line, i.e., E = (R,T) where B = {[a,r[: a R and r Q} is a base for T. Then let a E be any point, r Q such that a < r and X 1 = [a,r[. Now E = X 1 (E \ X 1 ). 4.2 Product spaces Definition 4.2. Let {X α } α I be a family of non-empty sets. The cartesian product of the sets X α is α I X α = {x : I α I X α such that x(α) X α for all α I} Usually we write X α instead of X α and x α for x(α). α I We also say that x α is the α:th coordinate of x. The function (map) π β : X α X β, defined by π β (x) = x β, is the β:th projection. Definition 4.3. Let X be a topological space with topologies T 1 and T 2. We say that T 1 is weaker than T 2 or T 2 is stronger than T 1 when T 1 T 2. Example ) When I = {1,2,3,...,n}, it is customary to write X α = α I n X α = {(x 1,x 2,...,x n ) : x k X k,k = 1,2,3,...,n}. α=1

28 4.2 Product spaces 26 2) Also in the case I = N, we may write X α = X α = {(x 1,x 2,...) : x k X k for every k N}. α N α=1 3) If X α = X for all α I, then X α is the set of all functions from I to α I X. The notation can be simplified by writing X α = X I. Warning. The statement that an arbitrary product space X α is non-empty is equivalent to the Axiom of choice. Definition 4.4. The Tychonoff topology or the product topology on obtained by taking as a base the sets of the form A α, where a) A α is open in X α for each α I, b) A α = X α for all but finitely many α I. α I α I α I α I X α is Remark. Condition b) says that A α = X α except for α 1,α 2,...,α n, so we can write A α = ( A 1 ) ( α A 2 ) ( α A n α), α I α I α I α I where A k α = X α for all but α k (k = 1,2,...,n). In other words, α I A α = π 1 α 1 (A α1 ) π 1 α 2 (A α2 ) π 1 α n (A αn ). This says that the family C = { π 1 α (A α ) : A α open in X α } is a subbase for the Tychonoff topology. Example Let S 1 be the unit circle in R 2. Then S 1 [0,1] is a cylinder and S 1 S 1 is a torus. 2. For each α I, let X α be discrete. Then X α is discrete if and only if I is finite. 3. Let X = R R = α R α I R be the set of all real-valued functions on R, i.e., X = {f : R R}. Let f X, and proceed to inspect what a neighbourhood of f looks like. Let U be a neighbourhood of f. Then f U = A α, where A α = R for all α I, but for some α 1,α 2,...,α n. Then for each k = 1,2,...,n, there exists r k > 0 such that f(α k ) A αk =] r k + f(α k ),r k + f(α k )[. We can give the neighbourhood of f in R R as α R U(f;α 1,α 2,...,α n ;r 1,r 2,...,r n ) = α R V α, where V α = R except for α k,k = 1,2,...,n, for which V αk =] r k + f(α k ),r k + f(α k )[.

29 4.2 Product spaces 27 Perhaps a more illustrative way of describing U is to give it as the set {g R R : g(α k ) f(α k ) < r k,k = 1,2,...,n}. A simpler neighbourhood is obtained by letting r = min{r 1,r 2,...,r n } and F = {α 1,α 2,...,α n }. Then ] r + f(α),r + f(α)[ ] r k + f(α k ),r k + f(α k )[ for each k = 1,...,n, and so f α R W α α R V α, where W α = R for all but α F. For α F, W α =] r + f(α),r + f(α)[. Again, this neighbourhood can be written simply as W α = W(f;F,r) = {g R R : g(α) f(α) < r,α F } α R f α α α α n Figure 4.1: The neighbourhood of f in Example 4.3 part 3. Theorem 4.3. Each projection π β : X α X β is continuous and open, but need not to be closed. Furthermore, the Tychonoff topology is the weakest topology on X α for which each projection is continuous. Proof. Let A β X β be open. Then π 1 β (A β) = U α, where U β = A β and U α = X α for all α β, is open in X α. This shows that π β is continuous. Let πγ 1 (A γ ) be an element in the subbase of the Tychonoff topology (in other words, A γ is open in X γ ). Then πγ 1 (A γ ) = U α, where U γ = A γ and U α = X α for all α γ. The projection π β (πγ 1 (A γ )) = π β ( U α ) = A γ if β = γ or X β if β γ. In both cases π β ( U α ) is open, and so π β is an open function. Proceed to show that the Tychonoff topology T is the weakest topology for which each π β is continuous. Let T be another topology on X α with this property. We need to show that T T. Let πα 1 (A α ) be an arbitrary element from the subbase of T. Then A α is open in X α, and, since π α is continuous with respect to T, πα 1 (A α ) is open in ( X α,t ). This shows that each member of T is also in T, i.e., T T.

30 4.2 Product spaces 28 Theorem 4.4. Let X be a topological space and let X α be equipped with the Tychonoff topology. Then f : X Xα π α f π α is continuous if and only if π α f is continuous for each α I. Proof. Necessity is immediate, since composition of continuous functions is continuous (see Exercise 62). Conversely let πα 1 (A α ) be an arbitrary element from the subbase of the Tychonoff topology. Then f 1 (πα 1 (A α )) = (π α f) 1 (A α ) is open in X since A α is open in X α and π α f is continuous by assumption. It follows that f is continuous. Remark. 1) Let X be a set and {X α } α I be a family of topological spaces, and let f α : X X α. The weak topology induced on X by the family {f α } α I of functions is the smallest topology on X making each f α continuous. A subbase for this topology is X α {f 1 α (A α ) : A α is open in X α,α I}. Theorem 4.2 says that the Tychonoff topology is the weak topology on Xα induced by {π α } α I. 2) Dually to the weak topology, we also have the notion of strong topology (or the largest topology ) induced on Y by the family of functions f α : Y α Y, where Y α is a topological space for each α. This is called the quotient topology. Details on the quotient topology can be found in most books on general topology. Let f : X Y, f continuous. Is the quotient space Y open or closed? 3) Another way of constructing topological spaces is with the so-called inverse systems and their limits.