A Glimpse into Topology

Transcription

1 A Glimpse into Topology by Vishal Malik A project submitted to the Department of Mathematical Sciences in conformity with the requirements for Math 430 (203-4) Lakehead University Thunder Bay, Ontario, Canada

2 Abstract This report covers the basics of Topology, including topological spaces, density, continuity, compactness and nets. It is followed by a short discussion of a few topics in metric space theory, such as metrizable spaces and completeness. Finally, the concepts of Topology are applied to metric spaces to derive three interesting results: Cantor s Intersection Theorem, Bourbaki s Mittag-Leffler Theorem and Baire s Theorem. i

3 Acknowledgements Dr. Monica Ilie acted as my advisor and provided guidance in learning and preparing the material. Dr. Adam Van Tuyl was the course coordinator and helped with learning L A TEX and preparing this report. ii

4 Abstract Acknowledgements Contents Chapter. Introduction Chapter 2. Topology 2. Topological Spaces 2 2. Closure 2 3. Bases and Subbases 3 4. Density 4 5. Subspaces 4 6. Neighbourhoods 4 7. Continuity 5 8. Nets 6 9. Separation 7 0. Compactness 7. Urysohn s Lemma 8 Chapter 3. Topics in Metric Spaces 0. Metric Spaces 0 2. Metrizable Spaces 0 3. Continuity in Metric Spaces 4. Sequences and Completeness 2 Chapter 4. Application of Topology to Metric Spaces 3 Bibliography 9 i ii iii

5 CHAPTER Introduction Topology was first introduced by German mathematician Felix Hausdorff in 94. It is a theory that is applied to several branches of mathematics, including Algebra and Analysis. The goal of this project was to study the basics of Topology and its applications in Analysis, including popular results like Baire s Theorem.

6 CHAPTER 2 Topology In this chapter, we will discuss basic concepts in Topology.. Topological Spaces We begin with definitions of topology and other related terms. Definition 2.. Let τ be a collection of subsets of a set X. Then τ is a topology on X if, () for any A τ, A τ, (2) for any A, A 2 τ, A A 2 τ, (3) φ τ, (4) X τ. Example 2.2. If X is the set of colours, then the following is a topology on it: τ = {{red}, {green}, {red, green}, φ, C} Definition 2.3. A tuple (X, τ) is a topological space if τ is a topology on X. Definition 2.4. Let (X, τ) be a topological space. A subset U of X is open if U τ. Definition 2.5. Let (X, τ) be a topological space. X F τ. A subset F of X is closed if 2. Closure Definition 2.6. Let (X, τ) be a topological space and A X. Then the closure of A, denoted as A, is a closed subset of X with the following properties: () A A. (2) If F is a closed subset of X and A F, then A F. Theorem 2.7. Let (X, τ) be a topological space and A, B X. A B. Proof. If A B, then A B. Therefore A B. 2 If A B, then

7 Chapter 2. Topology 3 3. Bases and Subbases Sometimes we don t need the entire collection of open sets to work with a topology. Bases and subbases are smaller subcollections that can suffice. Definition 2.8. Let (X, τ) be a topological space and B τ. Then B is a base of τ if τ = { B C B C B}. Theorem 2.9. Let (X, τ) be a topological space. A collection of sets B is a base of τ if and only if the following two conditions are satisfied: () X = B. (2) For any B, B 2 B, if p B B 2, then there exists a B 3 B such that p B 3 B B 2. Proof. Let B be a base of τ. Therefore τ = { B C B C B}. () Since X τ, there exists a C B such that X = B C B. Therefore X B. Also, since B τ, we have B X. Therefore X = B. (2) If B, B 2 B, then B, B 2 τ. Therefore B B 2 τ. Therefore, there exists a C B such that B B 2 = B C B. This means that if p B B 2, then there exists a B 3 C such that p B 3. Also B 3 B C B. Therefore we have p B 3 B B 2. Conversely, let B be a collection of sets that satisfies the two conditions. Let τ = { B C B C B}. We will prove that τ is a topology on X i.e. it has each of the four properties of Definition 2.. () Let A τ. Each set in A is a union of a collection of sets in B. Therefore A is a union of a collection of sets in B. Therefore A τ. (2) Let A, A 2 τ. There are C, C 2 B such that A = X C X and A 2 = Y C 2 Y. Therefore ( ) ( ) A A 2 = = (( ) ) Y X C X Y C 2 Y Y C 2 = Y C 2 X C X X C (X Y ) Since X, Y B, there exists, for every p X Y, a B p B such that p B p X Y (from the second condition). Therefore X Y = p X Y B p. Therefore A A 2 is a union of a collection of sets in B, which proves that A A 2 τ. (3) Since φ B and B φ B = φ, we have φ τ. (4) From the first condition, X τ. Definition 2.0. Let (X, τ) be a topological space and S τ. Then S is a subbase of τ if the set { S S S and S is finite} is a base of τ.

8 Chapter 2. Topology 4 4. Density Definition 2.. Let (X, τ) be a topological space and D X. Then D is dense in X if, for every nonempty open subset U of X, U D in nonempty. Theorem 2.2. If (X, τ) is a topological space and U, U 2 τ are dense in X, then U U 2 is dense in X. Proof. Let U be a nonempty open subset of X. Since U is dense in X, U U φ. Since U U is open and U 2 is dense in X, U U U 2 φ. Since U is arbitrary, this proves that U U 2 is dense in X. 5. Subspaces Theorem 2.3. Let (X, τ) be a topological space and Y X. Let τ be a collection of sets with the following property: V τ if and only if V = Y U for some U τ. Then τ is a topology on Y. Proof. We will prove that τ has each of the four properties of Definition 2.. () Let A τ. For each V A, V = Y U for some U τ. Therefore A = Y ( B) for some B τ. Since B τ, A τ. (2) Let A, A 2 τ. Therefore, there exist B, B 2 τ such that A = Y B and A 2 = Y B 2. Therefore A A 2 = (Y B ) (Y B 2 ) = Y (B B 2 ). Since B B 2 τ, A A 2 τ. (3) Since φ = Y φ and φ τ, φ τ. (4) Since Y = Y X and X τ, Y τ. Definition 2.4. Let (X, τ) be a topological space and Y X. Then (Y, τ ) is a subspace of (X, τ) if τ is a topology derived using Theorem 2.3. Theorem 2.5. Let (X, τ) be a topological space and D Y X. If D is dense in X, then it is also dense in Y. Proof. Let V be a nonempty open subset of Y. Then V = Y U for some nonempty open subset U of X. Therefore D V = D Y U = D U. Since D is dense in X, D U is nonempty. Therefore D V is nonempty. 6. Neighbourhoods Definition 2.6. Let (X, τ) be a topological space, x X and N X. Then N is a neighbourhood of x if there exists a set U τ such that x U N. The following theorem provides a characterization of open sets in terms of neighbourhoods.

9 Chapter 2. Topology 5 Theorem 2.7. Let (X, τ) be a topological space and U be a subset of X. Then U is open if and only if, for every x U, there is a neighbourhood N x of x such that N x U. Proof. If U is open, then it is a neighbourhood of every x U. Conversely, suppose, for every x U, there is a neighbourhood N x of x such that N x U. Then, clearly, for every x U, there is a U x τ such that x U x U. Moreover U = U x, which proves that U τ. x 7. Continuity Definition 2.8. Let (X, τ ) and (Y, τ 2 ) be two topolological spaces. A function f : X Y is continuous at x X if, for every neighbourhood V of f(x), there is a neighbourhood U of x such that f(u) V. The function f is continuous on the set X if it is continuous at every x X. Theorem 2.9 provides a much more practical definition of continuity of a function on a set. Theorem 2.9. Given two topological spaces (X, τ ) and (Y, τ 2 ), and a function f : X Y, the following are equivalent. () f is continuous on X. (2) For every open subset V of Y, f (V ) is an open subset of X. (3) For every closed subset F of Y, f (F ) is a closed subset of X. (4) For every subset A of X, f(a) f(a). Proof. () = (2) Let f be continuous on X. For any open subset V of Y, if x f (V ), then there is an open neighbourhood U of x such that f(u) V. However, this implies that U f (V ), which proves that f (V ) is open. (2) = (3) If F is a closed subset of Y, then Y F is an open subset of Y. Therefore f (Y F ) is an open subset of X. Therefore f (F ) = X f (Y F ) is a closed subset of X. (3) = (4) For any subset A of X, since f(a) is closed, f (f(a)) is closed. Since A f (f(a)), we have A f (f(a)). Therefore f(a) f(a). (4) = () For any x X, let V be an open neighbourhood of f(x). Let A = X f (V ) and U = X A. We will prove that U is a neighbourhood of x and f(u) V. Since f(a) Y V and Y V is closed, f(a) Y V. Since f(x) / Y V, we have f(x) / f(a). This implies that f(x) / f(a), since f(a) f(a). Therefore x / A, which proves that x U. Since U is open, it is a neighbourhood of x. Now, for any y U, we have y / A, which implies that y / A. Therefore y f (V ) (or f(y) V ). Therefore f(u) V.

10 Chapter 2. Topology 6 8. Nets Definition A set Λ is directed if it has a relation on it such that, () for any λ Λ, λ λ, (2) for any λ, λ 2, λ 3 Λ, if λ λ 2 and λ 2 λ 3, then λ λ 3, (3) for any λ, λ 2 Λ, there exists a λ 3 Λ such that λ λ 3 and λ 2 λ 3. Example 2.2. N is a directed set. Example The set N x of neighbourhoods of an element x of a topological space is a directed set with the relation defined as follows: If N, N 2 N x, then N N 2 if N 2 N. Definition A function f : Λ X is a net in X if Λ is directed. For any λ Λ, f(λ) is denoted as x λ. The net itself is denoted as (x λ ). Definition A net (x λ ) is eventually in a set U if there exists a λ such that, for every λ λ, x λ U. Definition Let (X, τ) be a topological space and x X. A net (x λ ) converges to x if it is eventually in every neighbourhood of x. The convergence of (x λ ) to x is denoted as x λ x. Definition Let f : Λ X be a net. Let φ be a directed set such that the net g : φ Λ has the following properties: () For every µ, µ 2 φ, if µ µ 2, then g(µ ) g(µ 2 ). (2) For every λ Λ, there is a µ φ such that λ g(µ). Then f g is a subnet of f. For any µ φ, f g(µ) is denotes as x λµ. The subnet itself is denoted as (x λµ ). x. Theorem If a net (x λ ) converges to x, then its every subnet also converges to Proof. Let (x λµ ) be a subnet of (x λ ). Let N be a neighbourhood of x. If (x λ ) converges to x, then there exists a λ such that, for every λ λ, x λ N. Now, there exists a µ such that λ µ λ. Moreover, for every µ µ, λ µ λ µ. Therefore, for every µ µ, x λµ N. Since N is arbitrary, (x λµ ) is eventually in every neighbourhood of x. The following theorem characterizes continuity in terms of convergence of nets. Theorem A function f : X Y is continuous at x X if and only if, for every net (x λ ) in X such that x λ x, f(x λ ) f(x). Proof. Suppose f is continuous at x. If V is an open neighbourhood of f(x), then f (V ) is an open neighbourhood of x. If x λ x, then (x λ ) is eventually in f (V ). Therefore f(x λ ) is eventually in V, which proves that f(x λ ) f(x).

11 Chapter 2. Topology 7 Conversely suppose f(x λ ) f(x) for every net (x λ ) in X such that x λ x. Suppose f is not continuous at x. Therefore, there exists a neighbourhood V of f(x) such that, for every neighbourhood U of x, there exists an x U U such f(x U ) / V. This implies that the net (x U ) converges to x (Example 2.22), but (f(x U )) does not converge to f(x). This, however, is is a contradiction. Therefore f is continuous at x. Theorem Let (X, τ) be a topological space, F be a closed subset of X and (x λ ) be a net in F. If x λ x, then x F. Proof. Let x X F. Since X F is open, there is a neighbourhood N of x such that N X F (Theorem 2.7). Also, since x λ x, (x λ ) is eventually in N. This, however, is contradiction, since (x λ ) is in F. 9. Separation In this section, we will define topological spaces whose elements are separated from one another in a certain manner. Definition A topological space (X, τ) is a T -space if, for every x, y X such that x y, there exist open subsets U and V of X such that x U but y / U, and y V but x / V. Definition 2.3. A topological space (X, τ) is Hausdorff if, for every x, y X such that x y, there exist disjoint open subsets U and V of X such that x U and y V. It can be observed that every Hausdorff space is also a T -space. Example Every metrizable space (Definition 3.6) is Hausdorff. Definition A T -space (X, τ) is normal if, for any disjoint closed subsets F and F 2 of X, there exist disjoint open subsets U and U 2 of X such that F U and F 2 U Compactness Definition Let (X, τ) be a topological space and A X. A collection of sets U is an open cover of A if U τ and A U. Definition A set A is compact if every open cover of A contains a finite subcover of A. Theorem Let (X, τ) be a topological space and A X. If X is compact and A is closed, then A is compact. Proof. Let U be any open cover of A. If A is closed, then X A is open. Therefore U {X A} is an open cover of X. Since X is compact, there exist U,..., U n U {X A} such that X U U n. If X A / {U,..., U n }, then {U,..., U n } is a finite subcover of A in U. Otherwire, {U,..., U n } {X A} is a finite subcover of A in U. Therefore A is compact.

12 Chapter 2. Topology 8 Theorem Let (X, τ) be a Hausdorff space and A X. If A is compact, then it is closed. Proof. Consider an arbitrary x X A. Since (X, τ) is Hausdorff, for any y A, there are U y, V y τ such that x U y, y V y and U y V y = φ. Now V = {V y : y A} is an open cover of A. Since A is compact, there are V y,..., V yn V such that A V y V yn. Let U = U y U yn. Then U is a neighbourhood of x. Also, (U A) U (V y V yn ) = φ. Therefore U X A. Since x is arbitrary, X A is open (Theorem 2.7), which proves that A is closed. The following theorem shows that a continuous function maps a compact set to a compact set. Theorem Let (X, τ) and (Y, τ) be topological spaces, f : X Y be a continuous function and A X. If A is compact, then f(a) is also compact. Proof. Let U be any open cover of f(a). Since f is continuous, f (U) is open for every U U. Therefore, U A = {f (U) : U U} is an open cover of A. Since A is compact, there are f (U ),..., f (U n ) U A such that A f (U ) f (U n ). However, this implies that f(a) U U n, which proves that f(a) is compact.. Urysohn s Lemma Urysohn s Lemma shows the existence of certain continuous functions on a normal topological space. Lemma Let (X, τ) be a normal topological space, U be an open subset of X, F be a closed subset of X and F U. Then there exists an open subset V of X such that F V V U Proof. Since U is open and F U, X U is closed, and F and X U are disjoint. Since (X, τ) is normal, there exist disjoint open sets V and W such that F V and X U W. Since V and W are disjoint, V X W. Since X W is closed, V X W. However, since X U W, X W U. Therefore V U. This proves that F V V U. Theorem 2.40 (Urysohn s Lemma). Let (X, τ) be a normal topological space. For every pair of disjoint closed subsets, F and G, of X, there exists a continuous function f : X [0, ] such that f(f ) = 0 and f(g) =. Proof. Let D be the set of all rational numbers of the form m 2 n, where m, n N and m < 2 n. We construct a function g : D τ in the following manner. Since G is closed and F and G are disjoint, X G is open and F X G. Therefore, from Lemma 2.39, there exists an open set, say U, such that 2 F U 2 U 2 X G

13 Chapter 2. Topology 9 Let g( ) = U 2. Similarly, there exist open sets, say U 2 4 F U 4 U 4 U 2 U 2 U 3 4 U 3 4 and U 3, such that 4 X G Let g( ) = U 4 and g( 3) = U And so on. 4 The function g constructed above has the following property. For any d, d 2 D such that d < d 2, F g(d ) g(d ) g(d 2 ) g(d 2 ) X G Now let a function f : X [0, ] be defined as follows. For any x X, { sup(d D; x / g(d)) x / d D f(x) = g(d) 0 x d D g(d) Here f(f ) = 0 and f(g) =. We will prove that f is continuous. Consider an arbitrary a (0, ). For any x X, if f(x) < a, then there exists a d D such that d < a and x g(d). Conversly, for any d D such that d < a, if x g(d), then f(x) < a. Therefore f ([0, a)) = d<a g(d), which implies that f ([0, a)) is open. For any x X, if f(x) > a, then there exists a d D such that d > a and x / g(d). Conversly, for any d D such that d > a, if x / g(d), then f(x) > a. Therefore f ((a, ]) = d>a (X g(d)), which is open. Since a is arbitrary and the set of intervals of the form [0, a) or (a, ] is a subbase of [0, ], f (V ) is open for any open subset V of [0, ]. Therefore f is continuous (Theorem 2.9). It can be observed that, in Urysohn s Lemma, the interval [0, ] can be replaced with any closed interval of real numbers.

14 CHAPTER 3 Topics in Metric Spaces In this chapter, we will look at a few concepts in metric space theory, which will be needed to discuss the application of Topology to metric spaces in the next chapter.. Metric Spaces We begin with the definition of a metric space. Definition 3.. Given a set X, a function d : X X R is a metric on X if, for every x, y, z X, () d(x, y) 0, (2) d(x, y) = 0 if and only if x = y, (3) d(x, y) = d(y, x), (4) d(x, y) d(x, z) + d(z, y). Example 3.2. The function d : R R R, defined as d(x, y) = x y for every x, y R, is a metric on R. This is the usual metric on R. Definition 3.3. A tuple (X, d) is a metric space if d is a metric on X. 2. Metrizable Spaces The theories of topology and metric spaces can be linked together with the concept of open balls. Definition 3.4. Let (X, d) be a metric space, x X and ɛ R. Then the set of all y X such that d(x, y) < ɛ is an open ball of radius ɛ centered at x. It is denoted as B d (x, ɛ). Theorem 3.5. Let B be the set of open balls in a metric space (X, d). Then B is a base of a topology on X. Proof. If B is the set of open balls in (X, d), then X = B. Now let B d (x, ɛ ), B d (x 2, ɛ 2 ) B. For any x B d (x, ɛ ) B d (x 2, ɛ 2 ), let ɛ = min(ɛ d(x, x ), ɛ 2 d(x, x 2 )). For any y B d (x, ɛ), d(y, x ) d(y, x) + d(x, x ) < ɛ + d(x, x ) ɛ d(x, x ) + d(x, x ) = ɛ 0

15 Chapter 3. Topics in Metric Spaces Therefore y B d (x, ɛ ). Similarly it can be proven that y B d (x 2, ɛ 2 ). Since x and y are arbitrary, B d (x, ɛ) = B d (x, ɛ ) B d (x 2, ɛ 2 ). Therefore B is a base of a topology on X (Theorem 2.9). Definition 3.6. A topological space (X, τ) is metrizable if τ has been derived from a metric space (X, d) using Theorem 3.5. From this point on, whenever we talk about a topological property of a metric space, such as convergence, continuity, density etc., we would mean the topological property of the corresponding metrizable space. Theorem 3.7. If U is an open set in a metric space (X, d), then, for every x U, there is an ɛ > 0 such that B d (x, ɛ) U. Proof. If U is open, then x U implies that x B d (x 0, δ) U for some x 0 U and δ > 0. Let ɛ = δ d(x, x 0 ). Therefore, if y B d (x, ɛ), then Therefore B d (x, ɛ) B d (x 0, δ) U. d(y, x 0 ) d(y, x) + d(x, x 0 ) < ɛ + d(x, x 0 ) = δ d(x, x 0 ) + d(x, x 0 ) = δ 3. Continuity in Metric Spaces Continuity, a topological property, was discussed in the previous chapter. Here, we provide a few additional results as they relate to metric spaces. Theorem 3.8. Given metric spaces (X, d ) and (X 2, d 2 ), a function f : X X 2, and x X, the following are equivalent: () f is continuous at x. (2) For every ɛ > 0, there exists a δ > 0 such that f(b d (x, δ)) B d2 (f(x), ɛ). Proof. () = (2) For any ɛ > 0, B d2 (f(x), ɛ) is open. Therefore it is a neighbourhood of f(x). Since f is continuous, there exists an open neighbourhood U of x in X such that f(u) B d2 (f(x), ɛ). Since U is open, there is a δ > 0 such that B d (x, δ) U and, therefore, f(b d (x, δ)) B d2 (f(x), ɛ). (2) = () Let V be an open neighbourhood of f(x). Therefore, there exists an ɛ > 0 such that B d2 (f(x), ɛ) V. There also exists a δ > 0 such that f(b d (x, δ)) B d2 (f(x), ɛ) and, therefore, f(b d (x, δ)) V. Since B d (x, δ) is open, it is a neighbourhood of x, which proves that f is continuous. Definition 3.9. Let d and d be metrics on a set X. Then d is equivalent to d if the identity map on X is continuous both from (X, d) to (X, d) and from (X, d) to (X, d).

16 Chapter 3. Topics in Metric Spaces 2 Theorem 3.0. Let d and d be metrics on a set X, and τ and τ be topologies derived from (X, d) and (X, d) respectively using Theorem 3.5. If d is equivalent to d, then τ = τ. Proof. Let U τ. Since d is equivalent to d, the identity map on X is continuous from (X, d) to (X, d). Therefore U τ (Theorem 2.9). This implies that τ τ. It can be proven in an identical manner that τ τ. 4. Sequences and Completeness Definition 3.. A net is a sequence if its domain is N. Theorem 3.2. If (x n ) is a sequence in a metric space (X, d) and x X, then the following are equivalent: () (x n ) converges to x. (2) For every ɛ > 0, (x n ) is eventually in B d (x, ɛ). Proof. () = (2) For every ɛ > 0, B d (x, ɛ) is open. Therefore it is a neighbourhood of x. Since x n converges to x, (x n ) is eventually in B d (x, ɛ). (2) = () Let N be a neighbourhood of x. Therefore, there is an open set U such that x U N. Since U is open, there is an ɛ > 0 such that B d (x, ɛ) U (Theorem 3.7). Since (x n ) is eventually in B d (x, ɛ), it is eventually in N. Theorem 3.3. Let (X, d ) and (Y, d 2 ) be metric spaces and x X. Then a function f : X Y is continuous at x if and only if the following statement is true: For every sequence (x n ) in X such that x n x, f(x n ) f(x). Proof. If f is continuous at x, then the statement is true (Theorem 2.28). Conversely, let the statement be true. Suppose f is not continuous at x. Therefore, there exists an ɛ > 0, such that, for every δ > 0, f(b d (x, δ)) B d2 (f(x), ɛ) (Theorem 3.8). Therefore, for every n N, there exists an x n B d (x, ) such that f(x n n) / B d2 (f(x), ɛ). For every δ > 0, the sequence (x n ) is eventually in B d (x, δ). However, f(x n ) is never in B d2 (f(x), ɛ). This is a contradiction. Definition 3.4. A sequence (x n ) in a metric space (X, d) is Cauchy if, for every ɛ > 0, there exists an N ɛ N such that, for every m, n N ɛ, d(x m, x n ) < ɛ. Definition 3.5. A metric space (X, d) is complete if every Cauchy sequence converges in it. Example 3.6. R with its usual metric (Example 3.2) is a complete metric space.

17 CHAPTER 4 Application of Topology to Metric Spaces In this chapter, we will apply the topological concepts discussed in Chapter 2 to metric spaces to derive some interesting results, namely Cantor s Intersection Theorem, Bourbaki s Mittag-Leffler Theorem and Baire s Theorem. We begin with a few essential definitions. Definition 4.. A metric space (X, d) is bounded if there exists an r R such that, for every x, y X, d(x, y) r. Definition 4.2. If (X, d) is a bounded metric space, then sup d(x, y) is its diameter. x,y X It is denoted as diam(x). Theorem 4.3 (Cantor s Intersection Theorem). Let (X, d) be a complete metric space and (F n ) be a sequence of nonempty closed subsets of X such that diam(f n ) 0 and, for every n N, F n+ F n. Then there is one and only one element in F n. Proof. For every n N, let x n F n. Let ɛ be a positive real number. Since diam(f n ) 0, there exists an N ɛ N such that, for every n N ɛ, diam(f n ) B d (0, ɛ) (Theorem 3.2). Therefore diam(f Nɛ ) < ɛ. Now we know that, for every n N, F n+ F n. Therefore, for every m, n N ɛ, x m, x n F Nɛ. This implies that d(x m, x n ) < ɛ. Since ɛ is arbitrary, this proves that (x n ) is Cauchy. Since (X, d) is complete, there exists an x X such that x n x. For any n N, let (x nk ) be the subsequence of (x n ) such that n k = n + k. Since x n x, x nk x (Theorem 2.27). Moreover, (x nk ) F n. Since F n is closed, x F n (Theorem 2.29). Since n is arbitrary, x F n. Let x F n such that x x. Therefore d(x, x) > 0. Since diam(f n ) 0, there exists an n N such that diam(f n ) < d(x, x) (Theorem 3.2). However, since x, x F n, d(x, x) diam(f n ). This is a contradiction. Therefore x is the only element in F n. The following lemma is needed to prove Bourbaki s Mittag-Leffler Theorem. Lemma 4.4. Let (X, d ) and (X 2, d 2 ) be metric spaces, (X, d ) be complete, and f : X X 2 be a continuous function. Then a metric d can be defined on X as follows: For any x, y X, d (x, y) = d (x, y) + d 2 (f(x), f(y)). Moreover, () d is equivalent to d, and (2) (X, d ) is complete. 3

18 Chapter 4. Application of Topology to Metric Spaces 4 Proof. First, we will prove that d is a metric on X i.e. it has each of the four properties of Definition 3.. Let x, y, z X. () It is obvious that d (x, y) 0. (2) If d (x, y) = 0, then d (x, y) = 0, which implies that x = y. Conversely, if x = y, then d (x, y) = 0. (3) It is obvious that d (x, y) = d (y, x). (4) The fourth property can be shown as follows: d (x, y) = d (x, y) + d 2 (f(x), f(y)) d (x, z) + d (z, y) + d 2 (f(x), f(z)) + d 2 (f(z), f(y)) = (d (x, z) + d 2 (f(x), f(z))) + (d (z, y) + d 2 (f(z), f(y))) = d (x, z) + d (z, y) Now we will prove that d is equivalent to d. For any x, y X, d (x, y) d (x, y). Therefore, for any x X and ɛ > 0, B d (x, ɛ) B d (x, ɛ). Therefore, the identity map on X is continuous from (X, d ) to (X, d ) (Theorem 3.8). Let ɛ be an arbitrary positive real number. Since f is continuous, there exists a δ x for every x X such that f(b d (x, δ x )) B d2 (f(x), ɛ ) (Theorem 3.8). Let δ = min(δ 2 x, ɛ ). 2 If y B d (x, δ), then y B d (x, δ x ), which implies that f(y) B d2 (f(x), ɛ ). Therefore 2 d (x, y) = d (x, y) + d 2 (f(x), f(y)) < δ + ɛ 2 ɛ 2 + ɛ 2 = ɛ Therefore B d (x, δ) B d (x, ɛ). Therefore the identity map on X is continuous from (X, d ) to (X, d ) (Theorem 3.8). Finally, we will prove that (X, d ) is complete. Let (x n ) be an arbitrary Cauchy sequence in (X, d ). Since, for every x, y X, d (x, y) d (x, y), (x n ) is Cauchy in (X, d ). Since (X, d ) is complete, (x n ) is convergent in it. Since d is equivalent to d, (x n ) is convergent in (X, d ) (Theorem 3.0), which proves the completeness of (X, d ). Theorem 4.5 (Bourbaki s Mittag-Leffler Theorem). Let (X n, d n ) be a sequence of metric spaces and (f n : X n+ X n ) a sequence of functions with the following properties for every n N: () (X n, d n ) is complete. (2) f n is continuous. (3) f n (X n+ ) is dense in X n.

19 Chapter 4. Application of Topology to Metric Spaces 5 Then, is dense in X. D = n N(f f n )(X n+ ) Proof. We begin by using Lemma 4.4 to inductively replace, for every n N, the metric d n with the metric d n defined as follows: d is just d and, for any x, y X n+, d n+ (x, y) = d n+ (x, y) + d n (f n (x), f n (y)) The lemma assures that, for every n N, (X n, d n ) is still complete, and, since d n is equivalent to d n, f n is still continuous and f n (X n+ ) still dense in X n. But most importantly, for every x, y X n+, d n (f n (x), f n (y)) d n+ (x, y) This property is the reason why we replaced the metrics. We will need it later in the proof. Now, if U 0 is an arbitrary non-empty open set in X, then we have to show that U 0 D φ. This can be done by constructing a sequence of open sets in the following manner. Since f (X 2 ) is dense in X, U 0 f (X 2 ) φ. Therefore, there exists an x 2 X 2 such that f (x 2 ) U 0. Since U 0 is open, it is a neighbourhood of f (x 2 ), and, therefore, there exists an ɛ > 0 such that B d (f (x 2 ), ɛ) U 0 (Theorem 3.7). Since f is continuous, there exists a δ > 0 such that f (B d2 (x 2, δ)) B d (f (x 2 ), ɛ ) and δ (Theorem 3.8). 2 In this case, f (B d2 (x 2, δ)) B d (f (x 2 ), ɛ ) B (f 2 d (x 2 ), ɛ) U 0 (Theorem 2.7). Let U = B d2 (x 2, δ). Therefore, diam(u ) and f (U ) U 0. Since U is an open set in X 2, we can similarly obtain an open set U 2 in X 3 such that diam(u 2 ) 2 and f 2(U 2 ) U. Thus, we can obtain a sequence (U n ) of open sets such that diam(u n ) n and f n(u n ) U n. For any m 0, let (Y m n ) be a sequence of sets in U m such that This sequence has the following properties: Y m n = (f m+ f m+n )(U m+n ) () For every n N, Yn m is closed. (2) For every n N, Yn+ m Yn m (Theorem 2.7). (3) We know that, for every n N and every x, y X n+, dn (f n (x), f n (y)) d n+ (x, y). This implies that, for every n N, diam(yn m ). Therefore m+n diam(yn m ) 0. Therefore, there exits a unique y m n N Y m n (Theorem 4.3). For every m 0, f m+ is continuous. Therefore f m+ (Yn m+ ) Yn+ m (Theorem 2.9). Since, for every n N, y m+ Yn m+, f m+ (y m+ ) Yn+. m Moreover y m+ U m+, which

20 Chapter 4. Application of Topology to Metric Spaces 6 implies that f m+ (y m+ ) f m+ (U m+ ). Therefore f m+ (y m+ ) f m+ (U m+ ) n N Y m n+ = Y m Yn+ m n N = n N Y m n This shows that, for every m 0, y m = f m+ (y m+ ). Therefore, there exists a y 0 U 0 such that, for every n N, y 0 (f f n )(X n+ ). Therefore y 0 U 0 D. We need another definition and a couple of lemmas before we can prove Baire s Theorem. Definition 4.6. Let (X, d) be a metric space, x X and A X. Then, inf d(x, y) y A is the distance of x to A. It is denoted as dist(x, A). Lemma 4.7. Let (X, d) be a metric space, A X and (x n ) a sequence in X. x n x, then dist(x n, A) dist(x, A). If Proof. For every n N and a A, Similarly it can be proven that Therefore d(x n, a) d(x n, x) + d(x, a) which proves that dist(x n, A) dist(x, A). = inf d(x n, y) d(x n, x) + d(x, a) y A = inf d(x n, y) d(x n, x) d(x, a) y A = inf d(x n, y) d(x n, x) inf d(x, y) y A y A = inf d(x n, y) inf d(x, y) d(x n, x) y A y A = dist(x n, A) dist(x, A) d(x n, x) dist(x, A) dist(x n, A) d(x n, x) dist(x n, A) dist(x, A) d(x n, x) Lemma 4.8. Let (X, d) be a complete metric space, U be an open subset of X, and d U : U U R be defined as follows: For x, y U, d U (x, y) = d(x, y) + dist(x, X U) dist(y, X U) Then, () d U is a metric on U,

21 Chapter 4. Application of Topology to Metric Spaces 7 (2) d U is equivalent to d, and (3) (U, d U ) is complete. Proof. First we will prove that d U is well defined i.e. dist(x, X U) 0 for every x U. If dist(x, X U) = 0 for some x U, then, for every ɛ > 0, there is a y X U such that d(x, y) < ɛ. However, since U is open, there is an ɛ such that B d (x, ɛ) U (Theorem 3.7). This is a contradiction. Now we will prove that d U has each of the four properties of Definition 3.. Let x, y, z U. () It is obvious that d U (x, y) 0. (2) If d U (x, y) = 0, then d(x, y) = 0, which implies that x = y. Conversely, if x = y, then d U (x, y) = 0. (3) It is obvious that d U (x, y) = d U (y, x). (4) The fourth property can be shown as follows: d U (x, y) = d(x, y) + dist(x, X U) dist(y, X U) d(x, z) + d(z, y) + dist(x, X U) dist(z, X U) + dist(z, X U) dist(y, X U) = d(x, z) + dist(x, X U) dist(z, X U) + d(z, y) + dist(z, X U) dist(y, X U) = d U (x, z) + d U (z, y) Therefore d U is a metric on U. Now we will prove that d U is equivalent to d. For any x, y U, d(x, y) d U (x, y). Therefore, for any x U and ɛ > 0, B du (x, ɛ) B d (x, ɛ). Therefore, the identity map on U is continuous from (U, d U ) to (U, d) (Theorem 3.8). For any x U, let (x n ) be a sequence in U such that x n x in (U, d). For any ɛ > 0, there exists an N N such that, for every n N, d(x n, x) < ɛ (Theorem 2 3.2). Also, from Lemma 4.7, dist(x n, X U) dist(x, X U), which implies that. Therefore, there exists an N dist(x n,x U) dist(x,x U) 2 N such that, for every n N 2, < ɛ (Theorem 3.2). Let N = max(n 2, N 2 ). For every n N, dist(x n,x U) dist(x,x U) d U (x n, x) = d(x n, x) + dist(x n, X U) dist(x, X U) < ɛ 2 + ɛ 2 = ɛ

22 Chapter 4. Application of Topology to Metric Spaces 8 Therefore x n x in (U, d U ) also (Theorem 3.2). This proves that the identity map on U is continuous from (U, d) to (U, d U ) (Theorem 3.3). Now we will prove that (U, d U ) is complete. Let (x n ) be a Cauchy sequence in (U, d U ). Since d(x, y) d U (x, y) for every x, y U, (x n ) is Cauchy in (X, d) and, therefore, convergent. Let x n x. First we will prove that x U. Suppose x X U. This implies that dist(x n, X U) 0 (Lemma 4.7). Since (x n ) is Cauchy in (U, d U ) and, for every m, n N, dist(x m,x U) dist(x n,x U) d U (x m, x n ), ( ) ( ) is a Cauchy sequence in R. Therefore, is convergent in R. dist(x n,x U) dist(x n,x U) This, however, is is a contradiction since dist(x n, X U) 0. Therefore x U. Since d U is equivalent to d, x n x in (U, d U ) also, which proves that (U, d U ) is complete. Theorem 4.9 (Baire s Theorem). Let (X, d) be a complete metric space and (U n ) a sequence of dense open subsets of X. Then, n= U n is also dense in X. Proof. For any n N, let V n = U U n. The sequence (V n ) has the following properties: () For every n N, V n+ V n. (2) n= V n = n= U n. Let f : V X be defined in the following manner: For any x V, f (x) = x. For n >, let f n : V n V n be defined in a similar manner. Therefore, for every n N, V n = (f f n )(V n ), which implies U n = (f f n )(V n ) n= n= V n = n N For any n N, let d Vn be the metric on V n as defined in Lemma 4.8. Then the sequence of metric spaces (V n, d Vn ) and the sequence of functions (f n ) have the following properties: () For every n N, (V n, d Vn ) is complete (Lemma 4.8). (2) The function f is continuous from (V, d) to (X, d). Since d V is equivalent to d, f is also continuous from (V, d V ) to (X, d) (Theorem 3.0). Similarly f n is continuous from (V n, d Vn ) to (V n, d Vn ) for every n >. (3) Since U n is dense in (X, d), V n is also dense in (X, d) for every n N (Theorem 2.2). Therefore f(v ) is dense in (X, d). For n >, V n is dense in (V n, d) (Theorem 2.5). Since d Vn is equivalent to d, V n is dense in (V n, d Vn ) (Theorem 3.0), and, therefore, f n (V n ) is dense in (V n, d Vn ). Therefore n N (f f n )(V n ) is dense in (X, d) (Theorem 4.5). Baire s Theorem is very popular in functional analysis. For instance, it can be used to prove the Open Mapping Theorem and the Uniform Boundedness Principle.

23 Bibliography [] Stephen Willard, General Topology, Dover Publications (2004) [2] Volker Runde, A Taste of Topology, Springer (2005) 9