Folland: Real Analysis, Chapter 4 Sébastien Picard

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1 Folland: Real Analysis, Chapter 4 Sébastien Picard Problem 4.19 If {X α } is a family of topological spaces, X α X α (with the product topology) is uniquely determined up to homeomorphism by the following property: There exist continuous maps π α : X X α such that if Y is any topological space and f α C(Y,X α ) for each α, there is a unique F C(Y,X) such that f α π α F. First, we show that if X α A X α with the product topology, then the coordinate maps π α : X X α are such that if Y is any topological space and f α C(Y,X α ) for each α, there is a unique F C(Y,X) such that f α π α F. Indeed, define F : Y X in the following way: for any y Y, let F(y) α A X α be the map F(y) : A α A X α such that (F(y)) α f α (y) for each α A. It then follows from the definition of the coordinate maps that for any y Y, π α F(y) (F(y)) α f α (y). Since the f α are continuous, π α F is continuous for each α, hence F is continuous by Proposition For uniqueness, let F C(Y,X) be such that f α π α F. Suppose there exists y Y such that F (y) F(y). Then there is an α A such that (F (y)) α (F(y)) α. But then f α (y) π α F (y) (F (y)) α (F(y)) α π α F(y) f α (y). This contradiction shows that F F. Next, we show that up to homeomorphism, the product topology is the only space with this property. Let X be a topological space such that there exists continuous maps π α : X X α such that if Y is any topological space and f α C(Y,X α ) for each α, there is a unique F C(Y,X ) such that f α π α F. Let G : X X be the continuous function such that π α π α G and G : X X be the continuous function such that π α π α G. To complete the proof, we will show that G G G G id. We have π α (G G) (π α G ) G π α G π α, hence π α π α id π α (G G). Since π α C(X,X α ), by uniqueness we must have id G G. Similarly, π α (G G ) (π α G) G π α G π α id, and by uniquenes we must have id G G. 1

2 Problem 4.22 Let X be a topological space, (Y,ρ) a complete metric space, and {f n } a sequence in Y X such that sup x X ρ(f n (x),f m (x)) 0as m,n. Thereis a unique f Y X suchthat sup x X ρ(f n (x),f(x)) 0 as n. If each f n is continuous, so is f. For each x X, we have ρ(f n (x),f m (x)) sup z X ρ(f n (z),f m (z)) 0 as m,n. Since (Y,ρ) is complete, {f n (x)} is a Cauchy sequence, hence converges to a point in Y which we may call f(x). Therefore {f n } converges pointwise to a function f Y X. For any ǫ > 0, there exists a positive integer N such that when m,n > N we have ρ(f n (x),f m (x)) ǫ for all x X. If we fix n and let m, we obtain ρ(f n (x),f(x)) ǫ. Since this holds for all x X, we have sup x X ρ(f n (x),f(x)) 0 as n. Suppose there exists another function g Y X such that sup x X ρ(f n (x),g(x)) 0 as n. If for some x X, f(x) g(x), then lim f n(x) f(x) g(x) lim f n (x), n n which contradicts uniqueness of the limit of a sequence in a metric space. Suppose each f n is continuous. Let ǫ > 0 and let x X. Choose a positive integer N such that ρ(f N (z),f(z)) < ǫ/3 for all z X. Choose δ > 0 such that ρ(f N (x),f N (y)) < ǫ/3 for all y X such that ρ(x,y) < δ. Then we have ρ(f(x),f(y)) ρ(f(x),f N (x))+ρ(f N (x),f N (y))+ρ(f N (y),f(y)) ǫ/3+ǫ/3+ǫ/3 ǫ, for all y Y such that ρ(x,y) < δ. This proves continuity of f. Problem 4.28 Let X be a topological space equipped with an equivalence relation, X the set of equivalence classes, π : X X the map taking each x X to its equivalence class, and T {U X : π 1 (U) is open in X}. a. T is a topology on X. (It is called the quotient topology.) b. If Y is a topological space, f : X Y is continuous iff f π is continuous. c. X is T1 iff every equivalence class is closed. a. Since π 1 ( ), then T, and since π is a surjection, π 1 ( X) X hence X T. Next, suppose {U α } α J T. Then π 1 ( α J U α ) α Jπ 1 (U α ), hence α J U α T. Similarly, if U i T for i {1,...,n}, then n n π 1 ( U i ) π 1 (U i ). i1 i1 2

3 b. The quotient topology is defined such that π is continuous. Suppose f : X Y is continuous. Then f π is the composition of continuous functions, hence is continuous. Conversely, suppose f π is continuous. Then for all open sets V Y, π 1 f 1 (V) is open in X, so f 1 (V) is in T. Therefore f is continuous. c. Suppose X is T 1. Then {x} is closed for every x X (Proposition 4.7). By continuity, π 1 ({x}) is closed. Therefore equivalence classes are closed. Conversely, suppose every equivalence class is closed. Then π 1 ({x}) is closed for every x X. Then (π 1 ({x})) c π 1 ({x} c ) is open, hence {x} c T, so {x} is closed in X. Therefore X is T 1 (Proposition 4.7). Problem 4.32 A topological space X is Hausdorff iff every net in X converges to at most one point. Suppose X is Hausdorff and there is a net x α α A that converges to two points x,y. There exists disjoint open sets U,V such that x U,y V. But then there exists α 1 A such that x α U for all α α 1 and α 2 A such that x α V for all α α 2. There exists a γ A such that γ α 1 and γ α 2, so α γ U V. This contradiction shows that every net in X converges to at most one point. Conversely, suppose X is not Hausdorff. Then there exists two points x, y X such that every pair of open sets U,V such that x U,y V has a non-empty intersection. Consider the directed set N x N y, where N x,n y are the families of open neighbourhoods of x,y and (U 1,V 1 ) (U 2,V 2 ) iff U 1 U 2 and V 1 V 2. Define the net x α α Nx N y by mapping (U,V) N x N y to a point z U V. For any open set Ũ containing x, let Ṽ be an arbitrary open set containing y. It follows that for every (U,V) (Ũ,Ṽ) we have x (U,V) Ũ. Hence x α α Nx N y converges to x. Similarly, for any open set Ṽ containing y, let Ũ be an arbitrary open set containing x. It follows that for every (U,V) (Ũ,Ṽ) we have x (U,V ) Ṽ. Hence x α α Nx N y converges to y. Therefore there exists a net in X which converges to more than one point. Problem 4.34 If X has the weak topology generated by a family F of functions, then x α converges to x X iff f(x α ) converges to f(x) for all f F. Suppose x α converges to x X. Since every f F is continuous, f(x α ) converges to f(x) for all f F by Proposition Conversely, suppose f(x α ) converges to f(x) for all f F. Let U be an open set such that x U. By the definition weak topology generated by F, there exists an open set U U such that x U and U n i1 f 1 i (V i ) 3

4 where f i F and V i are open sets in the target space of f i. Then for each integer i between 1 and n, there exists an α i such that f i (x α ) V i for all α α i. Take γ α i for all integers i between 1 and n. Then when α γ, we have x α f 1 i (V i ) for all integers i between 1 and n. Hence x α U U for all α γ. Problem 4.52 The one-point compactification of R n is homeomorphic to the n-sphere S n {x R n+1 : x 1}. Denote N (0,...,0,1) S n R n+1. Then the stereographic projection σ : S n \{N} R n defined by σ(x 1,...,x n+1 ) (x1,...,x n ) 1 x n+1 is a homeomorphism. It is easy to see that the mapping is continuous. Its inverse is given by the continous map σ 1 (u 1,...,u n ) (2u1,...,2u n, u 2 1). u 2 +1 Indeed, σ σ 1 id: ( ) (2u σ σ 1 (u 1,...,u n 1,...,2u n, u 2 1) ) σ u 2 +1 (2u 1,...,2u n ) ( u 2 +1)(1 ( u 2 1)( u 2 +1) 1 ) u 2 +1 ( u 2 +1)( u 2 +1 ( u 2 1) (2u1,...,2u n ) Also, σ 1 σ id: ( ) (x σ 1 σ(x 1,...,x n+1 ) σ 1 1,...,x n ) ( 1+ 1 x n+1 n i1 (u 1,...,u n ) (x i ) 2 ) 1 ( 2x 1 2x n n (x i ) 2 1 x n+1,..., 1 x n+1, (1 x n+1 ) 1) 2 + 2x1 2x n n (x i ) 2 n i1 (xi ) 2( 1 x n+1,..., 1 x n+1, ) i1 2x 1 2x n 1 (xn+1 ) 2 +1 (x n+1 ) 2( 1 x n+1,..., 1 x n+1, ) (1 xn+1 ) 2 2 2x ( 2x 1 2x n 2xn+1 2(x n+1 ) 2 n+1 1 x n+1,..., 1 x n+1, ) (x 1,...,x n+1 ) Hence we have shown that R n is homeomorphic to S n \{N}. It follows that their one-point compactifications are homeomorphic. It is clear from the definition that the one-point compactification of S n \{N} is S n, hence the one-point compactification of R n is homeomorphic to S n. 4 i1

5 Problem 4.60 The product of countably many sequentially compact spaces is sequentially compact. Let X i N X i, and let {x j } j1 be a sequence in X. Then there exists a subsequence {x n j } j1 of {x j } such that {π 1 (x nj ))} j1 converges in X 1. Denote it by {x 1 j} j1. Proceeding inductively, for k N we obtain a subsequence {x k j} j1 of {xj k 1 } j1 such that {π k (x k j)} j1 converges is X k. Let y k x k k. Then {y k} k1 is a subsequence of {x j} j1 such that {π i (y k )} k1 converges in X i for all positive integers i. By Problem 4.34 (proved above), we can conclude that {y k } k1 converges in X. Problem 4.64 Let (X,ρ) be a metric space. A function f C(X) is called Holder continuous of exponent α (α > 0) if the quantity N α (f) sup x y ρ(x,y) α is finite. If X is compact, {f C(X) : f u 1 and N α (F) 1} is compact in C(X). Denote F {f C(X) : f u 1 and N α (F) 1}. It is clear from the definition that F is pointwise bounded. Let ǫ > 0,x X, and δ ǫ 1/α. Then for all y X such that ρ(x,y) < δ, we have ρ(x,y) α < ǫ for all f F. Hence F is equicontinuous. By Arzela-Ascoli, the closure of F in C(X) is compact. We show that F is equal to its closure in C(X) to complete the proof. Suppose {f n } F and f n f u 0. We will show that f F. For all ǫ > 0, there exists an n N such that f u f f n u + f n u 1+ǫ. Since this holds for all ǫ > 0, we can conclude that f u 1. Next, fix x,y X. Let ǫ > 0, and choose n N such that f n f u < ǫ/2. Then ρ(x,y) α f(x) f n(x) + f n (x) f n (y) + f n (y) f(y) ρ(x,y) α ǫ ρ(x,y) + f n(x) f n (y) α ρ(x,y) α ǫ 1+ ρ(x,y) α 5

6 Since this holds for all ǫ > 0, we can conclude that ρ(x,y) α 1. This is true for all x,y X, therefore it follows that Hence f F, and therefore F is closed in C(X). sup 1. x y ρ(x,y) α 6