Math General Topology Fall 2012 Homework 11 Solutions

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1 Math General Topology Fall 2012 Homework 11 Solutions Problem 1. Let X be a topological space. a. Show that the following properties of a subset A X are equivalent. 1. The closure of A in X has empty interior: int(a) =. 2. For all non-empty open subset U X, there is a non-empty open subset V U satisfying V A =. A subset A X satisfying these equivalent properties is called nowhere dense in X. Solution. interior: Recall that a subset B X is dense if and only if its complement has empty B = X B c = = int(b c ). Now consider the following equivalent conditions. A has empty interior. A c is dense. But note A c = int(a c ). For all non-empty open subset U X, we have U int(a c ). For all non-empty open subset U X, there is a point x U A c and an open neighborhood W of x satisfying W A c, in other words W A =. (Taking V = U W ) For all non-empty open subset U X, there is a non-empty open subset V U satisfying V A =. b. Show that the following properties of the space X are equivalent. 1. Any countable intersection of open dense subsets is dense. In other words, if each U n X is open and dense in X, then U n is dense in X. 2. Any countable union of closed subsets with empty interior has empty interior. In other words, if each C n X is closed in X and satisfies int(c n ) =, then their union satisfies int ( C n) =. A space X satisfying these equivalent properties is called a Baire space. Solution. Consider the following equivalent conditions. If each U n X is open and dense in X, then U n is dense in X. If each U c n X is closed and has empty interior in X, then ( U n) c has empty interior in X. (Taking C n := U c n) If each C n X is closed and has empty interior in X, then C n has empty interior in X. 1

2 Definition. Let X be a topological space. A function f : X R is lower semicontinuous if for all a R, the preimage f 1 (a, + ) is open in X. Equivalently: For all x 0 X and ɛ > 0, there is a neighborhood U of x 0 satisfying f(x) > f(x 0 ) ɛ for all x U. This means that the values close to x 0 can suddenly jump up but not down. Problem 2. a. Let X be a topological space and f : X R a continuous real-valued function. Show that for every non-empty open subset U X, there is a non-empty open subset V U on which f is bounded. Solution. Pick a point x U. Since f is continuous at x, there is an open neighborhood W of x satisfying f(w ) (f(x) 1, f(x) + 1), in particular f is bounded on W. Now the subset V := W U is non-empty (since x V ), open, and f is bounded on V. b. (Willard Exercise 25C) Let X be a Baire space and f : X R a lower semicontinuous function. Show that for every non-empty open subset U X, there is a non-empty open subset V U on which f is bounded above. Solution. Note that for all a R, the preimage f 1 (, a] = (f 1 (a, + )) c is closed in X. Express X as the countable union of closed subsets, and likewise X = f 1 (R) ( ) f 1 (, n] f 1 (, n] =: U = A n (A n U). Since U is open (and non-empty) and X is Baire, U cannot be meager, so that for some m N, A m U is not nowhere dense. Let W X be a non-empty open subset satisfying W A m U A m U = A m U. Since W is open and satisfies W U, it also satisfies W U. This subset V := W U is non-empty, open, and contained in A m so that f is bounded above on V (by the upper bound m). 2

3 Problem 3. Show that a topological space X is of second category in itself if and only if any countable intersection of open dense subsets of X is non-empty. Solution. Consider the following equivalent conditions. X is of second category in itself, i.e. for any countable collection of nowhere dense subsets A n X, we have A n X. For any countable collection of closed nowhere dense subsets C n X, we have C n X. (This implies the previous condition since A being nowhere dense implies A being nowhere dense.) (Taking U n = Cn) c For any countable collection of open dense subsets U n X, we have U n. 3

4 Problem 4. (Uniform boundedness principle) (Willard Exercise 25D.5) (Munkres Exercise 48.10) (Bredon I.17.2) Let X be a Baire space and S C(X, R) a collection of real-valued continuous functions on X which is pointwise bounded: for each x X, there is a bound M x R satisfying f(x) M x for all f S. Show that there is a non-empty open subset U X on which the collection S is uniformly bounded: there is a bound M R satisfying f(x) M for all x U and all f S. Solution. For all n N, consider the subset of X C n = {x X f(x) n for all f S} = f S{x X f(x) n} = f S f 1 [ n, n] which is closed in X since each f S is continuous. Pointwise boundedness of the collection S yields x C n whenever n M x, or equivalently X = C n. Since X is Baire, it is in particular of second category in itself, so that for some m N, C m is not nowhere dense. Let U X be a non-empty open subset satisfying U C m = C m. Then the bound f(x) m holds for all x U and all f S. 4

5 Definition. Let X and Y be normed real vector spaces. A linear map T : X Y is bounded if there exists a constant C R satisfying for all x X. T x C x By linearity, this condition is equivalent to the following number being finite: T := T x sup x X\{0} x = sup T x x =1 = sup T x. x 1 The number T R { } is called the operator norm of T. Let L(X, Y ) := {T : X Y T is linear and T < } denote the vector space of bounded linear maps from X to Y. It is a vector space under pointwise addition and scalar multiplication. One readily checks that the assignment T T is indeed a norm on L(X, Y ). 5

6 Problem 5. Let T : X Y be a linear map between normed real vector spaces. Show that the following are equivalent. 1. T is continuous (everywhere). 2. T is continuous at some point x 0 X. 3. T is continuous at 0 X. 4. T is bounded. Solution. (1 2) X is non-empty since it contains 0 X. (2 3) Let ɛ > 0. By continuity of T at x 0, there is a δ > 0 satisfying T B δ (x 0 ) B ɛ (T x 0 ). For any x B δ (0), we have so that T is continuous at 0. d(t x, T (0)) = T x 0 = T x = T (x 0 + x x 0 ) = T (x 0 + x) T x 0 = d(t (x 0 + x), T x 0 ) < ɛ (3 4) Taking ɛ = 1, since T is continuous at 0, there is a δ > 0 satisfying T B δ (0) B 1 (T (0)) = B 1 (0). Thus for any x with x < 1, we have ( ) δ T x = T δ x = 1 δ T (δx) = 1 T (δx) δ < 1 δ (1) = 1 δ and linearity of T implies T x 1 δ whenever x 1. Therefore T is bounded: T = sup T x 1 x 1 δ. 6

7 (4 1) If T has bound C, then T is Lipschitz continuous with Lipschitz constant C, hence continuous. For all x, x X, we have d(t x, T x ) = T x T x = T (x x ) C x x = Cd(x, x ). 7

8 Problem 6. Consider the Banach space l = {x R N x < } with the supremum norm x = sup i N x i. Consider the linear subspace of lists that are eventually zero: X := {x l N N such that x i = 0 for all i > N} l. Consider the continuous linear maps T n : X R defined by T n (x) = nx n. a. Show that the collection {T n } n N is pointwise bounded but not uniformly bounded. Solution. Pointwise bounded. Let x X and let N N be large enough so that x i = 0 for all i > N. Then for all n > N, we have T n x = nx n = 0 and therefore sup T n x = max T nx <. n N 1 n N Not uniformly bounded. Consider the standard basis vectors e k X whose coordinates are { e k 1 if i = k i = 0 if i k and note that these are unit vectors: e k = 1. The equality T n (e n ) = n(e n n) = n(1) = n implies T n = T n x sup x X\{0} x T ne n e n = n 1 = n. It follows that the collection {T n } n N is not uniformly bounded: sup T n =. n N 8

9 b. Part (a) implies that X cannot be complete. Show explicitly that X is not complete by exhibiting a Cauchy sequence in X that does not converge in X. Solution. Let us denote the sequence index as a superscript. Consider the sequence (x (n) ) n N in X consisting of the following vectors: x (n) i = { 1 i if i n 0 if i > n. Note that each vector x (n) is eventually zero, hence a legitimate element of X. The sequence is Cauchy. For any N N and m, n N (with m n), the distance d(x (m), x (n) ) = x (m) x (n) 1 = max{ m + 1, 1 m + 2,..., 1 n } = 1 m + 1 converges to 0 as N. < 1 N The sequence does not converge in X. Let x X and let N N be large enough so that x i = 0 for all i > N. Then for all n > N, the distance d(x (n), x) = x (n) x = sup x (n) i x i i N sup x (n) i x i i>n = sup x (n) i i>n = x (n) N+1 = 1 N + 1 is bounded away from 0. Therefore the sequence (x (n) ) n N does not converge to x X. 9