Real Analysis Chapter 4 Solutions Jonathan Conder

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1 2. Let x, y X and suppose that x y. Then {x} c is open in the cofinite topology and contains y but not x. The cofinite topology on X is therefore T 1. Since X is infinite it contains two distinct points x and y. Suppose there exist disjoint open sets A and B (in the cofinite topology) such that x A and y B. Then A B c, which is finite so A is also finite. This is a contradiction because A c is finite but X = A A c is infinite. Hence, the cofinite topology on X is not T 2. Suppose that X is countable, and let x X. Choose a surjection q : N {x} c, and for each n N set A n := {q(k)} n k=1 and B n := A c n, so that x B n and B n is open. If U is open and x U then U c {x} c is finite and hence U c A N for some sufficiently large N N. This implies that B N U, so {B n } is a neighbourhood base at x. The cofinite topology on X is therefore first countable. Conversely, suppose that the cofinite topology on X is first countable. Choose x X and a countable neighbourhood base {A n } at x. If y {x}c then {y} c is an open neighbourhood of x, so there exists n N such that A n {y} c, and hence y A c n. It follows that {x} c Ac n, which is countable because each A c n is finite. Therefore X = {x} {x} c is also countable. 3. Let (X, ρ) be a metric space with closed subspaces A and B. Given x X and ε (, ), let y B ε (x). If a A then ρ(x, A) ρ(x, a) ρ(x, y) + ρ(y, a) < ρ(y, a) + ε, so ρ(x, A) ε ρ(y, A). Similarly ρ(y, A) ε ρ(x, A), so ε ρ(y, A) ρ(x, A) ε. This shows that x ρ(x, A) is continuous. Similarly x ρ(x, B) is continuous. It follows that x ρ(x, A) ρ(x, B) is continuous, so the preimages of (, ) and (, ) under this map are open. These preimages are {x X ρ(x, A) < ρ(x, B)} and {x X ρ(x, A) > ρ(x, B)}, which are clearly disjoint. The first set contains A and the second contains B because ρ(x, A) = iff x A and ρ(x, B) = iff x B. Indeed, if x X and ρ(x, A) = then every neighbourhood of x meets A, because no r (, ) is a lower bound for {ρ(x, a)} a A. Finally, (X, ρ) is T 1 because distinct points are separated by positive distance. Therefore (X, ρ) is normal. 4. Firstly, R T because = ( Q) and R = R ( Q). If {W α } α A T and W α has decomposition U α (V α Q) for each α A, then α A W α = ( α A U α ) (( α A V α ) Q) T. Moreover, if A = {1, 2} then W 1 W 2 = (U 1 (V 1 Q)) (U 2 (V 2 Q)) = (U 1 U 2 ) (U 1 (V 2 Q)) ((V 1 Q) U 2 ) ((V 1 Q) (V 2 Q)) = (U 1 U 2 ) (((U 1 V 2 ) (V 1 U 2 ) (V 1 V 2 )) Q) T. By induction it follows that T is closed under finite intersections, so T is a topology. Let x, y R be distinct. There exist disjoint open intervals U, V R such that x U and y V. Since U = U ( Q) and V = V ( Q), both these intervals are open in T. Therefore T is Hausdorff. Note that Q c is closed in T. Let W 1, W 2 T be neighbourhoods of and Q c respectively. Since W 1, there exists n N such that ( 1 n, 1 n ) Q W 1. If W 2 has decomposition U 2 (V 2 Q), then Q c U 2 because (V 2 Q) Q c =. In particular U 2 ( 1 n, 1 n ), because it contains, say, 2 2n. This implies that U 2 ( 1 n, 1 n ) Q, since Q is dense in R with respect to the usual topology. Therefore W 2 W 1, so T is not regular. 5. Let X be a separable metric space with a countable dense subspace A. Then B := {B 1/n (a) a A, n N} is countable because N N is countable. Clearly each member of B is open in X. Conversely, if U X is open and x U, then x B 1/n (x) U for some n N. If x A then B 1/n (x) B. Otherwise x acc(a) because 1

2 X = A = A acc(a). It follows that B 1/2n (x) contains some point a A, in which case x B 1/2n (a) B. By the triangle inequality B 1/2n (a) B 1/n (x) U. This shows that U is the union of a (possibly empty) subcollection of B. Therefore B is a base for the topology on X, so this topology is second countable. 6. (a) Clearly R = ( n, n]. Let a, b, c, d R and suppose that a < b and c < d. Then (a, b] (c, d] = or max{a, c} < min{b, d}, in which case (a, b] (c, d] = (max{a, c}, min{b, d}]. This shows that E is a base for a topology T on R. Clearly each member of E is open in T. If a, b R and a < b then is open in T, so its complement (a, b] is closed in T. ( (a n, a n + 1]) ( (b + n 1, b + n]) (b) Let x R and note that N := {(x 1 n, x]} is a collection of open neighbourhoods of x. If U T contains x then (since E is a base for T) there exist a, b R such that a < b and x (a, b] U. It follows that (x 1 n, x] U for some n N (with 1 n x a). Therefore N is a countable neighbourhood base at x, so T is first countable. Let B be a base for T, and let x R. Then B contains a neighbourhood base at x, so there exists U x B such that x U x (x 1, x]. In particular sup U x = x, which implies that x U x is injective. Therefore B is uncountable. This implies that T is not second countable. (c) Let x R and let U T contain x. Then x (a, b] U for some a, b R with a < b. There exists q (a, x) Q, since Q is dense in R with respect to the usual topology. Clearly q U \ {x}, so x acc(q) with respect to T. Therefore acc(q) = R, so Q is dense in R with respect to T. In particular T is separable. 7. Suppose (x n ) has a subsequence (x n k ) k=1 which converges to x. If U is a neighbourhood of x, there exists N N such that x nk U for all k N with k N, hence for infinitely many k. Therefore x is a cluster point of (x n ). Conversely, suppose that x is a cluster point of (x n ). Since X is first countable, there exists a nested countable neighbourhood base {U n } at x. Set n := and, for each k N, choose n k N inductively so that n k > n k 1 and x nk U k. This is possible because {m N x m U k } is infinite and hence {m N x m U k } \ [1, n k 1 ]. If U is a neighbourhood of x then U N U for some N N. It follows that x nk U k U N U for all k N with k N. This shows that the subsequence (x nk ) k=1 of (x n) converges to x. 1. (a) Suppose X is connected, and let A X be clopen. Then A and A c are disjoint open sets which cover X. Since X is connected, it follows that A = or A c =. Therefore and X are the only clopen subsets of X. Conversely, suppose that and X are the only clopen subsets of X. If U, V X are disjoint open sets which cover X, then U = V c and hence U is clopen. This implies that U {, X}, so U = or V =. Therefore X is not disconnected, i.e. it is connected. (b) Define E := α A E α and suppose that U, V E are non-empty open sets (relative to E) which cover E. Choose x α A E α, and without loss of generality assume x U. Also choose y V, so that y E α for some α A. Now x U E α and y V E α, so U E α and V E α are non-empty open sets (relative to E α ) which cover E α. Since E α is connected, these sets cannot be disjoint. Therefore U V, which shows that E is connected. (c) Let U, V A be disjoint open sets (relative to A) which cover A. Then U = U A and V = V A for some open sets U, V X which cover A such that U V A =. It follows that U A and V A are disjoint open sets (relative to A) which cover A. Since A is connected, U A = without loss of generality. This implies that U acc(a) =, because if x U acc(a) then (U \ {x}) A, which is impossible. Therefore U A =, because A = A acc(a), and hence A is connected. 2

3 (d) Let x X and set C := {A X A is connected and x A}. Then C := C is connected by part (b). If A X is connected and C A then x A and hence A C. This implies that A C, so C is maximal. Similarly, if C X is a maximal connected set containing x, then C C and hence C C, so C = C because C is maximal. Therefore C is unique. Since C is connected (by part (c)) and C C, it follows that C = C, which means C is closed. 13. Clearly U A U. Conversely, note that A is contained in the closed set U c (U A), and hence U c (U A) = X. This implies that U U A, so U U A. Therefore U = U A. 14. Suppose that f is continuous and let A X. Then f 1 (f(a)) is closed (as f(a) is closed) and A f 1 (f(a)) because f(a) f(a). It follows that A f 1 (f(a)), which implies that f(a) f(a). Now suppose that f(a) f(a) for all A X. If B Y, then f 1 (B) X and hence f(f 1 (B)) f(f 1 (B)). Since f(f 1 (B)) B, this implies that f(f 1 (B)) B, or equivalently f 1 (B) f 1 (B). Finally, suppose that f 1 (B) f 1 (B) for all B Y. If C Y is closed, then f 1 (C) f 1 (C) = f 1 (C). This implies that f 1 (C) = f 1 (C), so C is closed and hence f is continuous. 16. (a) Consider the following diagrams: Y id X h f id Y Y π 1 Y g Y Y π 1 Y π 2 π 2 Y Y There exist unique maps : Y Y Y and h : X Y Y which make the resulting diagrams commute, and they are both continuous by Proposition Note that (Y ) = {(y, y) y Y } is closed, because (Y ) c is open: if (a, b) (Y ) c then a b and there exist disjoint open neighbourhoods U and V of a and b respectively; in particular (a, b) U V (Y ) c. It follows that {x X f(x) = g(x)} = h 1 ( (Y )) is closed. (b) Since {x X f(x) = g(x)} is closed and contains a dense subset of X, it is equal to X. Therefore f = g. 17. Suppose that, for every x, y X with x y, there exists f F such that f(x) f(y). Given x, y X with x y, choose such a function f and disjoint open sets U, V R such that f(x) U and f(y) V. Then f 1 (U) and f 1 (V ) are disjoint members of T with x f 1 (U) and y f 1 (V ). This shows that T is Hausdorff. Now suppose that T is Hausdorff, and let x, y X be distinct. There exists U T such that x U and y / U. Clearly U X, so there exist U 1, U 2,..., U n R and f 1, f 2,..., f n F such that x n k=1 f 1 k (U k) U. It follows that y / f 1 k (U k) for some k {1, 2,..., n}, and hence f k (x) f k (y). 22. If x X, then ρ(f n (x ), f m (x )) sup x X ρ(f n (x), f m (x)) for all m, n N. Therefore (f n (x )) is a Cauchy sequence in the complete metric space (Y, ρ), which has a limit f(x ). This defines a map f : X Y. Given ε (, ) there exists N N such that sup x X ρ(f n (x), f m (x)) < ε 2 for all m, n N with m n N. If x X and n N with n N then ρ(f m (x), f(x)) < ε 2 for some m N with m n, so ρ(f n(x), f(x)) < ε. This implies that sup x X ρ(f n (x), f(x)) ε for all n N with n N, so (sup x X ρ(f n (x), f(x))) converges to. 3

4 Suppose g : X Y and (sup x X ρ(f n (x), g(x))) converges to. If ε (, ) and x X then ρ(f N(x), f(x)) < ε 2 and ρ(f N (x), g(x)) < ε 2 for some N N, which implies that ρ(f(x), g(x)) < ε. Therefore ρ(f(x), g(x)) =, so f(x) = g(x) and hence f = g. This shows that f is unique. Now suppose that f n is continuous for all n N. Let x X and suppose that U Y is a neighbourhood of f(x ). There exists ε (, ) such that B ε (f(x )) U, and there exists N N such that sup x X ρ(f N (x), f(x)) < ε 3. Since f N is continuous, V := f 1 N (B ε/3(f N (x ))) is an open neighbourhood of x. If x V then ρ(f(x), f(x )) ρ(f(x), f N (x)) + ρ(f N (x), f N (x )) + ρ(f N (x ), f(x )) < ε 3 + ε 3 + ε 3 = ε and hence f(v ) B ε (f(x )) U. This implies that f 1 (U) is a neighbourhood of x, so f is continuous. 23. Let A R be closed a R, b (a, ) and f C(A, [a, b]). We aim to extend f to some F C(R, [a, b]). Since A c is open, it is a disjoint union of open intervals {(a n, b n )}. At most two of these are unbounded, and on these we let F be constant, equal to the value of f at the finite endpoint. On bounded intervals (a n, b n ), we let F be the linear function from (a n, f(a n )) to (b n, f(b n )). By construction F has the same image as f, and is continuous on A c. Given x A, we will show that F is right continuous at x; the proof of left continuity is similar. For each ε (, ), there exists δ (, ) such that f(x) f(y) < ε for all y (x, x + δ) A. If (x, x + δ) A then we are done. Otherwise (x, x + δ) meets (a n, b n ) for some n N; this implies that a n = x or a n (x, x + δ). If a n = x we are done, since F is continuous on [a n, b n ]. Otherwise F (x) F (y) < ε for all y (x, a n ): if y A we know this already, and if y (a m, b m ) for some m N then F (y) is between f(a m ) and f(b m ), which are both within ε of f(x) = F (x) because x a m < b m < a n < x + δ and a m, b m A. Therefore F is right continuous at x, and we are done. 24. If X is normal, then by Urysohn s lemma and the Tietze extension theorem, it satisfies the conclusions of Urysohn s lemma and the Tietze extension theorem. Conversely, if X satisfies the conclusion of Urysohn s lemma, we claim that X is normal. To this end, let A and B be disjoint closed subsets of X. There exists f C(X, [, 1]) such that f A = and f B = 1. Note that f 1 ((, 1 37 )) and f 1 (( 37 42, )) are disjoint open sets which contain A and B respectively. Therefore X is normal. Finally, suppose that X satisfies the conclusion of the Tietze extension theorem. We need to show that X is normal; by the above it suffices to show that X satisfies the conclusion of Urysohn s lemma. To this end, let A and B be disjoint closed subsets of X. Note that A B is closed, and that A and B are open in A B. Therefore χ B C(A B, [, 1]), and there exists f C(X, [, 1]) extending χ B, as required. 26. (a) As is often the case, it is easier to talk about disconnected spaces. We need to show that, if f(x) is disconnected, then X is disconnected. Since f(x) is disconnected, it has a nonempty proper subset A which is both open and closed (clopen). Note that A = U f(x) for some open set U Y, and A = C f(x) for some closed set C Y, by definition of the relative topology on f(x). Therefore f 1 (A) = f 1 (U) = f 1 (C) is a nonempty proper clopen subset of X. This implies that X is disconnected. (b) Again we prove the contrapositive: if X is disconnected, then it is not arcwise connected (usually this is called path connected). Let A be a nonempty proper clopen subset of X. There exists x A and x 1 A c, but there is no path joining these points: if there was such a path, say f, then f([, 1]) A is a nonempty proper clopen subset of f([, 1]), which is impossible by part (a). (c) By part (a) the spaces X + := X ((, ) R) and X := X ((, ) R) are connected. If A is a clopen subset of X, we need to show that A = or A = X. Without loss of generality (, ) A (otherwise, take A c ). Clearly, there is a sequence in X + which converges to (, ). Since A is a neighbourhood of (, ), it follows that 4

5 A X + is a nonempty clopen subset of X +. This implies that A X + = X +. Similarly, A X = X. Therefore A = X, which shows that X is connected. If X is path connected, there should be a path f from (, ) to ( 1 π, ). Define t := sup f 1 ({(, )}) and note that f(t ) = (, ) because f 1 ({(, )}) is compact. Since f is continuous, there exists t 1 (t, 1) such that f(t) B 1 (, ) for all t [t, t 1 ]. Note that f(t 1 ) Y, so π 1 (f(t 1 )), where π 1 : R 2 R is the projection 2 onto the x-axis. In particular, there is a point (x, 1) of Y such that x is between π 1 (f(t 1 )) and. By part (a) π 1 (f([t, t 1 ])) is connected, so it contains x (if not, then (, x) π 1 (f([t, t 1 ])) is a nonempty proper clopen subset) and hence f([t, t 1 ]) contains (x, 1). This contradicts the fact that f([t, t 1 ]) B 1 (, ) If X = then X is connected. Otherwise choose x X and let C X be the connected component of x. Suppose that C X. Then there exists y C c, and C is closed (by the previous homework), so there exist α 1, α 2,..., α n A and U α1 X α1, U α2 X α2,..., U αn X αn open such that y n k=1 π 1 α k (U αk ) C c. Define z k X by π α (y), α {α 1, α 2,..., α k } π α (z k ) := π α (x), α / {α 1, α 2,..., α k } for each k {, 1,..., n}, so that z = x and z n n k=1 π 1 α k (U αk ). Given k {1, 2,..., n} define i k : X αk X by p, α = α k π α (i k (p)) :=, π α (z k ), α α k so that π α i k is continuous for all α A and i k is continuous. In particular i k (X αk ) is connected, and it contains both z k and z k 1. Since z = x this implies that i 1 (X α1 ) C, and by induction it follows that z n i n (X αn ) C. But z n C c, which is a contradiction. Therefore C = X, so X is connected. 32. Suppose X is Hausdorff and let x α α A be a net in X which converges to x X. If y X and x y, there exist disjoint open sets U, V X such that x U and y V. There exists β A such that x α U for all α A with α β. If γ A then there exists δ A such that δ β and δ γ, which implies that x α α A is not eventually in V because x δ / V. Therefore x is the only limit point of x α α A. Conversely, suppose that X is not Hausdorff, and choose x, y X distinct with no disjoint open neighbourhoods. For each pair (N x, N y ) N x N y we may choose x Nx,N y N x N y, because if N x and N y were disjoint, they would contain disjoint open neighbourhoods of x and y respectively. If N x N x, choose N y N y and note that x Ox,O y O x O y N x for all (O x, O y ) N x N y such that (O x, O y ) (N x, N y ). This implies that x Nx,N y (Nx,N y) N x N y converges to x, and a similar argument shows that it also converges to y. 34. If x α α A is a net in X which converges to x X, and f F, then f is continuous so f(x α ) α A converges to f(x). Conversely, let x α α A be a net in X and suppose there exists x X such that f(x α ) α A converges to f(x) for all f F. Given a neighbourhood U X of x, there exist f 1, f 2,..., f n F and U 1 f 1 (X), U 2 f 2 (X),..., U n f n (X) open such that x n k=1 f 1 k (U k) U. For each k {1, 2,..., n} there exists β k A such that f k (x α ) U k for all α A with α β k. By induction there exists an upper bound β for {β 1, β 2,..., β n }, and it follows that x α n k=1 f 1 k (U k) U for all α A with α β. Therefore x α α A converges to x. 36. Suppose there exists a topology T on X in which convergence corresponds to pointwise almost everywhere convergence. For each n N {} and k {, 1,..., 2 n 1} define f 2 n +k := χ [2 n k,2 n (k+1)]. Since it takes on each of the values 5

6 and 1 infinitely many times, (f n (x)) fails to converge for all x [, 1]. In particular, (f n) does not converge to in (X, T). Hence there exists U T such that U with the property that for each N N there exists n N such that n N and f n / U. Define n :=, and for each k N choose n k N so that n k > n k 1 and f nk / U. Clearly (f nk ) k=1 converges to in L1 (as does (f n ) ), so it has a subsequence (f n kj ) j=1 which converges to pointwise almost everywhere. This is impossible because f nkj / U for all j N. 38. Suppose that T T. If x, y X and x y, there exist disjoint sets U, V T T such that x U and y V. Therefore (X, T ) is Hausdorff. Suppose that (X, T ) is compact, and let V T. Then V c is closed in T, hence compact relative to T. The identity map from (X, T ) to (X, T) is continuous, so V c is compact relative to T. Since T is Hausdorff V c is closed in T, so V T and hence T = T. In particular, if T T then (X, T ) is not compact. Conversely, suppose that T T. The identity map from (X, T) to (X, T ) is continuous, so (X, T ) is compact. Therefore (X, T ) is not Hausdorff, by the previous argument. 39. Suppose a space X has a countable open cover {U n } with no finite subcover. For each n N there exists x n ( n k=1 U k) c. Let x nk k=1 be a subsequence of x k k=1 and let x X. Then x U m for some m N, and x nk / U m for sufficiently large k N. This implies that x nk k=1 does not converge, so X is not sequentially compact. 4. Let x n be a sequence in X, and for each n N define E n := {x k } k=n. If E n =, then {(E n ) c } is a countable open cover of X, which has a finite subcover {(E n1 ) c, (E n2 ) c,..., (E nm ) c }. It follows that m k=1 E n k m k=1 E n k = ( m k=1 (E n k ) c ) c = X c =, which is a contradiction because x N m k=1 E n k for N := max{n k } m k=1. Hence there exists x E n. If U X is a neighbourhood of x and n N then U \ {x } meets E n, so there exists k N such that k n and x k U. This implies that x is a cluster point of x n. By exercise 7 (see above), it follows that x n has a convergent subsequence provided that X is first countable. 43. Let a nk k=1 be a subsequence of a k k=1, and define x := k=1 (1+( 1)k )2 (nk+1). Then x [, 1) (in fact x [, 1 4 ]) and a nk (x) = 1 2 (1 + ( 1)k ) for all k N. The sequence a nk (x) k=1 =, 1,, 1,... does not converge, so a n k k=1 does not converge in [, 1] [,1) and hence a k k=1 has no convergent subsequence. 44. If {U n } is a (countable) open cover of f(x), then {f 1 (U n )} is an open cover of X, which has a finite subcover, say {f 1 (U n )} N. It follows that {U n} N covers f(x), which shows that f(x) is countably compact. 45. Suppose X is countably compact. If f C(X), then f(x) is countably compact by Exercise 44. Since C is first countable, it follows that f(x) is sequentially compact by Exercise 4, and hence bounded by Heine-Borel. Conversely, suppose that X has a countable open cover {U n } with no finite subcover. For each n N there exists x n ( n k=1 U k) c. Define C := {x n }, and let K C. If x X then x U n for some n N. Since X is T 1, for each k {1, 2,..., n} there is an open neighbourhood V k of x such that x k / V k \ {x } (if x k = x just set V k := X). It follows that U n ( n k=1 V k) is a neighbourhood of x which is disjoint from K \ {x }, so x / acc(k). Therefore acc(k) =, so every subset of C is closed. If n N then x n U m for some m N, and x k k=1 eventually lies outside U m. Hence we may define f : C N by f(c) := max{k N x k = c}. Note that f is continuous and unbounded, because every subset of C is closed and f(x n ) n for all n N. By Tietze s extension theorem, there exists an extension F C(X) of f, which is unbounded and hence C(X) BC(X). 6

7 49. (a) If x E, then x has a compact neighbourhood N E. This contains an open neighbourhood U N of x, and U = U E is relatively open, so N is a neighbourhood of x in the relative topology. It is actually a compact neighbourhood, because the topologies on N relative to E and X are the same. Therefore E is locally compact. (b) Let x E and choose a compact neighbourhood N E of x in the relative topology. As N is a compact subspace of the Hausdorff space X, it is closed. Let U N be a relatively open neighbourhood of x, so that U = V E for some open V X. By Exercise 13, x V V = V E N E, which shows that E is open. (c) Suppose E is locally compact in the relative topology. Since E X is closed, it is compact and E is a compact Hausdorff space. The relative topologies on E inherited from E and X are the same, so E is locally compact in the relative topology inherited from E. If C E is relatively closed and E C, then C = D E for some closed D X, so C is closed and hence C = E. This implies that E is dense in E, so E is relatively open in E by part (b) with X replaced by E. Conversely, suppose that E is relatively open in E. Again, note that E is a compact Hausdorff space, in which case E is locally compact in the relative topology inherited from E, by part (a). This is the same as the relative topology inherited from X, so E is also locally compact in that topology. 51. Suppose φ is proper, and extend it to a map X Y. Let U Y be open. If Y / U then φ 1 (U) is open, because φ X is continuous. Otherwise U c Y is compact, so φ 1 (U c ) is compact and hence φ 1 (U) = φ 1 (U c ) c is open (as X is Hausdorff, or because X φ 1 (U)). Therefore φ is continuous. Conversely, suppose that φ extends continuously to a map X Y. If K Y is compact, then K c is open in Y, so φ 1 (K c ) is open in X. Since X φ 1 (K c ), it follows that φ 1 (K) = φ 1 (K c ) c is compact. Thus φ is proper. 54. (a) If K Q is a neighbourhood of, then ( ε, ε) Q K for some ε (, ) Q. Clearly ε 2 K \ K, so K is not a closed subset of R; in particular it is not compact. Therefore has no compact neighbourhood in Q. (b) Define K := {} { 1 n }, and note that Kc = ( 1 n+1, 1 n ) (, ) (1, ) is open, so K is closed (in R) and bounded, hence compact. For each n N let f n : Q C be the indicator function of { 1 n }. The sequence f n converges to pointwise, but not uniformly on K. 56. (a) If t (, ) then Φ (t) = (t + 1) 2 >, so Φ is strictly increasing on [, ) by the mean value theorem. Since Φ(t) < 1 for all t [, ), it follows that Φ is strictly increasing. If s, t [, ) then Φ(t + s) = and the same clearly holds if s = or t =. t + s 1 + t + s = t 1 + t + s + s 1 + t + s t 1 + t + s = Φ(s) + Φ(t), 1 + s (b) If y Y then Φ(ρ(y, y)) = Φ() =. Conversely, if x, y Y and Φ(ρ(x, y)) = then ρ(x, y) = (as Φ is injective) and hence x = y. Clearly Φ ρ is nonnegative and bounded. If x, y, z Y then Φ(ρ(x, y)) = Φ(ρ(y, x)) and Φ(ρ(x, z)) Φ(ρ(x, y) + ρ(y, z)) Φ(ρ(x, y)) + Φ(ρ(y, z)). Therefore Φ ρ is a metric. If y Y and r (, ) then B ρ r (y) = {x Y ρ(x, y) < r} = {x Y Φ(ρ(x, y)) < Φ(r)} = B Φ ρ Φ(r) (y). Moreover, if r < 1 then r = Φ( r 1 r ) and hence Br Φ ρ (y) = {x Y Φ(ρ(x, y)) < Φ( r 1 r )} = {x Y ρ(x, y) < r Otherwise B Φ ρ r (y) = Y. This shows that ρ and Φ ρ define the same topologies on Y. 7 1 r } = Bρ r (y). 1 r

8 (c) The proof is the same as part (b), except that B 1 (g) = {f C X ρ(f, g) < 1} = {f C X sup f(x) g(x) < } x X for all g C X. These sets are still open in the topology of uniform convergence, as required. (d) It is routine to check that ρ is a metric. Let r (, ) and f C X. For each g B r (f) there exists ε (, ) such that B ε (g) B r (f). Choose m, N N so that Φ(m 1 ) < ε 2 and n=n 2 n < ε 2. Note that { ( ) } N 1 {h C X sup h(x) g(x) < m 1 } h C X 2 n Φ sup h(x) g(x) < ε B ε (g). x U N x U n 2 This shows that B r (f) is open in the topology of uniform convergence on compact sets. Conversely, let f C X and m, n N. For each g C X with sup x U n g(x) f(x) < m 1, there exists ε (, ) such that {h C X sup x U n h(x) g(x) < ε} {h C X sup x U n h(x) f(x) < m 1 }. It is clear that B 2 n Φ(ε)(g) is contained in the former set. This shows that the latter set is open in the metric topology, so the two topologies are equivalent. 57. (a) Let {U α } α A be an open cover of X, and choose a sequence E n of precompact open subsets of X such that E n E n+1 for all n N and E n = X. Set E := and for each n N define U n := {U α (E n+2 \E n 1 )} α A, so that U n is an open cover of the compact set E n+1 \ E n (or E 2 in the case n = 1), having a finite subcover V n. Write V n = {V n }, and note that this cover is a refinement of {U α} α A. We claim it is locally finite. Indeed, if x X then x E n+2 \ E n 1 for some n N, which is an open neighbourhood of x that meets no members of V m for all m N with m n 3. If n N then V n is a closed subset of E m+2 for some m N, and is therefore compact. For each x X there exists n N such that x V n, and x has a compact neighbourhood N x V n. Given n N, there is a finite set Y n X such that V n y Yn N y. From above it is clear that V n meets only finitely many members of {V m } m=1, say {V m} m Jn for some finite index set J n N. Define Z n := {y m Jn Y m N y V n } and W n := y Zn N y. Since Z n is finite W n y Zn N y V n. Moreover, if x X then there exists n N such that x V n y Yn N y, so x N y for some y Y n. By definition N y V m for some m N, and x V n V m so n J m, implying that y Z m and hence x W m. This shows that {W n } is an open cover of X. Since it is a refinement of {V n }, it is a locally finite refinement of {U α} α A. Note also that W n is precompact for each n N (as W n V n ). (b) For each n N, by Urysohn s lemma there exists f n C c (X, [, 1]) such that f n (W n ) = {1} and f n (V c n ) =, where {W n } and {V n} are the covers constructed in part (a). Since {V n} is locally finite, each x X has an open neighbourhood on which f := f n is well-defined and continuous. Note that f 1 because {W n } covers X, and in particular g n := f n /f is a well-defined member of C c (X, [, 1]) for each n N. It is clear that {g n } is a partition of unity subordinate to U. 58. Let K α A X α be closed, and suppose there exists x K. Then there exists a finite subset B A and open sets U β X β for each β B such that x β B π 1 β (U β) K. Since B is finite, X α is noncompact for some α A \ B. Hence there exists an open cover {V γ } γ Γ of X α which has no finite subcover. If {πα 1 (V δ )} δ covers K, for some Γ, then {V δ } δ covers X α. Indeed, if y α X α then the point y α A X α defined by π α (y) = y α 8

9 and π β (y) = π β (x) for all β A \ {α} lies in K, so y πα 1 (V δ ) and hence y α V δ for some δ. This implies that (V γ )} γ Γ is an open cover of K with no finite subcover. Therefore K is not compact, which shows that every {π 1 α closed compact subset of α A X α has empty interior. 59. Let X and Y be locally compact spaces. If (x, y) X Y, then x and y have compact neighbourhoods K X and L Y. By Tychonoff s theorem K L is a compact subset of X Y, and it is a neighbourhood of (x, y) (by the definition of the product topology). Therefore X Y is locally compact; by induction a finite product of locally compact spaces is locally compact. 6. Let {X n } be a collection of sequentially compact spaces, and let x n be a sequence in X n. Starting with x n := x n, define for each m N a subsequence xm n of xm 1 n such that π m(x m n ) converges to x m X m. Define x X n by π n (x) = x n for all n N, and let U X n be a neighbourhood of x. There exists N N and open subsets U n X n for each n {1, 2,..., N} such that x N π 1 n (U n ) U. Given m {1, 2,..., N}, x n n n=m is a subsequence of x m n, so there exists N m N such that π m (x n n) = x m for all n N with n N m. If n N and n max{n m } N m=1 then π m(x n n) = x m for all m {1, 2,..., N}, and hence x n n U. This shows that x n n (a subsequence of x n ) converges to x. Therefore X n is sequentially compact. 63. If f = then T f = C([, 1]). Otherwise f u >. Let ε (, ) and note that K is uniformly continuous (because [, 1] 2 is compact). Hence there exists δ (, ) such that K(z 1 ) K(z 2 ) < ε/ f u for all z 1, z 2 [, 1] 2 with z 1 z 2 < δ. It follows that T f(x 1 ) T f(x 2 ) = = = = ε K(x 1, y)f(y) dy K(x 2, y)f(y) dy (K(x 1, y) K(x 2, y))f(y) dy K(x 1, y) K(x 2, y) f(y) dy ε f u f(y) dy ε f u f u dy ε dy for all x 1, x 2 [, 1] with x 1 x 2 < δ. This shows that T f C([, 1]). Now let ε (, ) and choose δ (, ) such that K(z 1 ) K(z 2 ) < ε for all z 1, z 2 [, 1] 2 with z 1 z 2 < δ. Then K(z 1 ) K(z 2 ) < ε/ f u for all f C([, 1]) with < f u 1 and z 1, z 2 [, 1] 2 with z 1 z 2 < δ, so {T f f u 1} is equicontinuous by the above calculation and the fact that < ε. Moreover, if x [, 1] and f C([, 1]) with f u 1 then T f(x) = K(x, y)f(y) dy K(x, y) f(y) dy K(x, y) dy, so {T f f u 1} is pointwise bounded, and thus precompact by the Arzelà-Ascoli theorem. 64. Let ε (, ) and define δ := α ε. If f F := {f C(X) f u 1 and N α (f) 1} then f(x) f(y) N α (f)ρ(x, y) α ρ(x, y) α < δ α = ε 9

10 for all x, y X with ρ(x, y) < δ. This implies that F is equicontinuous. Moreover, if f F and x X then f(x) f u 1, so F is pointwise bounded and hence precompact, by the Arzelà-Ascoli theorem. Now let f n be a sequence in F which converges in C(X) to some f C(X). If x X then f n (x) is a sequence in [ 1, 1] which converges to f(x), so f(x) 1 and hence f u 1. Similarly, if x, y X then f n (x) f n (y) converges to f(x) f(y), and hence f(x) f(y) ρ(x, y) α. This implies that f F, so F is closed, thus compact. 68. Define E := {(x, y) f(x)g(y) f C(X), g C(Y )}. The collection A of finite sums of elements of E is the algebra generated by E, because this algebra contains A, and A is an algebra. Indeed, if f 1, f 2,... f n, h 1, h 2,..., h m C(X) and g 1, g 2,..., g n, k 1, k 2,..., k m C(Y ), then ( ) m f i g i h j k j = j=1 j=1 m f i g i h j k j = j=1 m (f i h j )(g i k j ) A. Note that A is closed under complex conjugation, because if f 1, f 2,... f n C(X) and g 1, g 2,..., g n C(Y ) then f i g i = f i g i = f i g i A. Because complex conjugation is continuous, A is also closed under complex conjugation. If (x 1, y 1 ), (x 2, y 2 ) X Y and (x 1, y 1 ) (x 2, y 2 ), then x 1 x 2 or y 1 y 2. In the former case, there exists f C(X) with f(x 1 ) f(x 2 ) because X is normal, and hence f 1 A separates (x 1, y 1 ) from (x 2, y 2 ). The latter case is similar, and we conclude that A separates points. Since X Y is compact and Hausdorff, and A contains the constant functions, the complex Stone-Weierstraß theorem implies that A = C(X Y ). 69. Let A and B be the subalgebras of C(X) and C(X, R), respectively, generated by the coordinate maps and the constant function 1. Then A = span C (B), because A contains B (note that A C(X, R) is a subalgebra of C(X, R)) and span C (B) is an algebra. Indeed, if a 1, a 2,..., a n, b 1, b 2,..., b m C and f 1, f 2,..., f n, g 1, g 2,..., g m B then ( ) m m m a i f i b j g j = a i f i b j g j = (a i b j )(f i g j ) span C (B). j=1 j=1 j=1 Therefore A is closed under complex conjugation, because if a 1, a 2,..., a n C and f 1, f 2,... f n B then a i f i = a i f i = a i f i A. Because complex conjugation is continuous, A is also closed under complex conjugation. If x 1, x 2 X and x 1 x 2, then π α (x 1 ) π α (x 2 ) for some α A, and hence A separates points. By Tychonoff s theorem X is compact, and it is also Hausdorff because [, 1] is Hausdorff. Since A contains the constant functions, the complex Stone-Weierstraß theorem implies that A = C(X). 7. (a) By definition h(i) = f I f 1 ({}), which is closed because {} is closed in R. (b) Clearly k(e). If f, g k(e) and a R then af(x) + g(x) = for all x E, so af + g k(e). Moreover, if f k(e) and g C(X, R) then f(x)g(x) = for all x E, so fg k(e). This shows that k(e) is an ideal of C(X, R). For each x E the coordinate map π x : C(X, R) R is (Lipschitz) continuous (relative to the uniform norm on C(X, R)), so k(e) = x E e 1 x ({}) is closed. 1

11 (c) Clearly E h(k(e)), and hence E h(k(e)). If x E c then, since X is normal, there exists f C(X, R) such that f(e) = and f(x) = 1. It follows that x / h(k(e)), because f k(e) but f(x). Thus h(k(e)) = E. (d) Clearly I k(h(i)), and hence I k(h(i)). Define U := X \ h(i), and let x U. Then {x} and U c are closed and disjoint, so there exist disjoint open neighbourhoods V, W X of x and U c. It follows that W c U is a compact neighbourhood of x, since X is compact and V W c. This shows that U is locally compact. Define J := {f U f I}. If f I and ε (, ) then {x U f U (x) ε} = {x X f(x) ε} because f(x) = for all x h(i). Since this set is closed (thus compact), f U C (U). Therefore J is a closed subalgebra of C (U, R) (because function restriction respects pointwise addition and multiplication, and uniform limits). If x, y U and x y, there exists f I such that f(x). Since X is normal there exists g C(X, R) such that g(x) = 1 and g(y) =. Then gf I and hence (gf) U J, so J separates points. Moreover, there is no x U such that f(x) = for all f J. By the Stone-Weierstraß theorem, it follows that J = C(U, R) or J = C (U, R), depending on whether U is closed or not. If f k(h(i)) then f U C (U, R) for the same reason that J C (U, R), so f U = g U for some g I, in which case f = g because f(x) = g(x) = for all x h(i). This shows that k(h(i)) I, and hence k(h(i)) = I. (e) The previous two exercises show that k is a bijection from the closed subsets of X onto the closed ideals of C(X, R). 76. Let B be a countable base for the topology on X. Since X is normal, for each U, V B such that V U, there exists f U,V C(X, [, 1]) which is on U c and 1 on V. Clearly F := {f U,V U, V B and V U} is countable. Let C X be closed and let x C c. There exists U B such that x U C c, and by normality there exist disjoint open sets U, V X such that U c U and x V. Then x V V for some V B, and V V (U ) c U. Now C U c, so f U,V (C) = {}, while f U,V (x) = 1 because x V. This shows that F separates points and closed sets. 77. Clearly ρ maps into [, 1]. If x X, then ρ(x, x) = by definition. Given x, y X with x y, there exists n N such that x n y n, in which case ρ(x, y) 2 n ρ n (x n, y n ) >. Clearly ρ(x, y) = ρ(y, x) for all x, y X. Moreover, ρ(x, z) = 2 n ρ n (x n, z n ) 2 n (ρ n (x n, y n )+ρ(y n, z n )) = for all x, y, z X. This shows that ρ is a metric. 2 n ρ n (x n, y n )+ 2 n ρ(y n, z n ) = ρ(x, y)+ρ(y, z) Let n N and U X n be open. Given x πn 1 (U), there exists r (, ) such that B r (x n ) U. If y B 2 n r(x) then ρ(x, y) < 2 n r, and in particular ρ n (x n, y n ) < r, which implies that y n U and hence y πn 1 (U). This shows that πn 1 (U) is open in (X, ρ), and hence every member of the product topology on X is open in (X, ρ). Conversely, let U X be open in (X, ρ). Given x U, there exists r (, ) such that B r (x) U. Choose N N so that 2 1 N r. For each n {1, 2,..., N} define U n := B r/2 (x n ), and set V := N π 1 n (U n ). If y V then ρ(x, y) = N 2 n ρ(x n, y n ) + n=n+1 2 n ρ(x n, y n ) < N 2 n 1 r + n=n+1 2 n < 2 1 r + 2 N r, so y U. Since x V, this shows that U is open in the product topology. Therefore (X, ρ) has the product topology. 11

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