Math General Topology Fall 2012 Homework 13 Solutions

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1 Math General Topology Fall 2012 Homework 13 Solutions Note: In this problem set, function spaces are endowed with the compact-open topology unless otherwise noted. Problem 1. Let X be a compact topological space, and (Y, d) a metric space. Consider the uniform metric d(f, g) := sup d (f(x), g(x)) x X on the set of continuous maps C(X, Y ). Show that the topology on C(X, Y ) induced by the uniform metric is the compact-open topology. Solution. Denote respectively by T co and T met the compact-open topology and the uniform metric topology on C(X, Y ). (T met T co ) It suffices to show that for any f C(X, Y ) and ɛ > 0, the open ball B(f, ɛ) C(X, Y ) is a neighborhood of f in the compact-open topology. Cover Y by open balls of radius ɛ 3 and pull this open cover back to X via f: X = f 1 (Y ) = f 1 ( y Y = y Y B(y, ɛ 3 ) ) f 1 ( B(y, ɛ 3 ) ). Since X is compact, there is a finite subcover X = f 1 ( B(y 1, ɛ 3 ) )... f 1 ( B(y n, ɛ 3 ) ) =: U 1... U n. The closures U i X are compact, since they are closed in X which is compact. Moreover they satisfy: f(u i ) f(u i ) since f is continuous = f (f (B(y 1 i, ɛ )) 3 ) B(y i, ɛ 3 ) B cl (y i, ɛ 3 ) := {y Y d(y i, y) ɛ 3 } B(y i, ɛ 2 ). 1

2 Therefore f satisfies f n i=1 V (U i, B(y i, ɛ )) =: N 2 where the latter subset N C(X, Y ) is open in the compact-open topology. Claim: N B(f, ɛ). Let g N and take any point x X. Since the U i cover X, we have x U i for some index 1 i n. The values f(x) and g(x) satisfy: f(x) f(u i ) B(y i, ɛ 3 ) g(x) g(u i ) g(u i ) B(y i, ɛ 2 ). By the triangle inequality, we obtain: d(f(x), g(x)) d(f(x), y i ) + d(y i, g(x)) < ɛ 3 + ɛ 2 = 5ɛ 6 and since x X was arbitrary, we conclude d(f, g) 5ɛ 6 < ɛ. [Note that the condition d(f(x), g(x)) < ɛ for all x X was enough to guarantee d(f, g) < ɛ, since the supremum sup x X d(f(x), g(x)) is achieved at a point x 0 X.] (T co T met ) Take a subbasic open V (K, U) C(X, Y ) and f V (X, U). We want to show that V (K, U) is metrically open, i.e. find a radius ɛ satisfying B(f, ɛ) V (K, U). The subset f(k) Y is compact, and moreover it satisfies f(k) U by assumption. Hence there is a number ɛ > 0 such that the ɛ-neighborhood of f(k) is contained in U, i.e. B(f(K), ɛ) U (c.f. Homework 5 Problem 5b). Claim: B(f, ɛ) V (K, U). Let g B(f, ɛ). For any x K, we have: g(x) B(f(x), ɛ) B(f(K), ɛ) U so that the map g : X Y satisfies g(k) U, i.e. g V (K, U). 2

3 Problem 2. Let X and Y be topological spaces. Let f, g : X Y be two continuous maps. Show that a homotopy from f to g induces a (continuous) path from f to g in the space of continuous maps C(X, Y ). More precisely, let F (X, Y ) denote the set of all functions from X to Y. There is a natural bijection of sets: ϕ: F (X [0, 1], Y ) = F ([0, 1], F (X, Y )) sending a function H : X [0, 1] Y to the function ϕ(h): [0, 1] F (X, Y ) defined by ϕ(h)(t) = H(, t) =: h t. Your task is to show that if a function H : X [0, 1] Y is continuous, then the following two conditions hold: 1. h t : X Y is continuous for all t [0, 1]; 2. The corresponding function ϕ(h): [0, 1] C(X, Y ) is continuous. Solution. 1. For t [0, 1], the slice inclusion map ι t : X X [0, 1] defined by ι t (x) = (x, t) is an embedding. Therefore the map h t : X Y is continuous, since it is the composite h t = H ι t, as illustrated here: X ι t X [0, 1] H Y. h t 2. Let V (K, U) C(X, Y ) be a subbasic open subset, with K X compact and U Y open. We want to show that the preimage ϕ(h) 1 (V (K, U)) [0, 1] is open in [0, 1]. This preimage is: ϕ(h) 1 (V (K, U)) = {t [0, 1] h t V (K, U)} = {t [0, 1] h t (K) U} = {t [0, 1] H(k, t) U for all k K} = {t [0, 1] K {t} H 1 (U)}. Since H : X [0, 1] Y is continuous, H 1 (U) is open in X [0, 1]. If the inclusion K {t} H 1 (U) holds, then since K and {t} are compact, there exist open subsets W X and J [0, 1] satisfying K {t} W J H 1 (U) by the tube lemma. In particular, every t J satisfies K {t } W {t } H 1 (U). Therefore the inclusion J ϕ(h) 1 (V (K, U)) holds, so that ϕ(h) 1 (V (K, U)) is open in [0, 1]. 3

4 Remark. If X is locally compact Hausdorff, then the converse holds as well: the two conditions guarantee that H : X [0, 1] Y is continuous. In that case, a homotopy from f to g is really the same as a path from f to g in the function space C(X, Y ). 4

5 Problem 3. a. Let X and Y be topological spaces, where Y is Hausdorff. Show that C(X, Y ) is Hausdorff. Solution. Let f, g C(X, Y ) be distinct elements, i.e. there is a point x X where f(x) g(x). Since Y is Hausdorff, the points f(x) and g(x) in Y can be separated by neighborhoods U, U Y, satisfying f(x) U, g(x) U, and U U =. Then we have f V ({x}, U) and g V ({x}, U ), where both subsets are open in C(X, Y ) since the singleton {x} X is compact. Moreover, their intersection is: V ({x}, U) V ({x}, U ) = V ({x}, U U ) = V ({x}, ) = so that f and g can be separated by neighborhoods in C(X, Y ). b. Assume there exists a topological space X such that C(X, Y ) is Hausdorff. Show that Y is Hausdorff. Solution. Let y, y Y be distinct points. Consider the constant functions f, f : X Y at y and y respectively, i.e. f y and f y. Note that f and f are continuous, i.e. elements of C(X, Y ). Since C(X, Y ) is Hausdorff, the distinct elements f and f can be separated by basic open neighborhoods f N = V (K 1, U 1 )... V (K n, U n ) f N = V (K 1, U 1)... V (K k, U k) in C(X, Y ). The condition f V (K i, U i ) means f(k i ) U i, or equivalently y U i since f is the constant function f y. Likewise, we have y U i for all i. Therefore we obtain neighborhoods of y and y in Y : y U := U 1... U n y U := U 1... U k. Claim: The neighborhoods U and U are disjoint. For any point y 0 U U, the constant function g : X Y with value y 0 is continuous and satisfies g V (K 1, U U )... V (K n, U U ) V (K 1, U)... V (K n, U) = V (K 1, U 1 )... V (K n, U n ) = N and likewise g N. This implies g N N =, and therefore U U =. 5

6 Problem 4. a. Let X, Y, and Z be topological spaces. Let g : Y Z be a continuous map. Show that the induced map postcomposition by g is continuous. g : C(X, Y ) C(X, Z) f g (f) = g f Solution. It suffices to show that the preimage of a subbasic open is open. Consider the subbasic open subset V (K, U) C(X, Z), where K X is compact and U Z is open. Its preimage in C(X, Y ) is (g ) 1 V (K, U) = {f C(X, Y ) g f V (K, U)} = {f C(X, Y ) g(f(k)) U} = {f C(X, Y ) f(k) g 1 (U)} = V (K, g 1 (U)) which is open in C(X, Y ) since K X is compact and g 1 (U) Y is open in Y. Remark. Alternate proof when X is locally compact Hausdorff. It suffices to show that the corresponding map g : C(X, Y ) X Z is continuous. But this map is the composite g e where e: C(X, Y ) X Y is the evaluation map: C(X, Y ) X g Z. e Y g Since X is locally compact Hausdorff, the evaluation map e is continuous, and so is the composite g = g e. 6

7 b. Let W, X, and Y be topological spaces. Let d: W X be a continuous map. Show that the induced map precomposition by d is continuous. d : C(X, Y ) C(W, Y ) f d (f) = f d Solution. Consider the subbasic open subset V (K, U) C(W, Y ), where K W is compact and U Y is open. Its preimage in C(X, Y ) is (d ) 1 V (K, U) = {f C(X, Y ) f d V (K, U)} = {f C(X, Y ) f(d(k)) U} = V (d(k), U) which is open in C(X, Y ) since d(k) X is compact and U Y is open in Y. Remark. Alternate proof when X is locally compact Hausdorff. It suffices to show that the corresponding map d : C(X, Y ) W Y is continuous. But this map is the composite e (id d) where e: C(X, Y ) X Y is the evaluation map: C(X, Y ) W d Y. id d C(X, Y ) X e Since X is locally compact Hausdorff, the evaluation map e is continuous, and so is the composite d = e (id d). 7

8 Problem 5. Let X and Y be topological spaces, where X is Hausdorff. Let S be a subbasis for the topology of Y. Show that the collection {V (K, S) K X compact, S S} is a subbasis for the compact-open topology on C(X, Y ). The notation above is V (K, S) = {f C(X, Y ) f(k) S}. Solution. Let T be the topology on C(X, Y ) generated by the collection above. We want to show the equality T = T co, where the latter denotes the compact-open topology. (T T co ) Since S S is open in Y, the subsets V (K, S) C(X, Y ) are open in the compactopen topology. (T co T ) It suffices to show that the generating open subsets V (K, U) are in T, where K X is compact and U Y is open. First, let B denote the collection of finite intersections of members of S, so that B is a basis for the topology on Y. The equality V (K, S 1 ) V (K, S 2 ) = {f C(X, Y ) f(k) S 1 and f(k) S 2 } = {f C(X, Y ) f(k) S 1 S 2 } = V (K, S 1 S 2 ) shows that subsets of the form V (K, B) C(X, Y ) are in T, where K X is compact and B B. Let f V (K, U), i.e. f satisfies f(k) U. We want to find a T -neighborhood of f contained in V (K, U). Since B is a basis for the topology on Y, the open subset U Y can be written as a union U = α A B α with B α B. Therefore K is covered by the preimages: K f 1 (U) = f 1 ( α A B α ) = α A f 1 (B α ) where each f 1 (B α ) X is open in X. Since X is Hausdorff, there exist compact subsets K 1,..., K n X satisfying K = K 1... K n and K i f 1 (B αi ) for some indices α i (c.f. Homework 8 Problem 4b). 8

9 The condition f(k i ) B αi says f V (K i, B αi ) for all i = 1,..., n. The conditions K = K 1... K n and B αi U imply: V (K 1, B α1 )... V (K n, B αn ) V (K 1, U)... V (K n, U) = V (K 1... K n, U) = V (K, U). Therefore V (K 1, B α1 )... V (K n, B αn ) is a T -neighborhood of f contained in V (K, U). 9

10 Problem 6. Consider the real line R and the rationals Q with their standard (metric) topology. Consider the evaluation map e: Q C(Q, R) R. Let f : Q R be a constant function (say, f 0), and let q Q. Show that the evaluation map e is not continuous at (q, f) Q C(Q, R). Hint: You may want to use the fact that all compact subsets of Q have empty interior (c.f. Homework 7 Problem 5), and the fact that Q is completely regular (since it is normal). Solution. Take ɛ = 1 and consider the open ball B 1 (0) of radius 1 centered at e(q, f) = f(q) = 0 R. We will show that on every neighborhood of (q, f) in Q C(Q, R), the evaluation map e takes values larger than 1, proving discontinuity of e at (q, f). Since open boxes form a basis of the product topology on Q C(Q, R), it suffices to consider a neighborhood of (q, f) of the form V N where V Q is an open neighborhood of q Q and N C(Q, R) is a basic open neighborhood of f C(Q, R), of the form N = V (K 1, U 1 )... V (K n, U n ) for some compact subsets K i Q and open subsets U i R. The condition f N can be restated as the following equivalent conditions: f N f V (K i, U i ) for all i f(k i ) U i 0 U i for all i. for all i Therefore any continuous function g : Q R that vanishes on K := K 1... K n, i.e. satisfying g K 0, automatically satisfies g N. Moreover, K is a finite union of compact subsets of Q, thus itself compact. Therefore K Q has empty interior, which implies V K since V Q is open in Q. Pick a rational q V \ K. Since K is closed in Q (being compact in the Hausdorff space Q), and Q is completely regular, there exists a continuous function g : Q [0, 75] R satisfying g K 0 and g(q ) = 75. By construction, the pair (q, g) is sufficiently close to (q, f): and satisfies e(q, g) = g(q ) = 75 > 1. (q, g) V N 10

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