Suppose X, Y, and Z are metric space, and f: X rightarrow Y and g: Y rightarrow

Question

Suppose X, Y, and Z are metric space, and f: X rightarrow Y and g: Y rightarrow Z are continuous functions. Prove g.f: X rightarrow Z is also continuous.

Explanation / Answer

(cf. 4.3.9). Suppose X, Y , and Z are metric spaces, and f : X ! Y and g : Y ! Z are functions. Then: (i) If f is continous at a 2 X and g is continuous at f(a), then g  f is continuous at a. (ii) If f is continuous on X and g is continuous on Y , then g  f is continuous on X. Proof. (ii) We appeal to Theorem 6.8. Suppose W is an open subset of Z. Then V = g??1(W) is open in Y , and therefore f??1(V ) is open in X. Since f??1(V ) = f??1(g??1(W)) = (g  f)??1(W); this shows that g  f is continuous on X. A function f : X ! Y is continuous on X if and only if f??1(V ) is an open set in X whenever V is an open set in Y . 1 Proof. First suppose f is continuous on X, let V  Y be open, and x a 2 f??1(V ). Then f(a) 2 V , so there exists  > 0 such that V(f(a))  V . But f is continuous at a, so there exists  > 0 such that dY (f(x); f(a)) <  whenever dX(x; a) < . In other words, f(V(a))  V(f(a))  V , which means V(a)  f??1(V ). This shows that f??1(V ) is open. Now suppose f??1(V ) is open in X whenever V is open in Y , and x a 2 X. For any  > 0, the set V = V(f(a)) is open, so f??1(V ) = fx 2 X j f(x) 2 V(f(a))g = fx 2 X j dY (f(x); f(a)) < g is an open set in X which evidently contains a. Thus there exists  > 0 such that V(a)  f??1(V ), which means that dY (f(x); f(a)) <  whenever dX(x; a) < . This shows that f is continuous at a, and thus f is continuous on all of X

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