Uniform metrics define same topology on C(X,Y)

Kechris, p. 24
(B)
Let X be a compact metrizable space and Y a metrizable space. We denote by C(X,Y) the space of continuous functions from X into Y with the topology induced by the uniform metric: d_u(f, g) = \sup_{x \in X} d_Y( f(x), g(x) ), where d_Y is a compatible metric for Y. A simple compactness argument shows that this topology is independent of the choice of d_Y.

Proof.
Let d^1, d^2 be two compatible metrics for Y. I will show that the two corresponding uniform metrics, d^1_u, d^2_u give the same topology.

Suppose that d^1_u, d^2_u did not give the same topology.  That is, without loss of generality, there exists f, \epsilon, such that for all \eta, B^2(f, \eta) \nsubseteq B^1(f, \epsilon). (Here B^i(h, \nu) = \{ e \in C(X,Y) : d_u^i(h, e) < \nu \}.)  That is \forall \eta \exists g ( d^2_u(f,g) < \eta \text{ and } d_u^1(f,g) \ge \epsilon); that is, \sup_{x \in X} d^2( f(x), g(x) ) < \eta \text{ and } \sup_{x \in X} d^1(f(x),g(x)) \ge \epsilon, which is the same as

\sup_{x \in X} d^2( f(x), g(x) ) < \eta and there exists x such thatd^1(f(x),g(x)) \ge \epsilon.

Now, let n \in \mathbb{N}.  Let \eta_n = 1/n.  Then there exists g_n, x_n such that d^2(f(x_n), g_n(x_n)) < \eta_n and d^1(f(x_n),g_n(x_n)) \ge \epsilon.  Since f(X) is compact, the sequence (f(x_n)) must have a convergent subsequence (f(x_k)).  So f(x_k) \to \gamma \in f(X).

Let \delta > 0.  Choose K such that k \ge K implies d^2(f(x_k), \gamma ) < \delta / 2 and 1 / K < \delta / 2. Then k \ge K implies d^2(g_k(x_k), \gamma) \le d^2(g_k(x_k), f(x_k)) + d^2(f(x_k), \gamma) < \delta. So g_k(x_k) \to_{d^2} \gamma.

Therefore, (f(x_k), g_k(x_k)) converges to \gamma under d^2.  Thus, it must also do so under d^1.  But for all k, d^1(f(x_k),g_k(x_k)) \ge \epsilon.  So this is a contradiction.

Hence, both metrics give the same topology.

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