## 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 that$d^1(f(x),g(x)) \ge \epsilon$.

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.