Theorem 1

Let $X$ be a compact space. a) If $X$ is metrizable and if $Y$ is any metrizable uniform space of countable type (Chapter IX, § 2, no. 8), then the metrizable space $C_u(X; Y)$ of continuous mappings of $X$ into $Y$, endowed with the topology of uniform convergence, is of countable type. b) Conversely, if the metrizable space $C_u(X; \mathbf{R})$ is of countable type, then $X$ is metrizable.

a) Let $d$ (resp. $d'$) be a metric compatible with the topology of $X$ (resp. with the uniformity of $Y$); then $\delta(f, g) = \sup_{x \in X} d'(f(x), g(x))$ is a metric defining the uniformity of uniform convergence on the space $C(X; Y)$, the functions of $C(X; Y)$ being bounded because $X$ is compact (no. 1). For each pair of integers $m > 0, n > 0$, let $G_{mn}$ be the set of functions $f \in \mathscr{C}(X; Y)$ such that the relation $d(x, x') \leq 1/m$ implies $d(f(x), f(x')) \leq 1/n$. Every function $f \in \mathscr{C}(X; Y)$ is uniformly continuous (Chapter II, § 4, no. 1, Theorem 2) and therefore, for each $n > 0$, $\mathscr{C}(X; Y)$ is the union of the sets $G_{mn}$ ($m > 0$). Let $\{a_1, \ldots, a_{p(m)}\}$ be a finite subset of $X$ such that the open balls with centres $a_i$ and radii $1/m$ cover $X$ ($1 \leq i \leq p(m)$); and let $(b_r)_{r \in \mathbf{N}}$ be a countable sequence which is dense in $Y$. For each mapping $\varphi : [1, p(m)] \to \mathbf{N}$, let $H_\varphi$ be the set of those $f \in G_{mn}$ such that $d'(f(a_k), b_{\varphi(k)}) \leq 1/n$ for $1 \leq k \leq p(m)$. By the definition of the $b_r$, $G_{mn}$ is the union of the sets $H_\varphi$ for $\varphi \in \mathbf{N}^{p(m)}$; let $C_{mn}$ be the set of mappings $\varphi \in \mathbf{N}^{p(m)}$ such that $H_\varphi \neq \emptyset$, and for each $\varphi \in C_{mn}$ let $g_\varphi$ be an element of $H_\varphi$; finally, let $L_{mn}$ denote the countable set of $g_\varphi$ for $\varphi \in C_{mn}$. Let $f \in G_{mn}$, and let $\varphi$ be an element of $C_{mn}$ such that $f \in H_\varphi$; then it follows immediately from the definitions that we have $d'(f(x), g_\varphi(x)) \leq 4/n$ for all $x \in X$, i.e. $\delta(f, g_\varphi) \leq 4/n$. Hence the union of the sets $L_{mn}$ is dense in $\mathscr{C}_u(X; Y)$, because for every integer $n > 0$ and every $f \in \mathscr{C}(X; Y)$ there exists $m$ such that $f \in G_{mn}$, and we have just seen that the distance from $f$ to $L_{mn}$ is $\leq 4/n$.

b) Let $I = [0, 1]$. Since $\mathscr{C}_u(X; I)$ is a uniform subspace of $\mathscr{C}_u(X; \mathbf{R})$, it is of countable type. Let $(f_n)$ be a sequence which is dense in $\mathscr{C}_u(X; I)$. Consider the product space $K = \mathbf{N}$ and the mapping $\psi : x \to (f_n(x))$ of $X$ into $K$, which is obviously continuous. The mapping $\psi$ is injective; for, by definition of the sequence $(f_n)$, the relation $f_n(x) = f_n(x')$ for all $n$ implies, on passing to the limit, $f(x) = f(x')$ for every function $f \in \mathscr{C}(X; I)$; but this is impossible if $x \neq x'$ by virtue of Axiom (OIV) applied to the point $x$ and to a neighbourhood $V$ of $x$ which does not contain $x'$ (Chapter IX, § 1, no. 5, Theorem 2). It follows that the compact space $X$ is homeomorphic to the subspace $\psi(X)$ of $K$ (Chapter I, § 9, no. 4, Theorem 2, Corollary 2); since $K$ is metrizable and of countable type, so is $\psi(X)$ and therefore so is $X$.

Q.E.D.