Corollary

Let $X$ be a locally compact space whose topology admits a countable base, and let $Y$ be a metrizable uniform space of countable type.

a) The space $\mathscr{L}$ of continuous mappings of $X$ into $Y$ which have a limit at infinity, endowed with the topology of uniform convergence in $X$, is a metrizable space of countable type.

b) The space $\mathscr{C}_c(X; Y)$ of continuous mappings of $X$ into $Y$, endowed with the topology of compact convergence, is a metrizable space of countable type.

a) Let $X'$ be the compact space obtained by adjoining a point at infinity to $X$ (Chapter I, § 9, no. 8, Theorem 4); by definition, every function $f \in \mathscr{L}$ can be uniquely extended to a continuous function $\overline{f} : X' \to Y$, and $f \to \overline{f}$ is therefore a bijection of L onto $C(X'; Y)$; and this bijection is a homeomorphism of the space L onto $C_u(X'; Y)$ by Proposition 6 of § 1, no. 6. Since $X'$ is metrizable (Chapter IX, § 2, no. 9, Proposition 16, Corollary) the result follows from Theorem 1, applied to $X'$ and Y.

b) Let $(U_n)$ be a covering of X by relatively compact open sets, such that every compact subset of X is contained in some $U_n$ (Chapter I, § 9, no. 9, Proposition 15, Corollary 1). If $S$ is the set of the $\overline{U}_n$, the topology of compact convergence on $C(X; Y)$ is the same as the topology of $S$-convergence. Consequently ($§ 1,$ no. 2, Remark 3) the space $C_c(X; Y)$ is homeomorphic to a subspace of the product $\prod_n C_u(\overline{U}_n; Y)$; since each of the compact spaces $\overline{U}_n$ has a countable base, it is metrizable (Chapter IX, § 2, no. 9, Proposition 16); each of the $C_u(\overline{U}_n; Y)$ is therefore metrizable and of countable type by Theorem 1, and hence so is $C_c(X; Y)$.

Note that the space of all bounded continuous real-valued functions on $\mathbf{R}$, endowed with the topology of uniform convergence, is not of countable type (Exercise 4).