Remark 2

Let $\mathscr{S}$ be a subbase of the topology of $Y$, and let $\mathscr{K}$ be a set of compact subsets of $X$ with the following property:

(R) If $L$ is any compact subset of $X$ and $V$ is any neighbourhood of $L$, there exists a finite number of sets $K_i \in \mathscr{K}$ such that $L \subset \bigcup_i K_i \subset V$.

Then the sets $T(K, U)$, where $K \in \mathscr{K}$ and $U \in \mathscr{S}$, form a subbase for the compact-open topology on $C(X; Y)$. To prove this, we have to show that if $L$ is any compact subset of $X$ and $V$ any open subset of $Y$, and if $u \in T(L, V)$, then there exists a finite number of pairs $(K_i, U_i)$ such that $K_i \in \mathscr{K}$, $U_i \in \mathscr{S}$ and $u \in \bigcap_i T(K_i, U_i) \subset T(L, V)$. Note first that for every finite sequence $(s_k)$ of sets of $\mathscr{S}$ and every compact subset $M$ of $X$, we have $T\left(M, \bigcap_k S_k\right) = \bigcap_k T(M, S_k)$ by definition. We may therefore first of all replace $\mathscr{S}$ by the set of finite intersections of sets of $\mathscr{S}$, i.e. we may suppose that $\mathscr{S}$ is a base of the topology of $Y$. By hypothesis, $u(L)$ is quasi-compact and contained in $V$, hence there exists a finite number of sets $U_i \in \mathscr{S}$ contained in $V$ which cover $u(L)$.

The sets $\overline{u}^{-1}(U_i)$ are open in $X$ and cover $L$. For each $x \in L$ there is therefore a compact neighbourhood $N_x$ of $x$ in $L$, contained in some one of the $\overline{u}^{-1}(U_i)$. We can cover $L$ with a finite number of these sets $N_{x_j} = L_j$; for each $j$, let us denote by $i(j)$ one of the indices $i$ such that $L_j \subset \overline{u}^{-1}(U_i)$. This being so, for each index $j$ there exists [by (R)] a finite number of sets $K_{jk} \subset \overline{u}^{-1}(U_{i(j)})$, belonging to $\mathscr{K}$, which cover $L_j$. For each $v \in \bigcap_{j,k} T(K_{jk}, U_{i(j)})$ we have $\bigcup_k v(K_{jk}) \subset U_{i(j)}$ and therefore $v(L_j) \subset U_{i(j)}$, and $v(L) = \bigcup_j v(L_j) \subset \bigcup_j U_{i(j)} \subset V$; thus our assertion is proved.