Exercise 10, § 3

Let X, Y be two topological spaces. For each subset A of X and each subset B of Y let $T(A, B)$ denote the set of all $u \in \mathscr{C}(X; Y)$ such that $u(A) \subset B$.

Let $\mathscr{U} = (U_\alpha)$ be an open covering of X. Let $\mathscr{T}_\mathscr{U}$ denote the topology on $\mathscr{C}(X; Y)$ generated by the sets $T(F, V)$, where V runs through the set of all open sets of Y and F runs through the set of closed subsets of X contained in at least one $U_\alpha$.

a) Show that $\mathscr{T}_u$ is finer than the compact-open topology (use Theorem 3 of Chapter IX, § 4, no. 3). If $X$ is regular, the mapping $(u, x) \to u(x)$ of $C(X; Y) \times X$ into $Y$ is continuous when $C(X, Y)$ carries the topology $\mathscr{T}_u$.

b) Let $\mathscr{T}$ be a topology on $C(X; Y)$ with respect to which the mapping $(u, x) \to u(x)$ of $C(X; Y) \times X$ into $Y$ is continuous. Let $u_0$ be a continuous mapping of $X$ into $Y$, let $x_0$ be a point of $X$ and let $V$ be an open set in $Y$ which contains $u_0(x_0)$. Show that there exists an open neighbourhood $U$ of $x_0$ in $X$ such that $T(U, V)$ is a neighbourhood of $u_0$ with respect to $\mathscr{T}$.

c) Suppose that $X$ is completely regular. Show that if, among the topologies $\mathscr{T}$ on $C(X; \mathbf{R})$ for which $(u, x) \to u(x)$ is continuous, there is a topology $\mathscr{T}_0$ coarser than all the others, then $\mathscr{T}_0$ must be the compact-open topology. [Let $T(U, V)$ be a neighbourhood of $o$ in $C(X; \mathbf{R})$ with respect to $\mathscr{T}_0$. Let $(W_\alpha)$ be any open covering of $\overline{U}$; let $\mathfrak{W}$ be the covering of $X$ formed by the sets $W_\alpha$ and $C\overline{U}$. Using a) and arguing by contradiction, show that a finite number of the $W_\alpha$ cover $\overline{U}$.]

d) Show that, if $X$ is completely regular but not locally compact, the mapping $(u, x) \to u(x)$ of $C_c(X; \mathbf{R}) \times X$ into $\mathbf{R}$ is not continuous at every point. [Consider a point $x_0$ which has no compact neighbourhood and argue by contradiction, using b).]