Theorem 1

Let $X$ be a topological (resp. uniform) space, let $Y$ be a uniform space and let $H$ be an equicontinuous (resp. uniformly equicontinuous) subset of $\mathscr{C}(X; Y)$. Then the following uniformities on $H$ are identical: the uniformity of compact (resp. precompact) convergence, the uniformity of pointwise convergence and the uniformity of pointwise convergence in a dense subset $D$ of $X$.

It is enough to show that the last uniformity on $H$ is finer than the first; in other words that, given an entourage $V$ of $Y$ and a compact (resp. precompact) subset $A$ of $X$, there exists an entourage $W$ of $Y$ and a finite subset $F$ of $D$ such that the relation

implies

Suppose first that $A$ is compact and $H$ equicontinuous. Given a symmetric entourage $W$ of $Y$, every point $x \in X$ has a neighbourhood $U(x)$ such that the relation $x' \in U(x)$ implies $(u(x),\ u(x')) \in W$ for all $u \in H$. We can therefore cover the compact set $A$ by a finite number of open sets $U_i$ such that, for each pair of points $x',\ x''$ of the same set $U_i$, we have $(u(x'),\ u(x'')) \in \hat{W}$ for all $u \in H$. Let $a_i$ be a point of $D \cap U_i$, let $F$ be the set of the $a_i$, and suppose that (2) is true; then for each $x \in A$ there exists an index $i$ such that $a_i$ and $x$ belong to the same set $U_i$, so that we have $(u(x),\ u(a_i)) \in \hat{W}$ and $(v(a'),\ v(x)) \in \hat{W}$; thus (2) implies (3) provided that $W$ is chosen so that $\hat{W} \subset V$.

If $A$ is precompact and $H$ uniformly equicontinuous, we use Proposition 4 of no. 2; it is enough to note that $i(A)$ is compact in $\hat{X}$, $i(D)$ dense in $\hat{X}$ and that the entourages of $Y$ are the inverse images of those of $\hat{Y}$ under the mapping $j \times j$.