Proposition 1

Let T be a set, let $\mathscr{S}$ be a set of subsets of T, let Y be a uniform space, X a topological (resp. uniform) space, and let f be a mapping of $T \times X$ into $Y$. For each $A \in \mathscr{S}$, let $H_A \subset \mathscr{F}(X; Y)$ be the set of all mappings $x \to f(t, x)$ as $t$ runs through $A$. Then the mapping $x \to f(., x)$ of $X$ into $\mathscr{F}_\mathscr{S}(T; Y)$ is continuous at a point $x_0 \in X$ (resp. uniformly continuous) if and only if the set $H_A$ is equicontinuous at $x_0$ (resp. uniformly equicontinuous) for all $A \in \mathscr{S}$.

Consider first the particular case where $\mathscr{S} = \{ T \}$, i.e. $\mathscr{F}_\mathscr{S}(T; Y) = \mathscr{F}_u(T; Y)$. For each entourage $V$ of $Y$, the condition $(f(., x), f(., x')) \in W(V)$ signifies that $(f(t, x), f(t, x')) \in V$ for all $t \in T$. To say that $x \to f(., x)$ is continuous at $x_0$ (resp. is uniformly continuous) is therefore equivalent to saying that, for each entourage $V$ of $Y$, there is a neighbourhood $U$ of $x_0$ in $X$ (resp. an entourage $M$ of $X$) such that the relation $x \in U$ [resp. $(x, x') \in M$] implies $(f(t, x), f(t, x_0)) \in V$ [resp. $(f(t, x), f(t, x')) \in V$] for all $t \in T$, and the proposition follows from Definitions 1 and 2. In the general case, we have to express that, for each $A \in \mathscr{S}$, the mapping $x \to f(., x)|A$ of $X$ into $\mathscr{F}_u(A; Y)$ is continuous at $x_0$ (resp. uniformly continuous), by virtue of § 1, no. 2; from what has been said, this is equivalent to saying that, for each $A \in \mathscr{S}$, $H_A$ is equicontinuous at $x_0$ (resp. uniformly equicontinuous).

Proposition 1 allows us to translate Definitions 1 and 2 into forms which are sometimes useful, by applying it to the case where $T = H$ and $f$ is the mapping $(h, x) \to h(x)$ of $H \times X$ into $Y$; since $f(., x)$ is the mapping $h \to h(x)$ of $H$ into $Y$, we see that: