Theorem 2

Let $f$ be a function defined on a dense subspace $A$ of a uniform space $X$, taking its values in a complete Hausdorff uniform space $X'$, and suppose that $f$ is uniformly continuous on $A$. Then $f$ can be extended to the whole of $X$ by continuity, and the extended function $\overline{f}$ is uniformly continuous.

The existence of $\overline{f}$ is an immediate consequence of Propositions 3 and 11 of no. 1. Let us show that $\overline{f}$ is uniformly continuous. Let $V'$ be a closed symmetric entourage of $X'$, and let $V$ be an entourage of $X$ such that, when $x$ and $y$ are in $A$ and are $V$-close, then $f(x)$ and $f(y)$ are $V'$-close. We may assume that $V$ is the closure in $X \times X$ of an entourage $W$ of $A$ ($\S$ 2, no. 4, Proposition 6). We have $[ \bar{f}(x), \bar{f}(y) ] \in V'$ when $(x, y) \in W$; since $\bar{f} \times \bar{f}$ is continuous in $X \times X$ (Chapter I, $\S$ 4, no. 1, Proposition 1) we have also $[ \bar{f}(x), \bar{f}(y) ] \in V'$ when $(x, y) \in V = \overline{W}$, since $V'$ is closed (Chapter I, $\S$ 2, no. 1, Theorem 1).

Q.E.D.