Theorem 5.5.6 (Extension of Uniformly Continuous Functions, [Theorem 1.3.2, Bou13]). Let $(X, \fU)$ be a uniform space, $(Y, \fV)$ be a complete Hausdorff uniform space, $A \subset Y$ be a dense subset, and $f \in C(A; Y)$ be Cauchy continuous, then:
There exists a unique $F \in C(X; Y)$ such that $F|_{A} = f$.
If $f \in UC(A; Y)$, then $F \in UC(X; Y)$.
Proof. (1): By Proposition 5.4.6, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base for all $x \in X$. Thus the existence and uniqueness of the extension follows from Proposition 5.5.5.
(2): Let $V \in \fV$. Using Proposition 5.1.14, assume without loss of generality that $V$ is symmetric and closed. Since $f \in UC(A; Y)$, there exists $U \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in U$. Using Proposition 5.1.18, assume without loss of generality that $U = \overline{U}= \ol{U \cap (A \times A)}$. In which case, since $F$ is continuous,
by Proposition 4.5.3.$\square$