Proposition 6.1.5. Let $T$ be a set, $\mathfrak{S}\subset 2^{T}$ such that $\bigcup_{S \in \mathfrak{S}}S = T$, and $(X, \fU)$ be a complete uniform space, then $X^{T}$ equipped with the $\mathfrak{S}$-uniformity is complete.
Proof. Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \mathfrak{S}$ with $x \in S$, then $\pi_{x}(\fF) \subset 2^{X}$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_{x}(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \mathfrak{S}}S$.
Let $S \in \mathfrak{S}$ be non-empty and $V_{0}, V$ be symmetric entourages of $X$ with $V_{0} \circ V_{0} \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_{0})$-small set $F \in \fF$. Let $x \in S$, then $\pi_{x}(\fF) \to f(x)$ implies that $F_{x} = \pi_{x}^{-1}(V_{0}(f(x))) \in \fF$. Since $F \cap F_{x} \ne \emptyset$, $(f(x), g(x)) \in V_{0} \circ V_{0} \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.$\square$