Proposition 6.4.4.label Let $\seqi{X}$ be totally bounded uniform spaces, then $\prod_{i \in I}X_{i}$ is totally bounded.
Proof. Let $J \subset I$ be finite and $\seqj{U}$ such that $U_{j}$ is an entourage of $X_{j}$ for each $j \in J$. Since each $X_{j}$ is totally bounded, there exists $Y_{j} \subset X_{j}$ finite such that $X_{j} = U_{j}(Y_{j})$. Let $Y = \prod_{j \in J}Y_{j}$ and $\phi_{y} \in \pi_{J}^{-1}(y)$ for each $y \in Y$, then for any $x \in X$, there exists $y \in Y$ such that $\pi_{j}(x) \in U_{j}(\pi_{j}(y))$ for each $j \in J$. Therefore
\[X = \bigcup_{x \in Y}\braks{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)}(\phi_{x})\]
$\square$