Proof. For any Cauchy filter $\fF$ in $\prod_{i \in I}\wh X_{i}$, $\pi_{i}(\fF)$ is a Cauchy filter base by (1) of Definition 5.2.5 and Proposition 5.4.6. For each $i \in I$, $\wh X_{i}$ is complete, so there exists $x_{i} \in \wh X_{i}$ such that $\pi_{i}(\fF)$ converges to $x_{i}$. Let $x \in X$ such that $\pi_{i}(x) = x_{i}$ for all $i \in I$, then $\fF$ converges to $x$ by Proposition 4.7.2. Therefore $\prod_{i \in I}X_{i}$ is complete.$\square$