Let $E_\alpha$ be a family of complete topological vector spaces over $\Bbb K=\Bbb R\hbox{ or }\Bbb C$ for all $\alpha\in A$, that is to say, any Cauchy net is convergent. Then Helmut Shaefer claims in his book Topological Vector Spaces that $E:=\prod_{\alpha\in A}E_\alpha$ endowed with the box topology must then be complete (as a topological abelian group, since it is generally not a topological vector space). However, while this is easy for the product topology, I'm having difficulty proving that in this case, the different $\alpha$-terms converge sufficiently "uniformly".
More precisely, if we let $x^\lambda=(x_\alpha^\lambda)_{\alpha\in A}\subset\prod_{\alpha\in A}E_\alpha$ for all $\lambda\in\Lambda$ where $\Lambda$ is a directed poset, then if $(x^\lambda)_{\lambda\in\Lambda}\subset\prod_\alpha E_\alpha$ is Cauchy, then so is $(x_\alpha^\lambda)_\lambda\subset E_\alpha$ and therefore $(x^\lambda)_\lambda$ must converge pointwise to some $x\in\prod_\alpha E_\alpha$. Unfortunately, I'm not seeing precisely how to use the Cauchy criterion and the box topology to prove convergence in the box topology.
Let $U = \prod_{\alpha} U_{\alpha}$ a neighbourhood of $0$. For each $\alpha$, choose a symmetric neighbourhood $V_{\alpha}$ of $0$ in $E_{\alpha}$ such that $V_{\alpha} + V_{\alpha} \subset U_{\alpha}$. Let $V = \prod_{\alpha} V_{\alpha}$. Since $x^{\lambda}$ is a Cauchy net, there is a $\lambda_0$ such that for all $\alpha$ we have $x^{\lambda}_{\alpha} - x^{\mu}_{\alpha} \in V_{\alpha}$ for all $\lambda, \mu \geqslant \lambda_0$. Since $x^{\lambda}_{\alpha} \to x_{\alpha}$ it follows that $x_{\alpha} - x^{\mu}_{\alpha} \in \overline{V_{\alpha}} \subset U_{\alpha}$ for all $\alpha$ and all $\mu \geqslant \lambda_0$. But that means $x - x^{\mu} \in U$ for all $\mu \geqslant \lambda_0$.