Is the cartesian product of locally convex space again locally convex?

Question

Let $E_{1},...,E_{n}$ be locally convex spaces. Is the cartesian product $E_{1}\times \cdots \times E_{n}$ locally convex when equipped with the product topology? More generally, is the infinite/uncountable cartesian products $\prod_{n\in \mathbb{N}}E_{n}$ and $\prod_{\alpha \in I}E_{\alpha}$ of locally convex spaces also locally convex spaces when equipped with the product topology? I've been trying to find the answer to these questions but I couldn't find anywhere. I'd appreciate if the proof/sketch of the proof or references for the answers are provided.

Answer 1

It is a fact of general topology that if for all $\alpha$ and $x\in(X_\alpha,\tau_\alpha)$ we select $\mathcal B_{x,\alpha}$ a neighbourhood basis of $x$, then in the product topology $\prod_{\alpha\in A} (X_\alpha,\tau_\alpha)$ the family $$\mathcal B_y=\left\{\prod_{\alpha\in A}V_\alpha\,:\, \exists S\in \mathcal P_{f}(A),\, ((\forall \alpha\in S,V_\alpha\in \mathcal B_{y_\alpha,\alpha})\land(\forall \alpha\notin S, V_\alpha=X_\alpha))\right\}$$

is a neighbourhood basis of $y$ (for the record, by $\mathcal P_f(A)$ I mean the set of finite subsets of $A$).

Applying this to the case at hand, where $X_\alpha=E_\alpha$ and the $\tau_\alpha$-s are locally convex, if we select all the $\mathcal B_{0,\alpha}$-s to be families of convex subsets and we recall that product of convex subsets is convex in the product vector space, we obtain that $\mathcal B_{0}$ is a local basis of convex neighbourhoods of $0$.