# The product of an arbitrary family of locally convex spaces is locally convex.

Let $$\{E_\alpha\ : \ \alpha\in I\}$$ be a family of a locally convex sets, where $$I$$ is an index family. I want to prove that $$E:= \prod_{\alpha\in I}E_\alpha$$ is locally convex.

I know that, by definition, for each $$\alpha\in I$$, $$E_\alpha$$ is locally convex, that is, $$E_\alpha$$ is topological vector space such that there is a basis of neighborhoods in $$E_\alpha$$ consisting of convex sets. I also know that I must prove that there is a basis of neighborhoods in $$E$$ formed by convex sets, but I do not know how to prove it from hypotheses.

Let $$C$$ be a collection of convex spaces. Let $$a = \prod_{s \in C}a_s$$ and $$b = \prod_{s \in C}b_s$$ be two points in $$\prod C$$.
Show that the line from $$a$$ to $$b$$ is within $$\prod C$$.
• What is $a_s$? What means $s \in C$? – Guilherme de Loreno Jun 21 at 1:16
• @GuilhermedeLoreno $a_s$ is a point in s, one of the convex spaces of the collection C of convex spaces. – William Elliot Jun 21 at 7:28
$$E$$ has a base from the product topology, i.e. every open subset is a union of basic open sets which are of the form $$\prod_{\alpha \in I} O_\alpha$$ where all $$O_\alpha \subseteq E_\alpha$$ are open and there is a finite set $$F$$ of indices such that $$O_\alpha = E_\alpha$$ for all $$\alpha \notin F$$.
Now show that this remains a base if we take all non-trivial $$O_\alpha$$ to be convex open as well, and that the resulting basic open set is then also convex.