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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.

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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 $\prod C$.

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  • $\begingroup$ What is $a_s$? What means $s \in C$? $\endgroup$ – Guilherme de Loreno Jun 21 at 1:16
  • $\begingroup$ @GuilhermedeLoreno $a_s$ is a point in s, one of the convex spaces of the collection C of convex spaces. $\endgroup$ – William Elliot Jun 21 at 7:28
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$E$ 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.

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