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Let $X$ be a topological vector space. We recall that a set $A\subseteq X$ is algebraically if, for each $x,y \in X$, the set $\{t \in \mathbf{R}: x+ty \in A\}$ is open. Moreover, it is well known (and easy to prove) that every open set is algebraically open. Also, there are non-open sets which are algebraically open (see e.g. here).

Our question is about the converse, provided that $A$ is, in addition, convex.

Question. Let $A\subseteq X$ a convex algebraically open set. Is it true that $A$ is open?

(The answer is positive if $X=\mathbf{R}^n$.)

Paolo Leonetti
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Let $X=\prod_{i\in \omega}\Bbb R$. Under the product topology, an open set is the union of sets of the form $\prod_{i\in \omega} U_i$ where almost all $U_i=\Bbb R$. In particular, $A:=\prod_{i\in \omega}\left]0,1\right[$ is not open. But it is convex and algebraically open.

Hagen von Eitzen
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