Interior of the closure of a convex subset in topological vector space

Question

Let $E$ be a real topological vector space and $C\subset E$ be a convex subset. Is this true that $\mathrm{Int}(\bar{C})=\mathrm{Int}(C)$? Here $\bar{C}$ is the closure of $C\subset E$ and $\mathrm{Int}$ is the set of interior points.

We know that the statement holds for finite dimensional $E$. cf. Why does a convex set have the same interior points as its closure?

This proof in your linked question holds on any topological vector space. — Benjamin, Sep 19 '21 at 09:18

score 2 · Accepted Answer · edited Jun 25 '22 at 12:13

2

If $M$ is any dense proper subspace of a normed linear space then $Int (M)$ is empty and $int(\overline M)$ is the whole space.

Note that such a subspace $M$ does not exist in a finite dimensional case.

edited Jun 25 '22 at 12:13

Tian Vlasic

1,436

answered Sep 19 '21 at 09:20

Kavi Rama Murthy

359,332

Interior of the closure of a convex subset in topological vector space

1 Answers1