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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?

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    $\begingroup$ This proof in your linked question holds on any topological vector space. $\endgroup$
    – Benjamin
    Sep 19, 2021 at 9:18
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    $\begingroup$ Yes, it is true (provided that $int C$ is non empty). $\endgroup$
    – Evangelopoulos Phoevos
    Sep 19, 2021 at 9:20
  • $\begingroup$ But what if $\mathrm{Int}(C)$ is empty? $\endgroup$
    – Doug
    Sep 19, 2021 at 9:21

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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.

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