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How can one prove that $ \text{int}(A) = \text{int(cl}(A))$, where $ A \subseteq \mathbb{R}^n$ is convex?

Ashot
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Since ${\rm cl}(A)\supseteq A$, clearly ${\rm int}({\rm cl}(A))\supseteq {\rm int}(A)$. A point $p$ in ${\rm int}({\rm cl}(A))$ is surrounded by some open ball $B$ of points in ${\rm cl}(A)$. Pick $n+1$ points $q_1$, $\dots$, $q_{n+1}$ that form the vertices of a regular $n$-simplex, contained in $B\subseteq {\rm cl}(A)$, with $p$ at its center. For points $q'_1$, $\dots$, $q'_{n+1}\in A$ such that each $q'_i$ is sufficiently close to the corresponding $q_i$, $p$ will also be in the interior of the simplex with vertices $q'_1$, $\dots$, $q'_{n+1}$. Then, since $A$ is convex, $A$ completely contains the interior of this simplex, so $p\in {\rm int}(A)$. Therefore ${\rm int}({\rm cl}(A))\subseteq {\rm int}(A)$.

David Moews
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  • +1: Nice. I wonder if it is true in a topological vector space? – copper.hat Mar 01 '13 at 19:22
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    No. For example, let $\ell^1$ be the space of all sequences of reals whose sum is absolutely convergent, with the usual $\ell^1$ norm, and let $A\subseteq \ell^1$ be the set of all sequences which are eventually zero. Then $A$ is convex, dense in $\ell^1$, and has empty interior. – David Moews Mar 01 '13 at 23:14
  • Very nice example. – copper.hat Mar 01 '13 at 23:58
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The inclusion $\subseteq$ is trivial. For $\supseteq$, suppose $x\in \operatorname{int}(\operatorname{cl}(A))$. Then an open ball $B_r(x)$ of radius $r>0$ around $x$ is $\subseteq \operatorname{cl}(A)$. Then for any choice of signs $ e = (e_1,\ldots, e_n)\in\{-1,1\}^n$ and $0<c<\frac r{\sqrt n}$, the point $x+ce=x+(e_1c, \ldots , e_nc)$ is also in this ball, hence in $\operatorname{cl}(A)$. For each of these $2^n$ points we can find a point $y_e\in A$ with $|x+ce-y_e|<c$. Note that the $i$th component $(y_e)_i$ of $y_e$ is $>x_i,<x_i$ iff $e_i>0, e_i<0$, respectively. Let $d=\min_{e,i}\{|(y_e)_i-x_i|\}>0$. Consider $x'$ with $|x'-x|<d$. Assume that $x'$ is not in the convex hull of the $2^n$ points $y_e$, $e\in \{-1,1\}^n$. Then there exists a nonzero vector $v\in\mathbb R^n$ such that $\langle y_e-x',v\rangle\le0$ for all $e$. Consider $e$ with $e_i>0$ if $v_i\ge 0$ and $e_i<0$ if $v_i<0$. Then $\langle y_e-x',v\rangle>0$ because each product $((y_e)_i-x'_i)v_i$ is nonnegative and at least one is positive. From this contradiction, we see that $x'$ is in the convex hull of the $y_e$, hence $x'\in A$. Consequently, $x\in \operatorname{int}(A)$.

Hagen von Eitzen
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  • May I ask why such a nonzero $v$ must exists? – Alphie Aug 21 '21 at 14:17
  • I think this follows from the separating hyperplane theorem between a nonempty closed convex set $C$ (here the convex hull of the $y_e$s) and a point not in $C$ – Alphie Aug 21 '21 at 17:53