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(This is a cross-post from here.) Let $A,B\subseteq \mathbb R^d$ be non-empty, such that $B\subseteq \overline A.$ For $S\subseteq\mathbb R^d$ define the relative interior of $S$ by $$\text{ri}(S)=\{s\in S\mid \exists\varepsilon>0:B_\varepsilon(s)\cap\operatorname{aff}(S)\subseteq S\}.$$ Here, $B_\varepsilon(x)$ is the open $d$-dimensional ball with center $x$, $\operatorname{aff}$ denotes the affine hull, $\operatorname{ri}$ denotes the relative interior and $\operatorname{conv}$ denotes the convex hull.

  1. Do we have $\operatorname{conv}(A)\cap\operatorname{ri}(\text{conv}(B))\neq\{\emptyset\}?$
  2. Or do we even have $A\cap \text{ri}(\text{conv}(B))\neq\{\emptyset\}?$

My intuition is the following (I failed proving it this way): We know that $\text{ri}(\text{conv}(B))$ is non-empty, hence let $x\in\text{ri}(\text{conv}(B)).$ Now $x$ can be approximated by elements in $\text{conv}(\overline A)$, hence also by elements in $\text{conv}(A).$ Choosing a convex combination $a$ close enough to $x$, this should be our solution.

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    $\begingroup$ First of all, your assumption that $\operatorname{ri}(\operatorname{conv}(B))$ is non-empty is wrong, for example, when $B$ is a single point. Now, even if we assume that $\operatorname{ri}(\operatorname{conv}(B))$ is non-empty, consider the situation that $A$ is the interior of the unit square and that $B$ is an edge of the unit square. Neither (1) or (2) holds. $\endgroup$ – Yoav Kallus Nov 18 '13 at 17:27
  • $\begingroup$ I think you are mixing up interior with relative interior. The relative interior of a non-empty convex set is always nonempty!!! $\endgroup$ – andy teich Nov 18 '13 at 17:39
  • $\begingroup$ But your second argument seems to be plausible... $\endgroup$ – andy teich Nov 18 '13 at 17:49
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    $\begingroup$ I guess you're right that under your definition, the relative interior is non-empty. Sorry. But I think my second example still works. $\endgroup$ – Yoav Kallus Nov 18 '13 at 18:29
  • $\begingroup$ @Kallus: you want to give this as an answer? $\endgroup$ – andy teich Nov 19 '13 at 1:19
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As noted in the comments, the case where $A\subseteq\mathbb{R}^2$ is the interior of the unit square and $B$ is a side of that square satisfies the assumptions posed, but in that case neither (1) nor (2) holds. This example is easy to generalize, of course, and perhaps will lead you towards the claim you really want to make. Good luck.

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