Let $\Delta$ be an $n-1$-simplex in ${\mathbb R}^{n-1}$. For each vertex $v$ of $\Delta$ let $H_v$ be the hyperplane through $v$ and parallel to the opposite facet. By 2-neighborhood of a simplex I mean a simplex which is the intersection of the half-spaces $H_v^+$ containing $\Delta$. My question is:

Let $S$ be a collection of $2n - 4$ distinct points on ${\mathbb R}^{n-1}$ spanning ${\mathbb R}^{n-1}$. Does there exist $n$ points in $S$ so that the $2$-neighborhood of the simplex formed by these $n$ points containing $S$?

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    $\begingroup$ In the first sentence, did you intend that $\Delta$ is a simplex with $n$ vertices --- which would ordinarily be called an $(n-1)$-simplex? An $n$-simplex, in the usual meaning of the phrase, won't fit into $\mathbb R^{n-1}$. $\endgroup$ – Andreas Blass Nov 30 '13 at 14:44
  • $\begingroup$ Just take the $n$ points forming a simplex of the maximal volume. This works not only for $2n-4$ points. $\endgroup$ – Ilya Bogdanov Nov 30 '13 at 15:49
  • $\begingroup$ I modified as suggested by Andreas. Why the $2$-neighborhood of simplex with maximal volume contains all the points? That does not seem clear to me. $\endgroup$ – Thanh Vu Nov 30 '13 at 23:59

$\def\Vol{\mathop{\rm Vol}}$ Take $a_1,\dots,a_n\in S$ such that $\Vol(a_1,\dots,a_n)$ attains the maximum. Now, if some $b\in S$ lies outside, say, $H_{a_n}^+$, then $\Vol(a_1,a_2,\dots,a_{n-1},b)>\Vol(a_1,\dots,a_n)$ which is impossible. Thus $b$ lies in the intersection of all such halfspaces.

This works for every finite set $S$.

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  • $\begingroup$ No, this only true if all these remaining points lie on the same hyperplane as $a_n$. But certainly it might not be the case. Your proof only shows that the points lie in a slightly larger neighborhood, (maybe $2$-neighborhood of $2$-neighborhood.) $\endgroup$ – Thanh Vu Dec 1 '13 at 20:36
  • $\begingroup$ I meant it could happen that $b$ is on $H_{a_n}^-$, in this case, you cannot say much about the position of $b$ relative to the $2$-neighborhood of $\Delta$. $\endgroup$ – Thanh Vu Dec 1 '13 at 20:46
  • $\begingroup$ I do not understand, sorry. The halfplane $H_{a_n}$ passes through $a_n$; for every point $b\in H_{a_n}$, the volume of $(a_1,\dots,a_{n-1},b)$ is the same, and IF $b\in H_{a_n}^-$ THEN this volume becomes larger, doesn't it? $\endgroup$ – Ilya Bogdanov Dec 1 '13 at 21:35
  • $\begingroup$ Sorry, I imagine the other side of the facet $a_1...a_{n-1}$. My thought was that is this possible to find such a simplex, whose all the points are either in $\Delta$ or in one of its reflection via the facets. (It is the $2$-neighborhood of $\Delta$ in the plane, but not in higher space I think). That caused me problems in understanding your argument. $\endgroup$ – Thanh Vu Dec 1 '13 at 21:43

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