Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I wonder how to prove the following statement.

Let $V$ be a $d$-dimensional normed space with $d \geq 3$, let $P \subset V$ be a $(d-2)$-dimensional polytope. Then there is an $\epsilon > 0$ such that for every affine $(d-3)$-dimensional affine subspace $A \subset V$, there is a point $p_A \in P$ with $\Vert p_A - a \Vert \geq \epsilon$ for every $a \in A$.

Does anyone have an idea?


I'm left to show the following: Let $V$ be a $d$-dimensional normed space with $d \geq 3$, let $S \subset V$ be a $(d-2)$-dimensional affine subspace, let $B_r(s)$ be the closed ball with radius $r$ around $s \in S$, let $C = S \cap B_r(s)$. Then for every affine $(d-3)$-dimensional subspace $A$, there is a point $p_A \in C$ with $\Vert p_A - a \Vert \geq r$ for every $a \in A$.

Any idea how to show this?

share|improve this question
Did you mean $P\subset V$? It doesn't make sense to introduce $A$ and then say "for every $A$". –  joriki May 16 '12 at 12:15
Yes, thanks. I've edited my question accordingly. –  Tom Jonathan May 16 '12 at 13:23
Rough idea: (1) $P$ contains a $d-2$-dimensional simplex $C$; (2) Let $M$ be the affine span of $C$; (3) $C$ has nonempty interior relative to $M$, thus contains a ball $B(x,\epsilon)\cap M$; (4) given $A$ as in the problem, pick a unit vector $v\in A^\perp$ and consider $x+\epsilon v\in C$. –  user31373 May 16 '12 at 14:27
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.