Let $X=\{x_1,...,x_k\}\subset E^n$ be a finite subset in the Euclidean $n$-space, $r>0$ and $B(x_i,r)$ are open balls of radius $r$ centered at the points $x_i\in X, i=1,...,k$. Suppose that $$ \bigcap_{i=1}^k B(x_i,r)\ne \emptyset. $$ Is it true that the convex hull of $X$ is contained in $$ \bigcup_{i=1}^k B(x_i,r) ? $$ I needed a version of this in some Riemannian geometry argument and while I found a way around, the proof would have been a bit cleaner if this statement were true.
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$\begingroup$ By Caratheodory's theorem it is enough to consider only the case when $X$ has $n+1$ points. $\endgroup$– Paata IvanishviliOct 8, 2016 at 2:40
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2$\begingroup$ @PaataIvanisvili just a note: since dimension $n$ is arbitrary, this does not look very helpful, we may always assume that $|X|\leqslant n+1$ simply by embedding $E^n\subset E^{|X|-1}$. This is why we could expect a dimension-free proof, like that given below, a priori. $\endgroup$– Fedor PetrovOct 8, 2016 at 8:20
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$\begingroup$ Thanks. It was only helpful to verify the statement in $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$. But in general I agree with you. $\endgroup$– Paata IvanishviliOct 8, 2016 at 16:16
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Yes. Any point $y$ in the convex hull of $x$'s is a barycenter of some non-negative masses $m_i$ in $x_i$, $\sum m_i=1$, $y=\sum m_i x_i$. Point $y$ minimizes the moment of inertia $I(p)=\sum m_i |p-x_i|^2$, $p\in E^n$, just because $I(p)=I(y)+|p-y|^2$. In particular, it can not happen that all distances $|y-x_i|$ are at least $r$, but for some other point $p$ all distances $|p-x_i|$ are less than $r$.
answered Oct 8, 2016 at 6:08