# covering convex sets by round balls

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.

• By Caratheodory's theorem it is enough to consider only the case when $X$ has $n+1$ points. Oct 8, 2016 at 2:40
• @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. Oct 8, 2016 at 8:20
• Thanks. It was only helpful to verify the statement in $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$. But in general I agree with you. Oct 8, 2016 at 16:16

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 $$r$$.