# If a compact set is not convex, some ball is tangent to it at several points.

Let $$K\subseteq\mathbb{R}^n$$ be compact (probably $$K$$ being closed will be enough). If $$K$$ is not convex, is there necessarily an open ball disjoint with $$K$$ but tangent to it at more than one point?

Edit: I found a solution (it works with $$K$$ closed), although it is more complicated than I expected, simpler proofs would be welcome.

• A finite set $K$ is compact. (For french mathematicians all compact is closed; I know this is not so in USA) Dec 30, 2021 at 15:47
• I think compact sets in $\mathbb{R}^n$ are closed for anyone. But what happens with finite sets? If that is the misunderstanding, with 'tangent' I mean that the balls have no points of $K$ in the interior but they have points of $K$ in their boundary. Dec 30, 2021 at 16:01
• math.stackexchange.com/questions/274810/… Dec 30, 2021 at 21:10

After thinking a bit more, here is a proof that some open ball tangent to $$K$$ at several points has to exist.

Suppose no ball is tangent to $$K$$ at more than $$1$$ point. This implies that for every $$x\in\mathbb{R}^n$$, there is exactly one point $$y\in K$$ with $$d(x,y)=d(x,K)$$.

Now let $$D$$ be the convex closure of $$K$$, which we can suppose has interior (if not, we can change $$\mathbb{R}^n$$ by the affine subspace generated by $$K$$), and pick a point $$p$$ in the interior of $$D$$ but outside $$K$$.

Lemma: If $$p$$ is in the interior of $$D$$, then there is some $$R>0$$ such that every sphere of radius $$>R$$ and disjoint from $$K$$ does not contain $$p$$.

Proof of the lemma: Suppose $$R$$ does not exist, and take $$B_n$$ a sequence of balls of radius tending to infinity, with $$B_n$$ disjoint from $$K$$ and containing $$p$$. Let $$o_n$$ be the center of $$B_n$$ for each $$n$$. Taking a subsequence if necessary, we can suppose that $$p\neq o_n\;\forall n$$ and that the vectors $$\frac{\overrightarrow{po_n}}{|\overrightarrow{po_n}|}$$ converge to a unitary vector $$v$$. Then we will prove that the union of the balls $$B_n$$ contain the open half space of points $$y$$ with $$\overrightarrow{py}\cdot v>0$$. To check this, take a point $$y$$ with $$\overrightarrow{py}\cdot v>0$$. We want to see that for $$n$$ big enough, $$d(o_n,y), so that $$y$$ is in $$B_n$$. So, calling $$v_n=\overrightarrow{po_n}$$, we need to prove that $$|v_n|>|v_n-\overrightarrow{py}|$$ for big $$n$$. Squaring both sides, $$|v_n-\overrightarrow{py}|^2=|v_n|^2-2v_n\cdot \overrightarrow{py}+|\overrightarrow{py}|^2.$$ But $$v_n\cdot \overrightarrow{py}=|v_n|\frac{v_n}{|v_n|}\cdot\overrightarrow{py}$$, which tends to $$\infty$$, because $$|v_n|$$ tends to infinity and $$\frac{v_n}{|v_n|}\cdot\overrightarrow{py}$$ tends to $$v\cdot \overrightarrow{py}>0$$. So, for $$n$$ big enough, $$d(o_n,y), and we have proved that there is an open half space disjoint with $$K$$ and with $$p$$ in its boundary. This contradicts the fact that $$p$$ is in the interior of the convex closure of $$K$$, so we are done with the lemma. $$\square$$

So, pick the minimum $$R$$ such that any ball of radius $$>R$$ disjoint with $$K$$ cannot contain $$p$$, and for each $$n$$ pick an open ball $$B_n$$ of radius $$R-\frac{1}{n}$$ disjoint with $$K$$ and containing $$p$$. Taking a subsequence if necessary, we can suppose that the sequence $$o_n$$ of centers of $$B_n$$ is convergent to some point $$o$$, which will be at distance $$\geq R$$ from $$K$$ and at distance $$\leq R$$ from $$p$$. In fact $$o$$ has to be at distance $$R$$ from $$K$$, if not $$p$$ would be in a ball of radius $$>R$$ (and center $$o$$) disjoint from $$K$$.

Let $$q$$ be the unique point of $$K$$ at distance $$R$$ from $$o$$, and call $$B$$ the closed ball of center $$o$$ and radius $$R$$. Then, for $$\varepsilon>0$$ small enough, the closed ball $$B+\varepsilon\overrightarrow{qp}$$ is disjoint with $$K$$ and contains the point $$p$$. There is a positive distance from $$B+\varepsilon\overrightarrow{qp}$$ to $$K$$, so increasing its radius a bit, we find that there are balls of radius $$>R$$ disjoint from $$K$$ and containing $$p$$, a contradiction.