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.

  • $\begingroup$ A finite set $K$ is compact. (For french mathematicians all compact is closed; I know this is not so in USA) $\endgroup$
    – Piquito
    Dec 30, 2021 at 15:47
  • 2
    $\begingroup$ 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. $\endgroup$
    – Saúl RM
    Dec 30, 2021 at 16:01
  • $\begingroup$ math.stackexchange.com/questions/274810/… $\endgroup$
    – orangeskid
    Dec 30, 2021 at 21:10

1 Answer 1


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, $d(o_n,y)<d(o_n,p)$, 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, $d(o_n,y)<d(o_n,p)$, 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.


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