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Let $M\subset \mathbb{R}^{n+1}$ be a compact $n$-manifold. There exists, then, a smallest $n$-sphere containing $M$, and it must touch it in one point.

Must it touch it twice?

This seems quite intuitively right to me, but I've no idea how to prove it. It's easy to construct counterexamples where you can't have more than 2 (e.g. an ellipse which is not a circle).

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What about considering $x,y\in M$ such that $\|x-y\| = \mathrm{diam}(M)$? – Ilya Jan 31 '13 at 16:02
@Ilya: I'm sorry, I don't get your point... Could you elaborate? Thanks. – lentic catachresis Jan 31 '13 at 16:23
I'm not experienced in diff.geom so I can't come up with a formal proof. The idea I head is that there are two points $x,y$ on which the diameter $d$ of $M$ is attained. Then $M\subset B(x,d)$ and touches it at $y$ as it seems from here and vice-versa. I thought, that using such information (provided it is correct) you can show that there is a sphere in the intersection of $B(x,d) \cap B(y,d)$ which contains the whole $M$. – Ilya Jan 31 '13 at 16:40
Dear @Ilya, I really liked your idea, so I'm sorry, but in trying to complete your argument I detected a problem. Even the biggest sphere in the intersection, which is the sphere having $x$ and $y$ as antipodal points, can miss points of $M$. This sphere is the sphere around $(x+y)/2$ of radius $d/2$. But there may be points with strictly bigger distance than $d/2$. – Ben Jan 31 '13 at 18:30
up vote 7 down vote accepted

Yes. If not, let a smallest sphere containing $M$ touch $M$ at only $P$ and have center $O$. The sphere can be divided into two hemispheres one of which, the northern, say, has $P$ as north pole. The closed southern hemisphere, $S$, does not intersect $M$ so $\epsilon:=d(S,M)>0$. Translate the sphere $\epsilon/2$ in the direction $OP$. The translate of $S$ is still distance $\epsilon/2$ or more away from $M$, so it does not intersect $M$. The translate of the closed northern hemisphere is contained in the region formerly outside the sphere, so it does not intersect $M$ either. Therefore the translated sphere is not smallest, so neither was the original sphere.

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Very nice! (and other characters so the comment is long enough to be posted) – Jason DeVito Jan 31 '13 at 18:21

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