How does one prove that any closed, convex, and bounded subset of a Hilbert space is the intersection of the closed balls that contain it?

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    $\begingroup$ It seems we could start with separating hyperplanes and fit a ball around the convex set that misses a given external point. $\endgroup$
    – hardmath
    Commented Jan 20, 2013 at 22:00

1 Answer 1


Let $K$ be the convex, closed and bounded set and let $L$ be the intersection of the closed balls containing $K$. Clearly $K\subseteq L$.

If this inclusion is proper, then by translation we may assume that $0\in L\setminus K$.

$K$ is convex and closed, so we can choose a supporting hyperplane $\Pi$ orthogonal to $k\in \Pi\cap K$ so that $K\subseteq\Pi_+=k+\{h\in H:\mbox{Re}(h,k)\ge0\}$. By scaling, we may assume that $\|k\|=1$.

For $r>0$, consider the closed balls $B_r$ of radius $r$ and center $(\frac12+r)k$. We have $\bigcup_{r>0}B_r\supseteq\Pi_+\supseteq K$. Since $K$ is bounded, there is $r>0$ with $B_r\supseteq K$. Since $0\not\in B_r$ by construction, $0\not\in L$, a contradiction.


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