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I am working on the book "A Short course on Spectral Theory", written by William Arveson.

On page 28, there is Remark 1.10.2 stated that:

Every compact convex set $K \subseteq \mathbb{C}$ is the intersection of all closed half-space that contain it. It is also true that $K$ is the intersection of all closed disk that contain it. Equivalently, if $z_0 \in \mathbb{C}$ is any point not in the closed convex hull of $K$, then there is a disk $D = D_{a, R} = \left \{ z \in \mathbb{C} : > |z-a| \le R\right \}$ such that $K \subseteq D$ and $z_0 \neq D$

I tried to prove this remark in detail and sketch an illustration, however, I could not.

Thank you for your help.

Trần Linh
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  • Every subset of $\mathbb C$ is a subset of the intersection of all closed half spaces containing it. That part is easy. Next you have to prove the inclusion in the opposite direction. And that is where compactness and convexity will be used. – Michael Hardy Apr 07 '18 at 15:13
  • Actually, I got trouble in that direction. Could you please help me in more detail? – Trần Linh Apr 07 '18 at 15:15
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    The first is true for any closed convex set in any linear space, it is the intersection of all the half-spaces that contain it. Basic convex analysis. Circles with a large enough radius are almost lines. – Lutz Lehmann Apr 08 '18 at 06:26

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Hint: If $z_0\notin K$, let $z_1\in K$ be a point in $K$ closest to $z_0$. (It is unique, but you don't need that for the proof.) Look at the normal bisector of the line segment between $z_0$ and $z_1$.

Harald Hanche-Olsen
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