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If we should speak of a function f that is analytic in a set $S$ which is not open, it is to be understood that if is analytic in an open set $U$ containing $S$

Is it correct to illustrate this in a broken (dotted/dashed) circle with center $z_0$ (for example) that is inside another bigger circle with boundary not dashed/broken?

(I can't post my illustration because you know) Thanks

Malkin
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frstrtdmthmtcn
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  • Your question is quite unclear. Since $ {|z| \le 1}$ is compact then "$f$ is analytic on an open containing $ {|z| \le 1}$" is the same as "$f$ is analytic on $|z| < 1+\epsilon$ for some $\epsilon > 0$". More generally if a compact $\Omega \subset U$ open then the minimal distance between $\partial \Omega$ and $\partial U$ is $ \ge \epsilon > 0$ – reuns Nov 03 '17 at 05:33

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Yes, it is usual for an open set (in $\mathbb{R}^2$ or $\mathbb{C}$) to be sketched as a region with a dashed outline and a closed set to be sketched with a solid outline.

You do not need to use circles to sketch this setup, but if you wanted to, it might look like:pic

Malkin
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