I was reading statement of Cauchy integral formula.
In that there is assumption that f is holomorphic in open set that contain closure of a disc D.
I know that above assumption is important as we required compactness to have uniform continuty of function (i.e to exist suprema on that set).
But I thought this condition is directly done for every open set in which that function f is holomorphic.I do not find any counterexample except that open set itself as disk
f is holomorphic on open set such that there is disk completly inside whose closure does not in that open set.
I know that if That open set is itself disk then above fails but is there exist any other non trivial example.
Any help will be appreciated

  • $\begingroup$ I believe you are misinterpreting the statement. $\endgroup$ – Alan Muniz Sep 8 '18 at 15:38
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    $\begingroup$ Your statement in boldfont is quite convoluted, at least for my level of understanding of English language. You may want to rephrase it with more notation an less pronouns. $\endgroup$ – Saucy O'Path Sep 8 '18 at 15:38

Take the open set to be $U = \{z \mid |z| < 1\}$ and the open disc to be $D = \{ z \mid |z - 1/2| < 1/2\}$. The point $1$ is in the closure of $D$ but not in $U$.


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