0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I believe you are misinterpreting the statement. $\endgroup$ – Alan Muniz Sep 8 '18 at 15:38
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.