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In response to this question, is it true that for $f:\mathbb{C}-A\to \mathbb{C}$ analytic where $A$ is countable (but not closed!) that $f$ bounded implies $f=c$?

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  • $\begingroup$ oops, $A$ should be countable. $\endgroup$
    – edo arad
    Commented Jul 28, 2013 at 18:54
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    $\begingroup$ Analytic functions are defined on open sets, so $A$ needs to be closed. $\endgroup$ Commented Jul 28, 2013 at 18:55
  • $\begingroup$ Thanks! I should have figured it out.. $\endgroup$
    – edo arad
    Commented Jul 28, 2013 at 18:56
  • $\begingroup$ I think you could have left it as an answer $\endgroup$
    – edo arad
    Commented Jul 28, 2013 at 18:57
  • $\begingroup$ Edited and undeleted. My previous answer was for uncountable $A$. $\endgroup$ Commented Jul 28, 2013 at 18:58

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Analytic functions are defined on open sets, so $A$ needs to be closed.

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