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I want to ask a question of the maximal natural domain of an analytic function. We know that on $\mathbb{C}^n$, a domain $U$ is a domain of holomorphy if and only if $U$ is a pseudoconvex domain, and in particular in complex plane, we have that every open set is pseudoconvex.

Here I want to ask for the converse for $\mathbb{C}$:

Q: For an analytic function $f$, is the maximal set $W$ for which $f$ can be extended necessarily open? And moreover, does this generalize in $\mathbb{C}^n$?

Notice that for some particular cases, this does not raise a question. For instance, if $f$ is defined on some closed set $W$ with $C^1$ boundary(say, a Jordan curve), then an application of Caratheodory theorem and Schwartz reflection could be enough.

It might be a stupid question, but it seems that I could not find a counterexample or a proof yet.

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    $\begingroup$ What is your definition of analytic on non-open sets? $\endgroup$ – mrf Nov 21 '18 at 6:13
  • $\begingroup$ @mrf I guess for this case, I am considering $\pdv{f}{z}$ is still bounded when $z_j \to z$ for $z_j \in W$, where $W$ is a domain for $f$. $\endgroup$ – The Hong Kong Journalist Nov 21 '18 at 6:18
  • $\begingroup$ @TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $\mathbb{C}^*$ $\endgroup$ – Glougloubarbaki Nov 26 '18 at 12:12

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