Maximal Natural Domain of An Analytic Function on Complex Plane

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.

• What is your definition of analytic on non-open sets? – mrf Nov 21 '18 at 6:13
• @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 $f$. – The Hong Kong Journalist Nov 21 '18 at 6:18
• @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 $\mathbb{C}^*$ – Glougloubarbaki Nov 26 '18 at 12:12