Prove that every holomorphic function has a maximal continuation. I have to show that there exists a maximal holomorphic continuation for every $f : \Omega \rightarrow \mathbb{C}$ holomorphic, with $\Omega \subseteq \mathbb{C}$ open, connected.
We defined a holomorphic continuation as a tuple $(g,\Omega_g)$ where $\Omega \subseteq \Omega_g$ and $g|_\Omega = f$.
I think this is possible by applying Zorn’s Lemma to the set of all holomorphic continuations. I can proof that there exists a partial order on these continuations:
$(g_1, \Omega_1) \preceq (g_2,\Omega_2) \Leftrightarrow g_2|_{\Omega_1} = g_1 \ \land \ \Omega_1 \subseteq \Omega_2$
But I have no idea how to show that there exists an upper bound for every totally ordered subset of these continuations, which would be necessary to apply Zorn‘s Lemma.
 A: There is a maximal extension, but not always a largest extension. The existence of the maximal element follows from Zorn lemma.
$\bf{Added:}$
Your set consists of all $(g, \Omega_g)$ with $\Omega_g$ open connected, $\Omega_g \supset \Omega$ and $g_{|\Omega} = f$.  The ordering is in the OP.  To apply the Zorn lemma, we need to show that every totally  ordered subset
$((g_i, \Omega_i))_{i\in I}$ has an upper bound.  Indeed, consider $\Omega^*\colon = \cup_{i\in I} \Omega_i$. It will be open and connected, since all the subsets $\Omega_i$ contain $\Omega$, so they have a common point. Define now $g^*$ on $\Omega^*$ as follows: let $z \in \Omega^*$. Then $z\in \Omega_i$ for some $i \in I$.  Define $g^*(z) = g_i(z)$.  You need to check that if $z\in \Omega_j$ for another $j$ then $g_j(z) = g_i(z)$. This is because you have $\Omega_i\subset \Omega_j$, or $\Omega_j \subset \Omega_i$, and the way that the $g_i$ restrict. Therefore $g^*$ is well defined. Moreover, since $g^*$ is locally a holomorphic function, it will be holomorphic. Now check that $(\Omega^*, g^*)$ is an upper bound of the family $((\Omega_i, g_i))_{i\in I}$
Note: we need a totally ordered family, or at least a family that is directed.
The problem is  that you may not have a largest domain inside $\mathbb{C}$, think of the function $\sqrt{z}$. A maximal extension is $\sqrt{z}$ defined on $\mathbb{C}$ without a ray going from $0$ to $\infty$ ( a cut). But there are many of these, so no largest element.
