Complex Analysis proof question (from Markushevich text))

[imurme8]

There is a proof offered in the text "Theory of Functions of a Complex Variable" by Markushevich that I have a question about. Some of the definitions are a bit esoteric since it is an older book. Here "domain" is an open connected set (in $\mathbb{C}$, in this case.)

Theorem: Let $G$ be a domain with boundary $\gamma$, let $f(z)$ be a one-to-one function on $G$, and let $f(z)$ be continuous on $\overline{G}$. Then $f(\gamma)=\Gamma$, where $\Gamma$ is the boundary of the domain $f(G)$.

The proof that $f(\gamma)\subseteq \Gamma$ is clear to me. For the opposite containment, they offer the following proof:

Suppose $w \in \Gamma$. Then there exists a sequence $\{ w_n \}$ such that $w_n \in f(G)$ and $w_n \to w$ as $n \to \infty$. If $z_n$ is the inverse image of $w_n$, then $z_n \in G$, and from the sequence $\{z_n\}$ we can select a subsequence $\{z_{n_k}\}$ (*) converging to a point $z\in \overline{G}$, where

$f(z)= \lim\limits_{n\to \infty} f(z_{n_k}) = \lim\limits_{n\to \infty}w_{n_k}=w.$​

Thus $w$ is the image of a point $z\in \overline{G}$. But $z \notin G$, since, according to a theorem, the image of an interior point of $G$ is an interior point of $f(G)$ (since $f$ is one-to-one). Therefore $z$ is a boundary point of $G$ and $w=f(z)$ belongs to $f(\gamma)$. It follows that $f(\gamma)\supseteq \Gamma$, and the proof is complete.

My question is about the part marked with an asterisk. How do we know that $z_{n}$ has a convergent subsequence? If $\{z_n\}$ were bounded, then I see that we would have this result by the Bolzano-Weierstrass theorem. Also if we were working in the extended complex plane, then compactness of $\tilde{\mathbb{C}}$ would give us this result. But here, I don't see any indication that either $G$ or $f(G)$ is bounded, so I don't see how we have this.

Just reading this again, it strikes me that perhaps we can bound $\{z_n\}$ by the continuity of the inverse function $f^{-1}$. But I don't see that this continuity is assumed, so I'm not sure.

Thanks for any help!

 

[imurme8]

On second thought, I think the proof works as long as we suppose that we're working in the extended complex plane.