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imurme8
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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 [itex]\mathbb{C}[/itex], in this case.)
The proof that [itex]f(\gamma)\subseteq \Gamma[/itex] is clear to me. For the opposite containment, they offer the following proof:
My question is about the part marked with an asterisk. How do we know that [itex]z_{n}[/itex] has a convergent subsequence? If [itex]\{z_n\}[/itex] 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 [itex]\tilde{\mathbb{C}}[/itex] would give us this result. But here, I don't see any indication that either [itex]G[/itex] or [itex]f(G)[/itex] is bounded, so I don't see how we have this.
Just reading this again, it strikes me that perhaps we can bound [itex]\{z_n\}[/itex] by the continuity of the inverse function [itex]f^{-1}[/itex]. But I don't see that this continuity is assumed, so I'm not sure.
Thanks for any help!
Theorem: Let [itex]G[/itex] be a domain with boundary [itex]\gamma[/itex], let [itex]f(z)[/itex] be a one-to-one function on [itex]G[/itex], and let [itex]f(z)[/itex] be continuous on [itex]\overline{G}[/itex]. Then [itex]f(\gamma)=\Gamma[/itex], where [itex]\Gamma[/itex] is the boundary of the domain [itex]f(G)[/itex].
The proof that [itex]f(\gamma)\subseteq \Gamma[/itex] is clear to me. For the opposite containment, they offer the following proof:
Suppose [itex]w \in \Gamma[/itex]. Then there exists a sequence [itex]\{ w_n \}[/itex] such that [itex]w_n \in f(G)[/itex] and [itex]w_n \to w[/itex] as [itex]n \to \infty[/itex]. If [itex]z_n[/itex] is the inverse image of [itex]w_n[/itex], then [itex]z_n \in G[/itex], and from the sequence [itex]\{z_n\}[/itex] we can select a subsequence [itex]\{z_{n_k}\}[/itex] (*) converging to a point [itex]z\in \overline{G}[/itex], where
[itex]f(z)= \lim\limits_{n\to \infty} f(z_{n_k}) = \lim\limits_{n\to \infty}w_{n_k}=w.[/itex]
Thus [itex]w[/itex] is the image of a point [itex]z\in \overline{G}[/itex]. But [itex]z \notin G[/itex], since, according to a theorem, the image of an interior point of [itex]G[/itex] is an interior point of [itex]f(G)[/itex] (since [itex]f[/itex] is one-to-one). Therefore [itex]z[/itex] is a boundary point of [itex]G[/itex] and [itex]w=f(z)[/itex] belongs to [itex]f(\gamma)[/itex]. It follows that [itex]f(\gamma)\supseteq \Gamma[/itex], and the proof is complete.
My question is about the part marked with an asterisk. How do we know that [itex]z_{n}[/itex] has a convergent subsequence? If [itex]\{z_n\}[/itex] 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 [itex]\tilde{\mathbb{C}}[/itex] would give us this result. But here, I don't see any indication that either [itex]G[/itex] or [itex]f(G)[/itex] is bounded, so I don't see how we have this.
Just reading this again, it strikes me that perhaps we can bound [itex]\{z_n\}[/itex] by the continuity of the inverse function [itex]f^{-1}[/itex]. But I don't see that this continuity is assumed, so I'm not sure.
Thanks for any help!