I'm reading the textbook Function Theory of One Complex Variable third edition of Robert Greene and Steven Krantz and at page 164 at the beginning of the proof of the theorem 5.2.2 it says:
According to the lemma, there is a $\delta _1>0$ such
that every point of $D(P, \delta _1) \setminus \{P\}$ is a simple point of $f$. Now choose $\delta , \epsilon > 0$ as follows: Take $0 < \delta < \delta _1$ such that $Q \notin f (D(P, \delta ) \setminus \{P\})$. Choose $\epsilon > 0$ such that $D(Q, \epsilon ) \subset f (D(P, \delta ))$ and so that $\boldsymbol{D(Q, \epsilon )}$ does not meet $\boldsymbol{f (\partial D(P, \delta))}$; this choice is possible because $f (D(P,\delta ))$ is open by Theorem 5.2.1 and because $f$ is continuous.
Note: above we have that $Q=f(P)$ and $f$ is a nonconstant holomorphic function.
What I can't see is why we can choose a $\epsilon >0$ such that $D(Q, \epsilon )$ does not meet $f (\partial D(P, \delta ))$, I don't see why continuity of $f$ can ensure that, as much I can see that $f^{-1}(D(Q,\epsilon ))\subset f^{-1}(f (D(P,\delta )))$.
Best Answer
Ok, the answer was simple after all. Thank you to @CBBAM for the hint.
I think the book was wrong here as we need to ensure that $Q\notin f(\partial D(P,\delta ))$, otherwise the statement will be trivially false. But we can fix it easily choosing a smaller $\delta _2$, that is, for $0<\delta _2<\delta $ we have that $Q\notin f(\overline{D(P,\delta _2)}\setminus \{P\})$, therefore $Q\notin f(\partial D(P,\delta _2))$. And as $f(\partial D(P,\delta _2))$ is compact and $\mathbb{C}$ is a metric space, then its easy to show that the distance between $Q$ and $f(\partial D(P,\delta _2))$ is positive, so such $\epsilon >0$ exists.