(Proof Verification) Proof regarding connected open sets and analytic functions - Is my understanding correct?

Question

Here is the proof for Theorem 1.2 provided by Serge Lang and bellow I have written how I interpret it. I would be very grateful if any of you can check if I am understanding this correctly.

For clarifiction: I have added the statement and proof of 1.2 and other pages that could be relevant to the proof.

The way I understand this is the following,

Assuming that the zeroes of $f$ on $U$ are not discrete we have $S$ open. He afterwards goes to prove that $S$ is closed in $U$. Since $S$ is closed in U we have $U-S=V$ by Lemma 1.5 where $V$ is an open set closed in $U$. But by Theorem 1.6 we have $U$ topologically connected which implies that $U$ can not be expressed as $U=S ∪ V$ hence there doesn't exist $V$ s.t $U=S+V$ . Hence by contradiction we have $S$ not open and thus $S$ is discrete.

Is my understanding correct?

Answer 1

• “Assuming that the zeroes of $f$ on $U$ are not discrete we have $S$ open”: Actually, $S$ is always open. If the set of all zeroes of $f$ is discrete, then $S=\emptyset$, but it will still be an open set.
• “He afterwards goes to prove that $S$ is closed in $U$. Since $S$ is closed in $U$ we have $U−S=V$ by Lemma 1.5 where $V$ is an open set closed in $U$”: Right, except that there is no reason to suppose that $V$ is closed in $U$. And it is better to use the notation $U\setminus V$ instead of $U-V$.
• “But by Theorem 1.6 we have $U$ topologically connected which implies that $U$ can not be expressed as $U=S\cup V$ hence there doesn't exist $V$ s.t $U=S+V$”: We just say “connected”, not “topologically connected”. And you missed an important point here. Having $U$ written has the union of two open sets is not is not a problem, so to say. What is important here is that, since $U$ is connected, it cannot be written as the union of two disjoint and non-empty open sets. And you shouldn't write $U=S+V$.