Path connectedness of the complement of countable set

Question

Let $G$ be an open (path) connected subset of $\mathbb{C}$.

Let $f:G\rightarrow \mathbb{C}$ be a nonconstant anlytic function

I proved that the subspace topology on $f^{-1}(0)$ is discrete and $f^{-1}(0)$ is countable.

How do I conclude that $G\setminus f^{-1}(0)$ is (path) connected?

It is easy for the case $G=\mathbb{C}$, but in general it does not seem easy to prove

You do want to assume $f$ is not constant. – Robert Israel Oct 04 '15 at 19:26 — Robert Israel, Oct 04 '15 at 19:26
@RobertIsrael Oh right edited – Rubertos Oct 04 '15 at 19:26 — Rubertos, Oct 04 '15 at 19:26

score 1 · Accepted Answer · answered Oct 04 '15 at 19:29

1

Given two points $a, b \in G \backslash f^{-1}(0)$, start with the straight line segment joining them. This may contain points of $f^{-1}(0)$, but only finitely many. Make a sufficiently small semicircular detour around each of them.

answered Oct 04 '15 at 19:29

Robert Israel

448,999

Why does it contain only finitely many zeros?? – Rubertos Oct 04 '15 at 19:31
Aha it's because $[0,1]$ is compact – Rubertos Oct 04 '15 at 19:37
I presume you mean a path composed of straight line segments? $G$ is not necessarily convex... – copper.hat Oct 04 '15 at 19:42
@copper.hat Isn't it possible because zeros are isolated? – Rubertos Oct 04 '15 at 19:46
Take $G = {z | |z|>1}$ and $a=2, b=-2$. The straight line joining them doesn't lie in $G$. It is a nitpick. – copper.hat Oct 04 '15 at 19:49
@copper.hat I'm sure he meant polygonal lines. – Rubertos Oct 04 '15 at 19:53
@Rubertos: I am certain he did, I was just pointing out the omission... – copper.hat Oct 04 '15 at 19:55
Actually if $G$ is open and connected and $E\subset G$ is countable, then $G\setminus E$ is path connected. (The point being $E$ could be wild - dense in $G$ for example - compared with the zero set of an analytic function). – zhw. Oct 04 '15 at 20:43

Path connectedness of the complement of countable set

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