Why can the meromorphic function only have finitely many poles in the complex plane?

There is a proof in Complex Analysis by Stein-Shakarchi that I do not understand. I have highlighted it in red.

Why can they say that $f$ only has a finite number of points? What makes them able to exclude the countable number in the definition of a meromorphic function?

If $f\left(\frac{1}{z}\right)$ is holomorphic in $0 < \lvert z\rvert < r$, then $f(z)$ is holomorphic in $\frac{1}{r} < \lvert z\rvert < \infty$ (and vice versa). Thus $f$ can have poles - except maybe at $\infty$ - only in the compact set $\overline{D_{1/r}(0)} = \{z : \lvert z\rvert \leqslant 1/r\}$. Since the set of poles has no accumulation point, and every infinite subset of a compact set has an accumulation point, $f$ can have only finitely many poles in $\overline{D_{1/r}(0)}$, and possibly one at $\infty$, so finitely many altogether.
• I just have one follow-up question please. I want to prove that the set of poles have no accumulation points as you said, but I get a little problem. So if $z_0$ is a pole, then I know that the function $1/f(z)$ defined to be zero at $z_0$ is holomorphic in a neighbourhood. So I assume for contradiction that the set of poles has an accumulation point. Then I know that the function 1/f has an accumulation point for 0, so it must vanish everywhere in the domain. Now this says that f must have a pole at every point?, is this a contradiction?, if so why can not f have a pole et every point? Commented Sep 22, 2014 at 14:41
• A meromorphic function attains finite values except in at most countably many points by definition, @user119615. So since the assumption that the set of poles had an accumulation point (in $\Omega$; it can have accumulation points on the boundary) led to the conclusion that $1/f \equiv 0$, or in other words $f \equiv \infty$, the assumption contradicts the premise that $f$ is a meromorphic function. Commented Sep 22, 2014 at 14:52
• @user119615 No, $z_0$ is a pole of $f$, if there is a $k\in\mathbb{N}\setminus\{0\}$ such that we have the identity $$f(z) = \frac{h(z)}{(z-z_0)^k},$$ where $h$ is holomorphic (in a neighbourhood of $z_0$) with $h(z_0)\neq 0$. So since $h(z)$ is by continuity nonzero on a neighbourhood $U$ of $z_0$, and $\frac{1}{(z-z_0)^k}$ is holomorphic and nonzero on $\mathbb{C}\setminus \{z_0\}$, we have $f(z)$ finite and nonzero in $U\setminus \{z_0\}$. In particular, $U$ contains no other pole than $z_0$. Poles are by definition isolated singularities. Commented Sep 22, 2014 at 15:00