I've come across this statement:
If $f$ is a nonconstant meromorphic function, the module $M$ containing all its periods cannot have an accumulation point, since otherwise $f$ would be a constant. Therefore each point in $M$ is isolated. In other words, $M$ is a discrete module.
Now it is my understanding that
- an accumulation point $x$ of $M$ is a point for which every open neighbourhood contains infinitely many points of $M$.
- an isolated point $x$ of $M$ is a point for wich there is a neighbourhood containing only $x$.
I can't see how the absence of an accumulation point implies that every point in $M$ is isolated. I understand that it implies that a neighbourhood of $x$ cannot contain infinitely many points of $M$, but what prevents such a neighbourhood from containing finitely many?
If we have a neighbourhood $U$ of $x\in M$ such that $U\cap M$ is finite, consider $F := (M\cap U)\setminus \{x\}$. If $F = \varnothing$, we're done, otherwise note that $F$ is closed, since it's finite (and $\mathbb{C}$ is Hausdorff, hence $T_1$), and thus $V := U \setminus F$ is also a neighbourhood of $x$, and by construction we have $V\cap M = \{x\}$.
An argument using the metric is that then
$$\delta = \min \{ \lvert x-y\rvert : y\in F\} > 0,$$
since it is the minimum of finitely many (strictly) positive numbers. Then choose $r$ so small that $B_r(x) \subset U$ and set $\rho = \min \{r,\delta\}$. Then $B_\rho(x) \cap M = \{x\}$.