TITLE: Elliptic functions - proof non-constant function, finitely many zeros and poles.
QUESTION [3 upvotes]: I have questions on the attached lemma and proof.
$f(z)$ is an elliptic function here, $\Omega$ is a period lattice. 
So the idea behind the proof is this is a contradiction because the function was assumed to be non-constant but by the theorem that if f is analytic in a region $R$ with zeros at a sequence of points $a_i$ that tend to $a_0$ $\in R$, then $f$ is identically zero in $R$.
Questions - mainly I don't understand where the consruction of the sequence comes from

which of the conditions out of the three: bounded, closed, infinitely many zeros, means that a convergent sequence of zeros can be constructed? I don't understand the reasoning behind the sequence, and does it make use of the fact of the periodicity of $f(z)$?
I'm guessing this sequence would not be possible to construct if there were only finitely many zeros for the proof to work...?
When it argues that by continuity $f(a_0)=0$, we have that $\Omega$ has inifnitely many zeros as our assumption, but we haven't said anything about poles? We haven't disallowed poles, and this means the region is not analytic so not continous and so the limit may not neccessary exist?

Many thanks in advance

REPLY [2 votes]: which of the conditions out of the three: bounded, closed, infinitely many zeros, means that a convergent sequence of zeros can be constructed? I don't understand the reasoning behind the sequence, and does it make use of the fact of the periodicity of $f(z)$?

All of them. Bounded and closed implies compact for subsets of $\mathbf C$. For metric spaces, compact is the same as sequentially compact, which guarantees that infinite sequences have convergent subsequences. The periodicity of $f$ is only used to restrict the problem to a compact set, namely, a fundamental parallelogram.

I'm guessing this sequence would not be possible to construct if there were only finitely many zeros for the proof to work...?

You want an infinite sequence to be able to consider its limit.

When it argues that by continuity $f(a_0)=0$, we have that $\Omega$ has inifnitely many zeros as our assumption, but we haven't said anything about poles? We haven't disallowed poles, and this means the region is not analytic so not continous and so the limit may not neccessary exist?

You don't need to worry about poles for this argument. $f$ is continuous on its domain and, in particular, on its set of zeros. Recall that every continuous function is sequentially continuous. So, if a sequence of zeros $(a_n)$ converges to $a$, then $(f(a_n)) = (0)$ converges to $f(a)$.

As an exercise, you should use a similar argument to show that $f$ has finitely many poles. Would you know how to proceed?
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