Elliptic functions proof - finitely many zeros and poles

[binbagsss]

Homework Statement

Hi

I have questions on the attached lemma and proof.

$f(z)$ is an elliptic function here, and non-consant $\Omega$ 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 continuous and so the limit may not necessary exist?

Homework Equations

see above

The Attempt at a Solution

see above

 

[mighty2000]

Hello,
Thank you for your post. In order to better understand the proof and the construction of the sequence, it would be helpful to have more context and information about the specific lemma and theorem being referenced. Can you provide more details or a link to the attached lemma and proof? This will allow me to give a more thorough and accurate response.

In general, the construction of the sequence of zeros in the proof likely relies on the periodicity of the function and the fact that it is non-constant. This allows for the creation of a sequence that converges to a point in the region, which then implies that the function is identically zero in that region.

Additionally, the conditions of boundedness, closedness, and infinitely many zeros are all important in the construction of the sequence. Without all three of these conditions, the proof may not hold. If there were only finitely many zeros, the proof may not work because the sequence would not necessarily converge to a point in the region. If the region was not closed, there may be points outside of the region where the function is not zero, which would contradict the assumption that the function is identically zero in the region.

I hope this helps. Please provide more information about the specific lemma and proof so that I can give a more detailed response. Thank you.