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# Elliptic Functions - proof non-constant function has finitely many zeros and poles watch

1. Hi I have questions on the attached lemma and proof.

is an elliptic function here, 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 with zeros at a sequence of points that tend to
, then is identically zero in .*

**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 ?

- 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 , we have that 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?

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2. (Original post by xfootiecrazeesarax)
- which of the conditions out of the three: bounded, closed, infinitely many zeros, means that a convergent sequence of zeros can be constructed?
You need R to be bounded and contain infinitely many zeros to construct a convergent sequence of zeroes. If you want to ensure the sequence converges to a point in the set R you also need R to be closed.

You need an infinite number of zeroes because otherwise you can't even build a sequence of zeroes, let alone a convergent one (i.e. by definition a sequence of zeroes will have infintely many zeroes).

You need R bounded; otherwise take R to be , and the zeroes z_n to be 0, 1, 2, ... (i.e. z_n = n). Clearly there's no possible convergent subsequence (since each z_n is at least distance 1 from any other).

If R is bounded, then we can use the Bolzano-Weierstrauss theorem to constuct a convergent subsequence.

Note that this sequence doesn't have to converge to a point in R. e.g. take R to be the open unit disc, and z_n = 1-1/n. Then z_n clearly converges to 1, which is not in R. So if you want the sequence to converge to a point in R, you also want the condition that R is closed.

This is all true for any set R (that's a subset of , and any sequence z_n of points in R. So it doesn't depend on any properties of f at all.

- When it argues that by continuity , we have that has infinitely 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?
Yeah, I'm not keen on this bit.

But if you look at the theorem they're quoting, it doesn't actually require that f(a_0) = 0. So we don't need to appeal to continuity here. Of course, once we've shown the theorem applies, then f is identically zero on R (and so of course f(a_0) = 0).

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