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Theorem attached. enter image description here I know the theorem holds for a discrete subgroup of ${\mathbb C}$ more generally, ${\mathbb C}$ the complex plane, and that the set of periods of a non-constant meromorphic function is a discrete subset.

I have a question on part of the proof (showing the second type, integral multiples $nw$)

On proving the second case $\Omega$, the set of periods $\Omega = nw $ , $\Omega \neq \{0\}$ is considered and we show that exists $w_1 \in \Omega \setminus \{0\} $ with the least value of $|w_1|$. A step of doing this in my text is, and this is my problem:

Since $\Omega $ is discrete, there is some $\epsilon > 0 $ st any disk $\{z: |z| < \epsilon\} $ contains no elements of $\Omega \setminus \{0\}$, it follows that for any $w \in \Omega $, the disk $\{z: |z-w|< \epsilon \}$ contains no elements of $\Omega \setminus \{w\} $.

I don't understand why/ how this follows from the fact that is a group, why this 'translation' is possible with the same radius $\epsilon$ follows from the fact it is a group?

No idea where to start, any help much appreciated, but my guess would be that the proof will depend on what the operation of the group is, but this isn't specified in the theorem? or is just assumed to be addition?

1 Answers 1

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That's quite immediate: For every $w\in \Omega$, $a\in B(w, \epsilon)\cap \Omega$, $b=a-w\in \Omega$, since $\Omega$ is a group. Now, can you show that $|b-0|<\epsilon$? Here $B(p, r)$ is the open disk with the center $p$ and radius $r$.