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  1. The problem statement, all variables and given/known data

Hi,

As part of the proof that :

the set of periods $\Omega_f $ of periods of a meromorphic $f: U \to \hat{C} $, $U$ an open set and $\hat{C}=C \cup \infty $, $C$ the complex plane, form a discrete set of $C$ when $f$ is a non-constant

a step taken in the proof (by contradiction) is :

there exists an $w_{0} \in \Omega_{f} $ s.t for any open set $U$ containing $w_{0}$, there is an $w \in \Omega_{f} / {w_0} $ contained in $U$

Now the next step is the bit I am stuck on

By the standard trick in analysis, we can produce a sequence of periods $\{w_n\}$ such that $w_{n} \neq w_{0} $ and $\lim_{n\to \infty} w_{n} = w_0 $

  1. Relevant equations

  2. The attempt at a solution

It's been a few years since I've done analysis, and the 'trick' has no name so I am struggling to look it up and find it in google.

A proof of this to understand it's meaning is really what I'm after , what's the idea behind the construction / significance in the usual context it would arise

I am also confused with the notation, does $n=0$ ? So the sequence converges to it's first term, or is $n$ starting from one

Many thanks in advance

1 Answers 1

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The standard trick is: By assumption, there exists a period $\ne w_0$ in the open ball $B_{1/n}(w_0)$. Let $w_n$ be any such period. Then clearly $w_n\ne w_0$ for all $n$; and $|w_n-w_0|<\frac 1n$ implies $w_n\to w_0$.

Is suppose, one could do the whole proof simpler: As $f$ is assumed non-constant, we can pick $z_0$ such that $f$ is not constant in any neighbourhood of $z_0$. In particular, there exists $r>0$ such that $f(z)\ne f(z_0)$ for all $z$ with $0<|z-z_0|

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    okay thank you, why is it clear that $w_n\neq w_0 $ for the open ball $B_{1/n}(w_0)$ ? (my only guess would be because the ball is centred at $w_0$ and we can not take the radius of the ball to be zero? ) – 2017-01-24
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    I also don't understand why the $w_n$ generating by this are periods? So by our assumption we have $f(w_n)=f(w_0)$ , for a single value of $n$. We have defined the open ball $|w_n-w_0| < 1/n $, if $f$ was continous we could argue that we can find an $\epsilon$ such that $|f(w_n)-f(w_0)|<\epsilon$ and then taking the limit $\epsilon \to 0 $ would give $f(w_n)=f(w_0)$ the definition of a period, however holomorphic implies cts at every point but in this theorem the function f is mermorphic so this doesn't work? – 2017-01-24