# Homework Help: Q about a Proof -- periods meromorphic function form discrete set

1. Jan 22, 2017

### binbagsss

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 $\lim_{n\to \infty} w_{n} = w_0$

2. Relevant equations

3. 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

2. Jan 23, 2017

### RUber

Since you are using contradiction, you are assuming that the set is not discrete, right?
If the set is discrete, then there is a radius around $\omega_0$ where no members of the neighborhood are in the set $\Omega_f$.
Because this is not true, you should be able to find members in $\Omega_f$ for any neighborhood.
That is all you are doing with the sequence...letting your neighborhood contract around $\omega_0$ with each additional step.
To avoid confusion, I would not let n start at 0...you want to start at a known other member of the set, you can index that as 1, or use another variable name like x, and let that start at 0.

In the end, you should conclude that if the function is both periodic with period $\omega_0$ and period $\omega_0 + \epsilon$ for any epsilon >0, then
$f(x+\omega) = f(x+\omega+ \epsilon)$ for any choice of epsilon, thus contradicting your assumption that f is non-constant.

3. Jan 23, 2017

### binbagsss

Okay thank you, so I have just seen you take the radius of the open set to be $1/n$ , so you are contracting with increasing $n$, but , in order for the set to not need discrete you just need one more period in the open set right, that $\neq \omega_0$, we assume there is one more period, but what is there to suggest there is more than one other period $\neq \omega_0$ so that a sequence can be generated?

4. Jan 24, 2017

### RUber

There is no rule against a sequence being composed of one repeated entry...however, for some fixed $\omega_1$ and $\omega_2$ with $\omega_1 \neq \omega_2$, there is a fixed distance $|\omega_1-\omega_2|<0$, therefore there will, at some point be a high enough n, such that $\frac1n < |\omega_1-\omega_2|$. This implies that there can be a series constructed where each successive term is closer than the last. Thus, for a non-discrete set, you can say that for any period you find, there will be another period that is closer to, but not equal to $\omega_0$.

5. Jan 24, 2017

### binbagsss

I 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 can define 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?

6. Jan 25, 2017

### RUber

They are periods because they are elements of Omega, the set of periods.
So $f(x + \omega_0) = f(x)$.
There is no assumption about the continuity of the function, only the periodicity.
You are trying to show that $\Omega_f$ is discrete by proving (by contradiction) that it can only be non-discrete for constant functions.
Assume: $\Omega_f$ is not discrete and $f$ is non-constant.
--
By the assumption that $\Omega_f$ is not discrete, then for any period $\omega_0 \in \Omega_f$ there is no epsilon neighborhood around $\omega_0$ that does not contain another member of $\Omega_f$. That other member, by definition, is also a period of $f$.
The final piece is to use the idea that
$f(x) = f(x + \omega_0) =f(x+\omega_n), \, \forall n$
which will contradict the assumption that f is non-constant...not that $\omega_n = \omega_0$, which doesn't help you.
--
Does that make more sense?

7. Jan 28, 2017

### binbagsss

Sorry I don't think I am grasping this concept.
So the idea being that $w_1 - w_2$ is also a period, so each $w_n$ , in this case where there are only two periods (without trivial linear combinations that would generate another period), is really just a combination of $w1$ and $w2$ and the sequence terms just differ by changing the radius of the disc?

8. Jan 30, 2017

### RUber

Right. Your set $\Omega_f$ has elements which are periods of $f$.
Say you have one element that you know for sure, $\omega_0$.
Then, if your set $\Omega_f$ is not a discrete set, then you can form a sequence of subsets.
Assume there exists a sequence of open neighborhoods, radius = $r/n$, around $\omega_0$ that are entirely contained in your domain set $\Omega_f$. Call these neighborhood $B_n(\omega_0, r) = \{\omega: |\omega - \omega_0| < r/n \}$
Then, since the domain set is not discrete, for each $B_n$,there must be at least one member of $\Omega_f$ other than $\omega_0$ contained within $B_n$.
So for each step in the sequence, you are forcing the distance between one period and another to become closer and closer. Since this is not a discrete set, there is no limit to how close the periods can be.
As you pointed out, all linear combinations of periods, $\omega = c_1 \omega_0 + c_2 \omega_n$ will also be a period.

The only function that is periodic with infinitely small period is the constant function.