# Nxmod(1) is a discrete subset of [0,1] iff x is a rational?

1. Nov 24, 2012

### Silversonic

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

Let $\theta \in [0,1]$ and $n \in \mathbb{Z}$. Let $n\theta mod(1)$ denote $n\theta$ minus the integer part. Show $n \theta mod(1)$ is a discrete subset of [0,1] if and only if theta is rational.

3. The attempt at a solution

I'm having a bit of trouble with the "only if" part of the statement. The "if" part of the statement is simple. If theta is rational, then the subset is finite (because $n \theta mod(1)$ will repeat itself once $n$ is equal to or larger than $p$, where $\theta = q/p$). Since it is a finite subset, it is discrete.

But the only if part? Assume that it is a discrete subset, I need to show this is irrational. I'm having trouble with that. If I could show there was a surjective map from $n\theta mod(1)$, theta being irrational, to the subset [0,1] then that would prove it. Since no point would be isolated in that case, and the set wouldn't be discrete. I can't seem to get anywhere with that, though.

I know that $n \theta mod (1)$, for theta irrational, is an infinite set (not that this proves discreteness), so maybe I could go from there? It's infinite, because if it were not, then we would have that $k\theta mod(1)$ and $n\theta mod (1)$ would be mapped on to the same number, for different integers k and n.

i.e.

$k\theta = p + r$
$n\theta = q + r$

p,q are integers and r is the "remainder part".

equating shows

$\theta = (p-q)/(k-n)$

Maybe I could go somewhere with that fact that it's an infinite set? I haven't managed to though.

2. Nov 24, 2012

### Dick

Do you know that [0,1] is compact? If you have an infinite number of points there must be a limit point.

3. Nov 25, 2012

### Silversonic

Checking wikipedia quickly, "If one chooses an infinite number of distinct points in the unit interval, then there must be some accumulation point in that interval". I guess that would prove it. Compactness is not something I've dealt with before (I think it's relevant to metric spaces, a module I did not choose). The fact that my notes said this was meant to be an "easy proof" threw me off a bit though, it didn't seem like it would take such a result in order to prove it.

Is there any other way? If not, thanks. That will do.

Last edited: Nov 25, 2012
4. Nov 25, 2012

### Dick

There's a less abstract way. Take any N>0. Divide [0,1] up into N intervals of length 1/N. Since there are an infinite number of points there must be an interval [k/N,(k+1)/N] that contains at least two points. Think about their difference. It's less that 1/N, right? So?

Last edited: Nov 25, 2012