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

[Silversonic]

Homework Statement

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.

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)

Contradicting irrationality.

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

 

[Dick]

Homework Statement

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.

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)

Contradicting irrationality.

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

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

 

[Silversonic]

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

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.

 

[Dick]

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 throw 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.

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?