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

  • #1
Silversonic
130
1

Homework Statement



Let [itex] \theta \in [0,1] [/itex] and [itex] n \in \mathbb{Z} [/itex]. Let [itex] n\theta mod(1) [/itex] denote [itex] n\theta [/itex] minus the integer part. Show [itex] n \theta mod(1) [/itex] 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 [itex] n \theta mod(1) [/itex] will repeat itself once [itex] n [/itex] is equal to or larger than [itex] p [/itex], where [itex] \theta = q/p [/itex]). 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 [itex] n\theta mod(1) [/itex], 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 [itex] n \theta mod (1) [/itex], 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 [itex] k\theta mod(1) [/itex] and [itex] n\theta mod (1) [/itex] would be mapped on to the same number, for different integers k and n.

i.e.

[itex] k\theta = p + r [/itex]
[itex] n\theta = q + r [/itex]

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

equating shows

[itex] \theta = (p-q)/(k-n) [/itex]

Contradicting irrationality.

Maybe I could go somewhere with that fact that it's an infinite set? I haven't managed to though.
 
Physics news on Phys.org
  • #2
Silversonic said:

Homework Statement



Let [itex] \theta \in [0,1] [/itex] and [itex] n \in \mathbb{Z} [/itex]. Let [itex] n\theta mod(1) [/itex] denote [itex] n\theta [/itex] minus the integer part. Show [itex] n \theta mod(1) [/itex] 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 [itex] n \theta mod(1) [/itex] will repeat itself once [itex] n [/itex] is equal to or larger than [itex] p [/itex], where [itex] \theta = q/p [/itex]). 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 [itex] n\theta mod(1) [/itex], 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 [itex] n \theta mod (1) [/itex], 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 [itex] k\theta mod(1) [/itex] and [itex] n\theta mod (1) [/itex] would be mapped on to the same number, for different integers k and n.

i.e.

[itex] k\theta = p + r [/itex]
[itex] n\theta = q + r [/itex]

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

equating shows

[itex] \theta = (p-q)/(k-n) [/itex]

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.
 
  • #3
Dick said:
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.
 
Last edited:
  • #4
Silversonic said:
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?
 
Last edited:
Back
Top