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

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data

    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.


    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 [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.
     
  2. jcsd
  3. Nov 24, 2012 #2

    Dick

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    Do you know that [0,1] is compact? If you have an infinite number of points there must be a limit point.
     
  4. Nov 25, 2012 #3
    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
  5. Nov 25, 2012 #4

    Dick

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    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
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