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Proving rationals cannot have dense orbit in Cantor set

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Let C be the standard Cantor "middle third" set (ie Ck = {x:0[itex]\leq[/itex]T[itex]^{k}_{3/2}(x)[/itex][itex]\leq[/itex]1} and C = [itex]\bigcap[/itex][itex]^{inf}_{k=0}[/itex]Ck
    where T[itex]^{k}_{3/2} = 3x if x<1/2,
    = 3 - 3x if x \geq[/itex] 1/2)

    Show that a rational number x = p/q [itex]\in[/itex] C cannot have dense orbit in C.

    2. Relevant equations

    {T[itex]^{k}_{3/2}[/itex](x)}[itex]^{inf}_{k=0}[/itex] is the orbit of x.

    Def A subset A of the interval J is dense in J if A intersects every nonempty open subinterval of J.

    3. The attempt at a solution

    Want to show that the orbit A = {T[itex]^{k}_{3/2}[/itex](x)}[itex]^{inf}_{k=0}[/itex] of any rational number x = p/q is not dense in C. so we want to show that NOT every nonempty, open subinterval of C intersects with the orbit A (negation of definition of being dense). So what it comes down to (I think) is finding some element of C that is not in the orbit. the trouble is, I dont know exactly what will be in the orbit for any given rational number. The orbit will obviously contain all rational numbers, so do I just need to find some irrational element in C?

    for example, 2/3 is in C, and 2/3 = 0.2000... in base 3, so if we let y = 0.20200200020000200000... then a) its irrational (right?) and b) its not in the orbit A since its irrational, so the orbit can't be dense.

    Im having alot of difficulty with the concepts here (dense sets, dense orbit, the cantor set in general) so any insight is appreciated. thanks
     
    Last edited: Feb 28, 2012
  2. jcsd
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