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Sequence that has a subsequence that converges towards any value in R

  1. Oct 5, 2007 #1
    Is the sequence defined as the denumeration of Q the only such sequence?
     
  2. jcsd
  3. Oct 5, 2007 #2
    No, you can take that sequence, and remove a finite number of elements or add infinitely many elements and it will have the same property.

    A denumeration of the set of rationals of the form n/2^k for all n and [itex]k\geq 0[/itex] (ie all binary numbers with a finite number of digits) also works.
     
    Last edited: Oct 5, 2007
  4. Oct 5, 2007 #3
    True. I wonder then, is there such a sequence unrelated to the denumbering of Q?
     
  5. Oct 5, 2007 #4
    Depends on what you mean by related. The method of showing denseness and countability will probably be very similar, so much in fact that you'd be inclined to say it's the same thing.

    Another countable dense set of the reals is the set of all integer multiples of a sequence of reals converging to 0. Is that good enough? (I wonder if you could change "integer multiples" to multiples of any unbounded sequence*)

    Edit: * No you can't. For example, [tex]\{1/2^n|n \in \mathbb Z\}\{2^n|n \in \mathbb Z\} = \{2^n|n\in\mathbb Z\}[/tex] which isn't dense in R.
     
    Last edited: Oct 5, 2007
  6. Oct 5, 2007 #5

    morphism

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    I believe (tan(n)) works.
     
  7. Oct 5, 2007 #6

    mathwonk

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    there seem to be zilions of them. take any sequence ci=onverging to a given rational, and any sequence converging to another rational, and so on, and take them all.

    or instead of the rationals take any countable dense set, eaSILY OBTAINED BY REPEATEDLY SUBDIVIDING THE REALS AT INTERVALS of irrational length, and the union of sequences converging to each one.


    shoot, almost anything works.
     
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