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Cardinality of a subset of [0,1]

  1. Aug 5, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the cardinality of the set of all numbers in the interval [0, 1] which
    have decimal expansions with a finite number of non-zero digits?


    2. Relevant equations



    3. The attempt at a solution
    I say its still c? Am I correct, there is no way I can pair this set with the natural numbers.

    Define
    f(1) = 0.1
    f(2) = 0.02
    f(3) = 0.003
    f(4) = 0.0004
    ...
    But then we must pair elements of N to 0.2, 0.3, 0.4... and 0.03, 0.04, 0.05, and 0.004, 0.005,...

    This seems weak though no?
     
    Last edited: Aug 5, 2013
  2. jcsd
  3. Aug 5, 2013 #2

    micromass

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    Your set is a subset of ##\mathbb{Q}##, isn't it?
     
  4. Aug 5, 2013 #3

    verty

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    @Ribbon: Your argument doesn't make any sense to me. You seem to be arguing that there are a countable number of sets equinumerous with the naturals. Is this what you are doing and what are you trying to achieve by it?
     
  5. Aug 5, 2013 #4

    tiny-tim

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    hi ribbon! :smile:
    hint: can you pair the set of all numbers with two non-zero digits with the natural numbers? :wink:
     
  6. Aug 5, 2013 #5

    vela

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    Yes, your argument doesn't work. All you've shown is that you failed at your first attempt at finding a bijection between N and the set. Perhaps if you were more clever, you could find such a bijection.
     
  7. Aug 5, 2013 #6
    Could I go,
    f(1) = 0.11
    f(2) = 0.12
    f(3) = 0.13
    ...

    and skip numbers like 0.2, 0.3? But then what happens to all the numbers in between I'm missing?
     
  8. Aug 5, 2013 #7
    It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one, and if I am not mistaken with cardinal arithmetic, c would remain the cardinality. But it seems to be otherwise now.
     
  9. Aug 5, 2013 #8

    LCKurtz

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    Hey Tiny, what happened to your fish avatar? Did he try to swim through a screen?
     
  10. Aug 5, 2013 #9

    tiny-tim

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    but you're not even trying to count any numbers with more than 2 decimal places! :rolleyes:

    try again, but with binary numbers (instead of decimals), it's easier :smile:
    it wasn't a proper underwater camera! :redface:
     
  11. Aug 5, 2013 #10
    Hmm, ok well could the sequence be 0.1, 0.101, 0.010101... till I hit that n (n being the finite number of non zero digits?
     
  12. Aug 6, 2013 #11

    LCKurtz

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    So, have you just walked away from the thread where we were discussing that topic?
     
  13. Aug 6, 2013 #12

    pasmith

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    Your set is infinite: if [itex]x[/itex] is in that set, then so is [itex]10^{-n}x[/itex] for all [itex]n \in \mathbb{N}[/itex]. Adding any number of zeroes to the beginning of the decimal expansion does not change the number of non-zero digits in the expansion.
     
  14. Aug 6, 2013 #13

    HallsofIvy

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    You do not recognize this as the subset of all rational numbers whose denominators have only powers of 2 and powers of 5 as factors? If [tex]x= 0.a_1a_2\cdot\cdot\cdot a_n[/tex] has only "n" non-zero digits, then [tex]10^nx= a_1a_2\cdot\cdot\cdot a_n[/tex] and, finally, [tex]x= \frac{a_1a_2\cdot\cdot\cdot a_n}{10^n}[/tex]. You may be able to do a lot of "cancelling" to reduce the denominator but 10= 2(5) so there will never be any factors other than "5" or "2" in the denominator.


     
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