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Set of functions that are eventually zero

  1. Oct 14, 2012 #1
    The problem asks whether the set F of all functions from Z+ into {0,1} that are "eventually zero" is countable or uncountable.

    In my proof, I claimed countable. By definition, associated with each function f in F, there is a positive integer N such that f(n)=0 for n≥N. I took N' to be the largest of these N's. Hence, for each f in F, f(n)=0 for all n≥N'. Then, I defined a function g: F -> {0,1}N'-1 by

    g({(1,x1),....,(N'-1,xN'-1), (N',0),(N'+1,0)....}) = (x1,......,xN'-1)​

    [Note: I know there is a simplier way to denote elements in F]

    I then proved that this is a bijection. Since the range is a finite product of countable sets, it is countable. It follows that F is indeed countable.

    Looking back at my proof, I feel that I may have made an error in picking N'. Cannot I not just construct a function that sends 1 through, say, N'+1 to 1 and the rest to zero? In other words, what (if anything) allows me to pick such an N'?

    My proof may be correct. That is, I may be second guessing myself.
     
    Last edited: Oct 14, 2012
  2. jcsd
  3. Oct 14, 2012 #2

    mfb

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    What about f(n)=1 for all n in Z+?
    Or g(n)=1 for odd n, g(n)=0 for even n?

    If that N(f) would exist for all f in F, F would be countable. However, you would need a different proof:
    "The largest N" does not have to exist in infinite subsets of the natural numbers.

    If that N' would exist, F would be finite.
     
  4. Oct 14, 2012 #3
    Sorry. I posted the question wrong. The problem asks whether the set F of all functions from Z+ into {0,1} that are eventually zero is countable or uncountable.
     
  5. Oct 14, 2012 #4

    Dick

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    You can't choose a largest N' at all. Each function has a different N, no reason they should be bounded. Take the set of all functions such that f(n)=0 for n≥N. That's countable (in fact, it's finite). Now think about forming a union of sets.
     
  6. Oct 14, 2012 #5
    I knew that picking N' was wrong! :frown: It didn't sit right with me. What about this.

    Let AN be the set of all functions such that f(n)=0 for n≥N. For THIS set, I can construct a bijective correspondence between AN and {0,1}N-1 by defining g: AN->{0,1}N-1 by
    g({(1,x1),....,(N-1,xN-1), (N,0),(N+1,0)....}) = (x1,......,xN-1).
    Hence AN is countable for each N.
    Taking Z+ as our index set. The union of all AN for each NεZ+ is the countable union of countable sets ....
     
  7. Oct 14, 2012 #6

    Dick

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    Yes, that would do it.
     
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