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Countable set Proof.

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Assume B is a countable set. Thus, there exists [itex] f:\mathbb{N}→B [/itex]
    which is 1-1 and onto Let [itex] A{\subseteq}B [/itex] be an infinte subset of B.
    Show that A is countable.
    3. The attempt at a solution
    Lets assume for contradiction that A has an uncountable number of elements.
    This would imply that A has elements that are not in B. But this is a contradiction because all elements in A are in B. Therefore A is countable.
     
    Last edited: Feb 4, 2012
  2. jcsd
  3. Feb 5, 2012 #2

    tiny-tim

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    your reason??
     
  4. Feb 5, 2012 #3
    If all elements of A are in B, and if B is countable then A is countable.
     
  5. Feb 5, 2012 #4

    Deveno

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    try to look at the problem this way:

    you are given (for free) a bijection f:N→B.

    you want to leverage that to a bijection g:N→A.

    f looks like this:

    1→b1
    2→b2
    3→b3
    ....
    k→bk
    ....

    now A is a subset of B, so every element of A is already listed on the right. can you think of something to do with the "gaps"?
     
  6. Feb 5, 2012 #5
    can I just assign some function to N that takes it to A, so there is no gaps.
     
  7. Feb 5, 2012 #6

    Deveno

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    well, yes, that is what you want to do, but how? you need to indicate an algorithm of some sort.

    note that the bijection f linearly orders B.

    therefore it makes sense to speak of: the first element of B such that..., the second element of B such that...

    can you continue?
     
  8. Feb 6, 2012 #7
    why do i need N-->A why cant I just have A go to a subset of N, why does it need to be onto.
     
  9. Feb 6, 2012 #8

    Deveno

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    well, that depends.

    what are you trying to prove? what is the definition of what you are trying to prove?

    what does it MEAN to say "B is countable"?

    are you convinced, right down to your curly toes, that what you are saying is true? do you think the theroem is true? why, or why not?

    as for your suggestion, the map f:R→N given by f(x) = 1 certainly maps R to a subset of N, does this mean R is countable?
     
  10. Feb 6, 2012 #9
    A is also an infinite set and it is a subset of B, and B is countable.
    On your list that you gave on post #4 , why cant I just make the first natural go to the first element of A and then keep matching them up.
     
  11. Feb 6, 2012 #10

    Deveno

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    because until you know that A is countable, "the first element of A" doesn't even make sense.

    however, "the first element of B that is also in A" DOES make sense, because we already know that B is countable.
     
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