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Power of set of choice functions

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the power of set of all choice functions for P([itex]N[/itex]) - {0}.

    3. The attempt at a solution
    I really don't know how to start with that. Hope you will give some clues.
     
  2. jcsd
  3. Dec 4, 2011 #2
    Try to look at P(N)=2^n

    How many numbers are in N-{0}?
     
  4. Dec 4, 2011 #3
    There are [itex]\aleph[/itex] (aleph zero) numbers in N-{0}...
    So 2^[itex]\aleph[/itex] equals [itex]\Im[/itex] (continuum)... So it proves, that the power of P(N) equals [itex]\Im[/itex].. However, how to combine it with a choice function?
     
  5. Dec 4, 2011 #4
    Ok so you said P(N) has N amount of number except zero. Now what do you say for this?


    f(N)=??
     
  6. Dec 4, 2011 #5
    by "P(N) has N amount of number except zero" are you meaning, that P(N) has [itex]\aleph[/itex] one-element subsets excpet zero?

    f(N)? Do you mean one-element subsets by N as argument?
     
  7. Dec 4, 2011 #6
    No i mean that P(N) has an uncountable amount of subsets besides zero.

    You have to construct a choice function f and whose range is N-{0}, such that f(N) is an element of N.
     
    Last edited: Dec 4, 2011
  8. Dec 4, 2011 #7
    OK, but.. amount of subset of P(N) is more than natural numbers.. it is continuum, not aleph zero......
    edit::: wrong... I'm a little bit confused.. We need to find a amount of functions f: P(B) -> n, where B is a proper subset of N and n in B...
     
    Last edited: Dec 4, 2011
  9. Dec 4, 2011 #8
    So I think we are able to construct aleph zero these functions.. P(N) contains as subset the set of N so..
    We can build this function like this:
    F: P(N) -> N;
    F(N) = {case 1 - the power of N equals aleph zero: G(N);
    {case 2 - in other case (the power of N is less than aleph zer): max(N)

    where G: P(N) -> N.
    G(B) -> n, where P(N) containn B, and b in B. We can build aleph zero g functions, because we use it only if the argument of f functions has aleph zero power, so B als has alpeh zero functions - so the set of values of function G has also alpeh zero elements.

    It is correct?
     
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