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Transfinite Cardinals

  1. Feb 8, 2008 #1
    Let A,B,C be infinite sets. Define A^C as the set of all functions from C to A. Prove that if |A^C |=|B^C |, then |A|=|B|.

    So I assume |A|<|B|. Since |A^C |<=|B^C | is true (proved theorem), I need only show that |A^C | not=|B^C |. Assume |A^C |=|B^C |, then there is a bijection f:A^C -> B^C . But now I can't find a contradiction. Because |A|<|B|, then there is a bijection from A to a proper subset of B, but then what?
    Last edited: Feb 9, 2008
  2. jcsd
  3. Feb 9, 2008 #2


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    That's good; the thing you're trying to prove is not a theorem.
  4. Feb 9, 2008 #3
    Oh, it's not even true. Then can someone find a counter-example to the assertion?

    That is, find two infinite sets A and B such that |A|<|B| but |A^C |=|B^C | for some infinite set C?
  5. Feb 9, 2008 #4


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    Take A and C to be the integers, and B to be the reals.

    [tex]|A^C| = \aleph_0^{\aleph_0} = \mathfrak{c} = \mathfrak{c}^{\aleph_0} = |B^C|[/tex]
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