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Transfinite Cardinals

  • Thread starter mathboy
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  • #1
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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?
 
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Answers and Replies

  • #2
Hurkyl
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But now I can't find a contradiction.
That's good; the thing you're trying to prove is not a theorem.
 
  • #3
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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?
 
  • #4
morphism
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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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