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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?

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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