# Transfinite Cardinals

1. Feb 8, 2008

### mathboy

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. Feb 9, 2008

### Hurkyl

Staff Emeritus
That's good; the thing you're trying to prove is not a theorem.

3. Feb 9, 2008

### mathboy

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. Feb 9, 2008

### morphism

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

$$|A^C| = \aleph_0^{\aleph_0} = \mathfrak{c} = \mathfrak{c}^{\aleph_0} = |B^C|$$