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The question is to show that $ \mathcal{P}(A)$~$ \mathcal{P}(B) $ and $\mathcal{F}(A,C)$~$\mathcal{F}(B,D)$ where A,B,C, and D are sets, $A$~$B$, $C$~$D$, and $\mathcal{F}(A,C)$ and $\mathcal{F}(B,D)$ are the sets of all functions for their respective sets.

I don't exactly know how to prove the two power sets are equal in cardinality when the sets themselves are equal in cardinality. It seems intuitive, but I don't know how to really explain it.

The set of all functions question, I don't even know really how to begin

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Equal cardinality means there is a bijection. Take bijections between A and B, and between C and D, and use those to construct bijections of the power sets and the function sets.

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  • $\begingroup$ What do you mean by ZF and ZFC? I need to go back and read in my textbook about the axiom of choice. I remember we covered it, but I haven't really used it so I am a little forgetful. $\endgroup$
    – Tyler
    Apr 30, 2018 at 3:03
  • $\begingroup$ Actually, I goofed. Since you only have to make two choices (finitely many), you don't need the Axiom of Choice. $\endgroup$
    – C Monsour
    Apr 30, 2018 at 3:11
  • $\begingroup$ ZF and ZFC are "Zermelo-Frankel" and "Zermelo-Frankel with Choice", if you're still interested. But not particularly relevant to the question you asked. $\endgroup$
    – C Monsour
    Apr 30, 2018 at 3:12
  • $\begingroup$ How do you construct the bijections, though? $\endgroup$
    – Tyler
    Apr 30, 2018 at 3:18
  • $\begingroup$ I didn't mean to post that, but I just don't understand because I can't just arbitrarily construct them. $\endgroup$
    – Tyler
    Apr 30, 2018 at 3:18

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