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Cardinal power $\kappa^\kappa$. When is it equal to $2^\kappa$?

How can one prove that $|A^A|=|2^A|$ for infinite $A$? (summary of proof or providing link with proof will also suffice.) Thanks!

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marked as duplicate by Asaf Karagila, Zhen Lin, t.b., Chris Eagle, Gerry Myerson Aug 17 '12 at 12:41

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We have $|2^A|\leq |A^A|$ and by Cantor-Bernstein theorem we are reduced to show that $|A^A|\leq |2^A|$. To see that, we use the fact that $|A\times A|=|A|$, applying Zorn lemma to $S:=\{(B,f), B\subset A, f\colon B\times B\to B\}$ with the partial order $(B_1,f_1)\leq (B_2,f_2)$ if $B_1\subset B_2$ and $f_{2\mid B_1}=f_1$. We also have $$|2^A|=|2^{A\times A}|=|(2^A)^A|$$ which gives what we wanted.

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Is f 1-1 and onto? And we get what we wanted because |A^A|<=|(2^A)^A| using the fact that if A<=B then |A^C|<=|B^C|, right? –  Idan Aug 15 '12 at 19:17
    
That, and the fact that $\lvert 2^A\rvert>\lvert A\rvert$ in the first place. :) I don't really get the part about Zorn's lemma though. What do you use it for? –  tomasz Aug 15 '12 at 19:19
    
because $|A^A|\leq |(2^A)^A|$. –  Davide Giraudo Aug 15 '12 at 19:19
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@tomasz We use Zorn's Lemma to show that |AxA|=|A| –  Idan Aug 15 '12 at 19:20
    
@Idan I see. I figured it is either a known result or one which should be proven more carefully. Also, the $f$ in $S$ should be bijective. –  tomasz Aug 15 '12 at 19:24
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