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I am working on the following question :

Show that if $B$ is infinite and $|B| \leq |A|$ then $|^{A}B| = |^{A}2|$ by observing that $^{A}B \subseteq P(A\times B)$.

My thoughts are that $|B| \leq |A|$ means that $|B| < |A|$ or $|A| = |B|$ and also we know that for any set $A$, $|A| < |P(A)|$. However, I am struggling to conclude in a proof. Any help?

Thank you in advance.

Angela
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1 Answers1

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If you assume ZFC, then you can write:

$$|2^A|\leq |B^A| \leq \mathcal{P}(A\times B) \leq \mathcal{P}(A^2) = \mathcal{P}(A) = |2^A|.$$

The axiom of choice is used here to show that $\mathcal{P}(A)=\mathcal{P}(A^2)$ (see Cardinality of the Cartesian Product of Two Equinumerous Infinite Sets).

Thomas Delacroix
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  • Thank you for the response. Just one question, in your answer you assumed that $|A| = |B|$.Right? What happens when $|B| < |A|$? Does it work the same way since $A$ and $B$ are infinite? – Angela May 22 '17 at 23:38
  • No. I considered $|B|\leq|A|$ directly. There is no need to dissociate the two cases $|B|<|A|$ and $|B|=|A|$. – Thomas Delacroix May 23 '17 at 07:09