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.
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).