Cardinality of sets inequality

Question

Suppose A,B,C,D are sets with $|A|=|C|, |B|=|D|$.

I need to prove/disprove:

$$|A^B| \leq |P(C\times D)|$$

I know $|P(C\times D)|= 2^{|C\times D|}$, but I'm not sure what to do with the left side.

Help please :)

Answer 1

Using cardinal arithmetic, we know $2^{|C||D|}=\left(2^{|C|}\right)^{|D|}$.

Since $|D|=|B|$, $\left(2^{|C|}\right)^{|D|}=\left(2^{|C|}\right)^{|B|}$. Since $|A|=|C|$, $|A|<2^{|C|}$, and, again using cardinal arithmetic, $|A|^{|B|}\leq \left(2^{|C|}\right)^{|B|}=\left(2^{|C|}\right)^{|D|}$.

Answer 2

$|A^B|$ is a cardinality of a set of all functions $f:B\to A$. Each function is identified by its graph, which is a set of pairs $G=\{(b,a) \in B\times A\}\subseteq B\times A$ such that each $b\in B$ appears in exactly one pair. So each $f$ is identified by some subset of $B\times A$, that is by an element of $P(B\times A)$.

Therefore $|A^B|=|G|\le |P(B\times A)|$ and that's almost all.