Let $A, B$, two sets. $X$ is the set of all relations from $A$ to $B$, and $Y$ is the set of all functions from $A$ to $P(B)$ (power-set of $B$).
Prove that $\left|X\right|=\left|Y\right|$.
My Solution:
$\left|A \right| = a$,
$\left|B \right| = b$
$\left|Y\right| = \left| A\rightarrow P(B) \right| = \left| P(B) \right|^{\left|A \right|} = (2^b)^a = 2^{ab}$
$\left|X \right| = \left|P(A\times B \right)|$
$(*)$ Now, since every $\left<a,b\right>$ can be in the relation or not. We have $2^{ab}$ relations.
I've been told I didn't treat the infinite case (Maybe the meaning is I cannot use $(*)$ for the infinite case).
What should I do instead (or in addition to the finite case)?
Update:
I was guided not use functions. Instead, I should show it using cardinals arithmetics.
You should define a function $f:X\rightarrow Y$ and show that this function is a bijection. Think about how you could describe a relation between $A$ and $B$ by a function from $A$ to $\mathcal{P}(B)$.
Hint: for $x\in X$ and $a\in A$ you could set $f(x)(a) = \{b\in B\,\,|\,\,(a,b)\in x \}$.
Update: You seem to have solved the update already yourself. Anyway: $$ |Y| = |\mathcal{P}(B)|^{|A|} = 2^{|B|^{|A|}} = 2^{|A||B|} = |X|, $$ where the last equality follows, since we can choose to include/exclude each $(a,b)$ which gives two options per element.