Prove that if $X$ and $Y$ are sets where $\,\left|X\right|=\left|Y\right|,\,$ then $\,\left|P\left(X\right)\right|=\left| P\left(Y\right)\right|$.

Question

We are basically being asked to prove that if the cardinality of set $\,X\,$ and set $\,Y\,$ is the same, then how can we prove that the cardinality of their power sets is also the same.

I have considered using the Schroder Bernstein theorem (If $\,A\,$ and $\,B\,$ are sets with $\,\left\lvert A\right\rvert \le \left\lvert B\right\rvert,\,$ and $\,\left\lvert B\right\rvert \le \left\lvert A\right\rvert,\,$ then $\,\left\lvert A\right\rvert = \left\lvert B\right\rvert.\,$ In other words, if there are one-to-one functions $\,f\,$ from $\,\left\lvert A\right\rvert\,$ to $\,\left\lvert B\right\rvert\,$ and $\,g\,$ from $\,\left\lvert B\right\rvert\,$ to $\,\left\lvert A\right\rvert,\,$ then there is a one-to-one correspondence between $\,A\,$ and $\,B$ ) but I am unsure of how to employ it in a formal proof.

Answer 1

We have a bijection $f$ from $A$ to $B$. What function comes to mind from $P(A)$ to $P(B)$?

Send $S\subseteq A$ to $f(S)$ and prove this correspondence is injective and surjective.

Answer 2

$|A|=|B|$ means that there exists a bijection $f:A\rightarrow B$. This bijection induces a function $f:P(A)\rightarrow P(B)$ by setting $f(S)=\{ f(s) : s\in S \}$ for $S\subset A$. It is then a very easy exercise to see that this is a bijection...