Suppose X~Y, Prove that P(X) ~ P(Y)

Question

My attempt:

I imagined that if two sets are equivalent there would exist
$ f:X→Y$ that is bijective. If I conceptually create P(X) and apply the function defined for the first equivalence relation to every set in the power set one gets P(Y).

Let $X = \{x_1, x_2 , x_3 , ... \}$, thus $Y = \{f(x_n), \forall \quad x_n \in X, n=1,2,... \}$ Both X and Y will have power sets with $2^n$ elements. If the bijective relation from X to Y is applied to every subset in P(X) so that it is one-to-one and onto, it gives a new set B. Since X~Y, B = P(Y)

End of my solution

I am not convinced by my efforts, in my self-study in preparation for the fall I have found my own proofs difficult to "buy".

Specially, my issue is showing that the operation in the function from X to Y once applied to the subsets of X (elements of P(X)) will give P(Y).

PS P(.) is the operation for power set incase notation is of concern.

Question source: Foundations of Mathematical Analysis, Johnsonbaugh and Pfaffenberger.

Answer 1

Note that $f:X \rightarrow Y$ is a bijection iff there exists a $g:Y \rightarrow X$ such that $f \circ g = 1_Y$ and $g \circ f = 1_X$. Now define $\tilde{f}:P(X) \rightarrow P(Y)$ and $\tilde{g}:P(Y) \rightarrow P(X)$ as in Chris Caruvana's answer above:

$\tilde{f}(A) = \{f(x) \mid x\in A\}$ and $\tilde{g}(B) = \{g(y)\mid y\in B\}$

Then one checks $\tilde{f} \circ \tilde{g} = 1_{P(Y)}$ and $\tilde{g} \circ \tilde{f} = 1_{P(X)}$ - which is straight forward.

Answer 2

Given a bijection $f:X\to Y$, define the function $F: P(X) \to P(Y)$ by the rule $$F(A) = \{ f(x) : x \in A \}.$$ Then you should check that $F$ is, in fact, also a bijection.