I solved but don't know if it is correct, can you help me? Showing $P(X\cup Y)\approx P(X)\times P(Y)$ Question:
$X, Y$ are infinite sets that are not empty, and $X\cap Y=\emptyset$.
Show $P(X\cup Y)\approx P(X)\times P(Y)$

Hi! I tried to solve the question I wrote above, but I don't know if it is correct. Can you check if it is correct, and if not can you show me the correct one?
Thanks in advance.
$\approx$'s definiton: $n\in\mathbb N$, if $X\approx n$ for any $X$ sets, X is a finite set.
And P is Power Set.
Here is my solution:

$(\Rightarrow )$
Let $a=(X,Y)\in P(X\cup Y)$
$$\Rightarrow a\in (X\cup Y)$$
$$X\wedge Y=\emptyset \Rightarrow (a\in X\wedge a\not\in Y)\vee (a\not\in X\wedge a\in Y)$$
$$[a\in P(X)\wedge a\not\in P(Y)]\vee [a\not\in P(X)\wedge a\in P(Y)]$$
$$[a\in P(X)\times P(Y)]\vee [a\in P(X)\times P(Y)]$$
$(\Leftarrow)$
Let $a\in [P(X)\times P(Y)]$
$$\Rightarrow [a\in P(X)\wedge a\not\in P(Y)]\vee [a\not\in P(X)\wedge a\in P(Y)]$$
$$\Rightarrow [a\in X\wedge a\not\in Y]\vee [a\not\in X\wedge a\in Y]$$
$$\Rightarrow a\in X\cup Y\Rightarrow a\in P(X\cup Y)$$
 A: Here is just a remark.  I put it as an answer because it is too long for a comment.  I take it that $P \approx Q$ means the sets $P$ and $Q$ are equicardinal.
Let $X$ and $Y$ be arbitrary sets, which are not necessarily infinite, nonempty, or disjoint.  Then, there exists a bijection $f:\mathcal{P}(X\cap Y)\times \mathcal{P}(X\cup Y)\to \mathcal{P}(X)\times\mathcal{P}(Y)$.  This bijection can be defined as follows: for $A\subseteq  X\cap Y$ and $B\subseteq X\cup Y$, let
$$f(A,B):=\Big(A\cup (B\setminus Y),B\cap X\Big)\,.$$  The inverse $f^{-1}: \mathcal{P}(X)\times\mathcal{P}(Y)\to\mathcal{P}(X\cap Y)\times \mathcal{P}(X\cup Y)$ of $f$ is given by
$$f^{-1}(M,N):=\Big(M\cap Y,(M\setminus Y)\cap N\Big)$$
for all $M\subseteq X$ and $N\subseteq Y$.
Let $\sqcup$ denote disjoint union, which is usually defined as
$$P\sqcup Q:=\big(P\times\{1\}\big)\cup \big(Q\times\{2\}\big)$$
for all sets $P$ and $Q$. Define the bijection $\phi:(X\cap Y)\sqcup (X\cup Y)\to (X\sqcup Y)$, which sends

*

*$(t,i)$ with $t\in X\cap Y$ to $(t,i)$ for each $i\in\{1,2\}$,


*$(t,2)$ with $t\in (X\setminus Y)$ to $(t,1)$, and


*$(t,2)$ with $t\in (Y\setminus X)$ to $(t,2)$.
The inverse $\phi^{-1}: (X\sqcup Y) \to (X\cap Y)\sqcup (X\cup Y)$ sends

*

*$(t,i)$ with $t\in X\cap Y$ to $(t,i)$ for each $i\in\{1,2\}$,


*$(t,1)$ with $t\in (X\setminus Y)$ to $(t,2)$, and


*$(t,2)$ with $t\in (Y\setminus X)$ to $(t,2)$.
We can see that $f$ lifts the bijection $\phi$ in the sense that, if $f(A,B)=(M,N)$, then
$$\phi(A\sqcup B)=M\sqcup N$$
for all $A\subseteq X\cap Y$ and $B\subseteq X\cup Y$.
A: I have commented on what I believe your "$\approx$"-symbol means. Using this, I can help you answer your question:
Let $n:=|A|$ and $m:=|B|$ be the cardinalities of $A$ and $B$. Since they are finite sets, $n$ and $m$ are natural numbers.
Now, using that $A$ and $B$ are disjoint, what is the cardinality of $A\cup B$?
Next, for a set $S$ of cardinality $N$, its powerset has cardinality $2^N$. Can you find the cardinality of the powerset of $A\cup B$?
Finally, can you tell me what the cardinality of $\mathcal{P}(A)\times \mathcal{P}(B)$ is? Hint: If $S,T$ are two sets of cardinalities $N$ respectively $M$, then $S\times T$ has cardinality $NM$.
Can you conclude from this?
