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

Question:

$X, Y$ are infinite sets that are not empty, and $X\cap Y=\emptyset$. Show $P(X\cup Y)\approx P(X)\times P(Y)$

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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:

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$(\Rightarrow )$

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 $a\in [P(X)\times P(Y)]$ $$\Rightarrow [a\in P(X)\wedge a\not\in P(Y)]\vee [a\not\in P(X)\wedge a\in P(Y)]$$ $$\Rightarrow [a\in X\wedge a\not\in Y]\vee [a\not\in X\wedge a\in Y]$$ $$\Rightarrow a\in X\cup Y\Rightarrow a\in P(X\cup Y)$$

Answer 1

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 $B\subseteq X\cup Y$.

Answer 2

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?