# 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?

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

• how do you quantify $\approx$? I take it as $P$ means power set? Commented Jun 28, 2020 at 9:47
• Yes P means power set. And $\approx$ identifies finite set. Here is the definition: $n\in\mathbb N$, for any $X$ sets, If $X\approx n\space$ X is finite set. Commented Jun 28, 2020 at 9:50
• I believe your "$\approx$" symbol is defined as follows: If $A,B$ are two finite sets, we write $A\approx B$ when $A$ and $B$ have the same cardinality.
– Zuy
Commented Jun 28, 2020 at 10:23
• Well, I am not sure about that. I have the definition, which I have written above on my notes. Nothing more than that. Because if it is like you said, what does $X\approx n$ mean, as shown in the definition above? Commented Jun 28, 2020 at 10:27
• @Zuy Set theoretically, a natural number can be considered a set. For example, $0$ is the same as $\emptyset$, $1$ is $\{\emptyset\}$, $2$ is $\big\{\emptyset,\{\emptyset\}\}$, and so on and so forth. (You can define this inductively by setting $n+1:=\{\emptyset\}\cup\Big\{\{t\}\,\Big|\,t\in n\Big\}$ for each natural number $n$.) Thus, the notation $X\approx n$ where $X$ is a set and $n$ is a natural number can make sense. Commented Jun 28, 2020 at 10:54

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 $$B\subseteq X\cup Y$$.

• X and Y are infinite sets, aren't they? And... again, to be honest, I did understand nothing actually. :( Commented Jun 28, 2020 at 10:56
• They can be finite, or infinite. There is no restriction. Commented Jun 28, 2020 at 10:57
• But question says that "they are infinite sets that are not empty", why would we change it? OK, you know what, what about my solution? Is it not correct? There are no functions, no cardinalities (we haven't learned that)... is my solution wrong? Commented Jun 28, 2020 at 10:58
• If the bijection works for all cases (finite or infinite $X$ and $Y$), then it works when $X$ and $Y$ are both infinite. Commented Jun 28, 2020 at 10:59
• I think you missed it. The cartesian product is there. Note that the pair $(A,B)$ with $A\subseteq (X\cap Y)$ and $B\subseteq (X\cup Y)$ is an element of $\mathcal{P}(X\cap Y)\times\mathcal{P}(X\cup Y)$. Likewise with the pair $(M,N)$ such that $M\subseteq X$ and $N\subseteq Y$, for it is an element of $\mathcal{P}(X)\times\mathcal{P}(Y)$. Commented Jun 28, 2020 at 11:43

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?

• To be honest, I haven't been taught about cardinality or heard of it; and because of that, I am unable to understand it. I mean, when I google it, I learn its definition but I don't know what it means for a question or how to use it. Commented Jun 28, 2020 at 10:40
• A set's cardinality counts how many elements it contains. For example: $|\{5\}|=1$, $|\{1,2,50\}|=3$ etc.
– Zuy
Commented Jun 28, 2020 at 10:43