Is \mathcal{P}(A)\times\mathcal{P}(A) Equinumerous with \mathcal{P}(A)?

[Andrei1]

Suppose \(\displaystyle A\) is a set with at least two elements and \(\displaystyle A\times A\sim A.\) Then \(\displaystyle \mathcal{P}(A)\times\mathcal{P}(A)\sim\mathcal{P}(A).\)

My attempt: I know that \(\displaystyle \mathcal{P}((A\times A)\cup A)\sim\mathcal{P}(A\times A)\times\mathcal{P}(A)\sim\mathcal{P}(A) \times \mathcal{P}(A).\) How to prove that \(\displaystyle \mathcal{P}(A)\sim \mathcal{P}((A\times A)\cup A)\)? More generally, is it true that if \(\displaystyle X\) and \(\displaystyle Y\) are infinite and \(\displaystyle X\sim Y\), then \(\displaystyle X\cup Y\sim Y\)?

 

[Opalg]

Suppose \(\displaystyle A\) is a set with at least two elements and \(\displaystyle A\times A\sim A.\) Then \(\displaystyle \mathcal{P}(A)\times\mathcal{P}(A)\sim\mathcal{P}(A).\)

My attempt: I know that \(\displaystyle \mathcal{P}((A\times A)\cup A)\sim\mathcal{P}(A\times A)\times\mathcal{P}(A)\sim\mathcal{P}(A) \times \mathcal{P}(A).\) How to prove that \(\displaystyle \mathcal{P}(A)\sim \mathcal{P}((A\times A)\cup A)\)? More generally, is it true that if \(\displaystyle X\) and \(\displaystyle Y\) are infinite and \(\displaystyle X\sim Y\), then \(\displaystyle X\cup Y\sim Y\)?

Very often the easiest way to tackle problems like this is to use the Schröder–Bernstein theorem. If $A\times A \sim A$ then there is a bijective map $\phi:A\times A\to A.$ Given distinct points $x,y\in A$, the map $U\times V \mapsto \phi\bigl((U\times \{x\})\cup (V\times\{y\})\bigr)$ then gives you an injective map from $\mathcal{P}(A)\times\mathcal{P}(A)$ to $\mathcal{P}(A).$ On the other hand $U\mapsto U\times A$ is an injection in the reverse direction.

 

[Plato2]

Suppose \(\displaystyle A\) is a set with at least two elements and \(\displaystyle A\times A\sim A.\) Then \(\displaystyle \mathcal{P}(A)\times\mathcal{P}(A)\sim\mathcal{P}(A).\)

The fact is that if \(\displaystyle A\) is finite then \(\displaystyle (A\times A)\sim A\) is impossible.
If \(\displaystyle \|A\|=n\text{ then }\|A\times A\|=n^2\)

Suppose that \(\displaystyle \{C,D,G,H\}\subset \mathcal{P}(A)\)

Now if \(\displaystyle (C,D)\ne(G,H)\text{ then }\left( {C \ne G} \right) \vee \left( {D \ne H} \right)\)

What can you do with that?