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

[Andrei1]

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

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

 

[Opalg]

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

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

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

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

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

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

What can you do with that?