# equinumerous and infinite sets

#### Andrei

##### Member
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

##### MHB Oldtimer
Staff member
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.

#### Plato

##### Well-known member
MHB Math Helper
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?