Welcome to our community

Be a part of something great, join today!

equinumerous and infinite sets

Andrei

Member
Jan 18, 2013
36
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
Feb 7, 2012
2,721
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
Jan 27, 2012
196
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?