cardinality of infinite sets with cartesian product

Question

claim: $A,B,C,D$ are infinite if $|A\times B|=|C\times D|$ then $|A|=|C|$, $|B|=|D|$ , prove or give a counter example.

So imo, the claim is false, using $A=D=\mathbb{R}$ , $B=C=\mathbb{N}$ , is it correct to say that |$\mathbb{R}$x$\mathbb{N}$|=$\aleph$ ? what are the guidelines to prove that?

Answer 1

A slightly finer question might be whether $\lvert A \times B \rvert = \lvert C \times D \rvert$ implies $\lvert A \rvert = \lvert C \rvert$ and $\lvert B \rvert = \lvert D \rvert$, or $\lvert A \rvert = \lvert D \rvert$ and $\lvert B \rvert = \lvert C \rvert$.

However, assuming Choice, the cardinality of the product of two infinite sets is the largest cardinality of the two factors. So if one takes $A = C = 2^{\mathbb{R}}$, $B = \mathbb{N}$, $D = \mathbb{R}$, we see that even this fails.

So (again under Choice) all one can infer from $\lvert A \times B \rvert = \lvert C \times D \rvert$ is that one of $A, B$ will have the same cardinality as one of $C, D$.

Answer 2

For any sets X and Y (infinite or otherwise), $|X \times Y| = |Y \times X|$. This is easily demonstrated for non-empty sets by showing the existence of a bijection between $|X \times Y|$ and $|Y \times X|$, most obviously for example $(x, y) \leftrightarrow (y, x)$. On the other hand if either X or Y is empty, then $|X \times Y| = |Y \times X|$ = 0.

So, as in your example you can choose A = D = X, and B = C = Y and then $|A\times B|=|C\times D|$ as a consequence. If $|X| \ne |Y|$ then $|A| \ne |C|$ and $|B| \ne |D|$.