I want to find the cardinality of the cartesian product of N and P(Q), which is the power set of the rationals. I wish to compare it to the cardinality of P(NxQ), i.e, the power set of the cartesian product of N and Q.
My intuition say that it's equal to the cardinality of R, am I correct ? Why ?
Thank you
For infinite $A$ and $B$ we have $|A\times B| = \max(|A|,|B|),$ so $|\mathbb N \times \mathcal P(\mathbb Q)| = |\mathcal P(\mathbb Q)| = |\mathbb R|.$ Likewise $\mathbb N\times \mathbb Q$ is a countable set, so its power set has the cardinality of the reals.
Edit As a commenter notes, this general approach requires AC. But the core fact that $\mathbb N\times \mathbb R \simeq \mathbb R$ does not require choice. There is an easy to see bijection $\mathbb Z \times [0,1)\to \mathbb R$ and then we have $\mathbb N\simeq \mathbb Z$ and $[0,1)\simeq \mathbb R.$