What is the cardinality of $P(\mathbb N)\times \mathbb R$?

Question

Which set below has the same cardinality as $P(\mathbb N)\times \mathbb R$:

1. $P(\mathbb N\times \mathbb R)$

2. $\mathbb Q\times P(\mathbb R)$

3. $P(\mathbb R)$

4. $P(\mathbb Q)$

I think $|P(\mathbb N)\times \mathbb R|=|\mathbb R\times \mathbb R|=|\mathbb R|=c$. So $|P(\mathbb N)\times \mathbb R|=|P(\mathbb Q)|$.

Not sure if this is correct.

Answer 1

You are correct. You can biject $\mathbb{Q}$ with $\mathbb{N}$, biject $P(\mathbb{N})$ with $\mathbb{R}$, and biject $\mathbb{R} \times \mathbb{R}$ with $\mathbb{R}$.

In particular, you can biject $P(\mathbb{Q}) \times \mathbb{R}$ with $P(\mathbb{N}) \times \mathbb{R}$ with $\mathbb{R} \times \mathbb{R}$ with $\mathbb{R}$ with $P(\mathbb{N})$ with $P(\mathbb{Q})$.

This means that they have the same cardinalities, as you expected/asserted.

(The line above can be shortened a bit by observing $P(\mathbb{Q})$ bijects with $\mathbb{R}$.)

Each of the other sets has cardinality at least $|P(\mathbb{R})|$, which is strictly greater than $|\mathbb{R}|$.

(In fact, each of the remaining sets has cardinality $|P(\mathbb{R})|$.)

Answer 2

You're right. All other options are wrong due to having cardinality $2^c=\beth_2$.