Is there a set $\mathbb{X}$ such that $\mathcal{P}(\mathbb{X})$ has a cardinality of $\aleph_0$ ? Give a proof.
As for me, i think no, because $\aleph_0$ is a cardinality of a set of natural numbers, then suposing $\mathbb{X}$ is a set of natural numbers, then $\mathcal{P}(\mathbb{X})$ would be $2^{\aleph_0}$, which is equal to $\aleph_1$. Am i right or wrong? Please provide your thoughts.