How can the cardinalities of a set and its Power set be the same? How can the cardinalities of a set and its Power set be the same? If $|\mathcal P(\aleph_n)| = \aleph_{n+1}$, then $|\mathcal P(\aleph_{\aleph_0})| = \aleph_{\aleph_0 + 1}|$. But $\aleph_0 + 1 = \aleph_0$, thus $|\mathcal P (\aleph_{\aleph_0})| = \aleph_{\aleph_0}$?
 A: This is the reason why the index of $\aleph$ numbers is given by ordinals and not cardinals.
So $\aleph_{\aleph_0}$ is in fact $\aleph_\omega$ and since as ordinals, $\omega\neq\omega+1$ (even though they have the same cardinality), the cardinals $\aleph_\omega$ and $\aleph_{\omega+1}$ are distinct.
(And it should be mentioned that $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ is called the Generalized Continuum Hypothesis and it cannot be proved or disproved from the standard axioms of set theory. It is perfectly consistent that $2^{\aleph_n}=\aleph_{n+3}$ for all the finite $n$'s.)
A: First, you don't really intend anything here to be about alephs. The cardinals that play nice with the power set operation are the beth cardinals, and the only beth cardinal that is guaranteed to have anything to do with the alephs under ZFC is $\beth_0$ which is $\aleph_0$.
Second, the subscript in the alephs and the beths is not a cardinal number but an ordinal number. Thus the object you're thinking of is really $\beth_\omega$, where $\omega$ is the first countable ordinal. This is defined in a different way from the lower beth numbers (i.e. it is not actually the cardinality of the power set of some set). Unlike with infinite cardinals, $\omega+1$ is indeed greater than $\omega$ so it is self-consistent to say $\beth_{\omega+1}>\beth_\omega$ (which it is, by Cantor's theorem).
A: I’m assuming that you are operating in a system where the generalized continuum hypothesis is true.
A set and its power set can not have the same cardinality due to cantors theorem. The mistake you made is the subscripts on $\aleph$ are ordinals not cardinals.  So $\aleph_\aleph$ should be $\aleph_\omega$.
