# Does taking the power set give you the “next biggest cardinal”

I know that if you take the power set of a set, it has a higher cardinality. Therefore there are an infinite number of them as $P^{n}(\mathbb{N})$(the nth power set of the naturals) Let's say $$C_{n} = P^{n}(\mathbb{N})$$ My question is: Given $C_n$ and $C_{n+1}$ are there cardinal numbers between them? Or does taking a power set always give you the "next greatest cardinal".

NOTE: I don't know the formal notation for cardinals.

-
– Jeppe Stig Nielsen Jul 14 '14 at 14:19
@TylerHG: The notation is really $\beth_n$ in this case. – Asaf Karagila Jul 14 '14 at 14:22
True true. Beth numbers – TylerHG Jul 14 '14 at 14:33

Not necessarily.

The standard axioms of set theory, $\sf ZFC$ cannot prove nor disprove whether or not, given an infinite set $A$ there is a cardinal between that of $|A|$ and $\mathcal P(A)$.

To get to the "next cardinal" we use a different method, which is rooted in the fact that cardinals in modern set theory are ordinals, and that we can prove that there is a least ordinal whose cardinality is larger than a given cardinal. This hierarchy, which begins with $\aleph_0=|\Bbb N|$ is called the $\aleph$ numbers.

We can talk about the cardinals of power sets, and these define a hierarchy called $\beth$ numbers. Namely, $\beth_0=|\Bbb N|$ and $\beth_{\alpha+1}=2^{\beth_\alpha}$.

(Both hierarchies include "limit stages", and in both cases we take "the smallest cardinal possible". But that's not important right now.)

Perhaps it should also be noted, that if you consider $\bigcup\mathcal P^n(\Bbb N)$ you get a set whose cardinality is strictly larger than all the cardinalities of $\mathcal P^n(\Bbb N)$. The cardinality of this set is denoted by $\beth_\omega$, and it is a limit cardinal to which I referred before. The hierarchy then continues onwards, to $\beth_{\omega+1},\beth_{\omega+2}$ and so on. To understand better this hierarchy (as well the $\aleph$ numbers) you should learn more about ordinal numbers.

-
Oh dear. It looks like I have a lot of reading ahead of me. As usual, the answer is never as simple as you'd like it to be ^.^ – Cruncher Jul 14 '14 at 14:41
Your comment is right on the nose. If you search for "Continuum Hypothesis" on this site, you may find several answers that I have written which explain a little bit the phenomenon of unprovable statements in set theory. This is particularly relevant to this case, because as Hayden writes, the general statement is in fact called "The Generalized Continuum Hypothesis", and the 'particular' one is when considering the case of an intermediate cardinality between $|\Bbb N|$ and $|\mathcal P(\Bbb N)|$, or in better terms, the continuum hypothesis asserts $\aleph_1=\beth_1$. – Asaf Karagila Jul 14 '14 at 14:44

Not necessarily in ZFC; this is the Generalized Continuum Hypothesis when you are asking about infinite cardinals, which is shown to be independent of ZFC.

Note: the question "Does taking the power set give you the 'next biggest cardinal'" is the Generalized Continuum Hypothesis when restricted to infinite cardinals, but the claim that $\mathcal{P}^{n+1}(\mathbb{N})$ is the least cardinal greater than $\mathcal{P}^n(\mathbb{N})$ is strictly weaker.

However, it is important to note that it is false for finite cardinals; there are cardinals between $2$ and $2^2=4$, for example ($3$).

-
Strictly speaking, we only talk about the first $\omega$ power sets, so this is not $\sf GCH$, but rather a much weaker principle. But essentially yes. – Asaf Karagila Jul 14 '14 at 14:22
That is true, with the OP's formulation of the problem it is only the first $\omega$, but the question "does taking the power set given you the 'next biggest cardinal'" is GCH when restricted to infinite cardinals. I'll clarify that in my answer. – Hayden Jul 14 '14 at 14:25
Well, $\sf GCH$ is provably false if you include finite cardinals. – Asaf Karagila Jul 14 '14 at 14:26
@AsafKaragila Yes, as I indicate in my answer – Hayden Jul 14 '14 at 14:26
Yes, but you wrote in your comment, "is GCH when restricted to infinite cardinals". Which was why I wrote my second comment. – Asaf Karagila Jul 14 '14 at 14:27