Ordinal rank of inaccessible cardinals

Question

Maybe this question is still confused:

By Powerset applied to any countable infinite set, $V_{\omega+2}$ contains sets of uncountable cardinality, which means that the least ordinal equinumerous with those sets is uncountable. But since the rank of an ordinal is equal to itself, the least uncountable ordinal $\omega_1$ doesn't appear until much later in the hierarchy, namely, at $V_{\omega_1+1}$. How is this explained?

Also, given an inaccessible cardinal $\kappa$, what is the least rank of a set $A$ whose cardinality is $\kappa$?

Answer 1

The von Neumann hierarchy is very "wide", not very "tall". Namely, by taking power sets, we quickly rise in cardinality, but not in "depth" (which is what the notion of rank sort of models).

This is not a problem, and it's even kind of intuitive because for one, it also something that holds for the finite levels of the hierarchy. $V_4$ has an element of size $4$ (namely, $V_3$), but $4$ itself only appears in the next level. And things only get worse when you continue.

Now. Let's analyze something... define the following sequence of cardinals: $\alpha_0$ is an arbitrary fixed cardinal; $\alpha_{n+1}=|V_{\alpha_n}|$; and finally $\alpha=\sup\{\alpha_n\mid n<\omega\}$. I claim that any $\eta<\alpha$ satisfies that $|V_\eta|<\alpha$. And indeed, if $\eta<\alpha$, then $\eta<\alpha_n$ for some $n$, and therefore $V_\eta\subseteq V_{\alpha_{n+1}}$. Therefore, the first time a set of size $\alpha$ appears is exactly $V_{\alpha+1}$, where $\alpha$ itself is also a member.

But we can say even more. We can give a complete characterizations of all the $\alpha$'s which satisfy the last property. These are exactly those $\alpha$ such that $\beth_\alpha=\alpha$ (recall the $\beth$ numbers are defined by taking power sets at successor levels and limits at limits). This requires some calculation, but it's a good exercise.

Finally!

Where does that put us with inaccessible cardinals? If you mean strongly inaccessible, then $\kappa$ is a $\beth$ fixed-point, i.e. $\kappa=\beth_\kappa$, in which case $\kappa+1$ is the least rank of a set of size $\kappa$. But if we are only talking about weakly inaccessible cardinals, well, $2^{\aleph_0}$ can consistently be largrer than the least weakly inaccessible cardinals (in fact, it can be as big as we want), which would mean that $V_{\omega+2}$ already knows about some very, very large sets.

Answer 2

I don't understand your first question, but here's an answer to your second question. If $\kappa$ is inaccessible, then $|V_\alpha|<\kappa$ for all $\alpha<\kappa$. This is easy to prove by induction on $\alpha$: at successor steps you use the fact that $\kappa$ is strong limit and at limit steps you use the fact that $\kappa$ is regular. In particular if $A$ is a set of rank $\alpha<\kappa$, then $A\subseteq V_\alpha$ and so $|A|<\kappa$.