how many infinite cardinals are smaller than $\aleph$?

Question

What can be said in general (if we do not assume the continuum hypothesis) about the cardinality of the set of all infinite cardinals smaller than $\aleph$?

Answer 1

I assume by $\aleph$ you mean the continuum, the size of the reals, which is more commonly denoted $\mathfrak c$ or $2^{\aleph_0}$.

There is no bound, actually. Meaning, it is consistent that $2^{\aleph_0}=\aleph_\kappa=\kappa$ for some (necessarily limit) ordinal $\kappa$, so there are $\kappa$ (that is, continuum many) infinite cardinals below the continuum. Of course, we cannot have more than this. (That is, relative to its size, it can be as large as possible.)

More can be said: If $2^{\aleph_0}=\aleph_\beta$, then either $\beta$ is a successor ordinal, or it is a limit ordinal of cofinality larger than $\omega$. This follows from König's theorem.

Solovay showed, in the very early days of forcing, that in a sense this is all we can say: Starting with a model of $\mathsf{GCH}$, we can pick any ordinal $\beta$ satisfying those two restrictions, and force to obtain a model with the same cofinalities (and cardinals, and ordinals) as the ground model, and where $2^{\aleph_0}=\aleph_\beta$. (That is, in absolute terms, there is no bound.)

This was announced in

Robert M. Solovay. Independence results in the theory of cardinals. I, II (abstracts). Notices of the AMS, 10, (1963), 595,

and appeared as

Robert M. Solovay. $2^{\aleph_0}$ can be anything it ought to be, in The Theory of Models, Proceedings of the 1963 International Symposium at Berkeley. Amsterdam, North-Holland, 1965, Addison, Henkin, Tarski, eds., pg. 435.

What he needs to obtain in the extension that $2^{\aleph_0}=\aleph_\beta$ is that, in the ground model, $\aleph_\beta^{\aleph_0}=\aleph_\beta$, which of course follows from $\mathsf{GCH}$ if $\beta$ is successor or limit of uncountable cofinality.

So, to summarize: The continuum, in "absolute" terms, can be as large as you want. In relative terms as well: There can be continuum many cardinals below. And it can be any $\aleph_\beta$ at all, as long as the restriction given by König's theorem that $\mathrm{cf}(\mathfrak c)>\omega$ is satisfied.

The other question we could ask is whether $2^{\aleph_0}$ can be weakly inaccessible (that is, regular limit). The answer is positive (Solovay's argument does not change any cofinalities), but more is possible: It can even be real-valued measurable, which implies that $2^{\aleph_0}=\aleph_\kappa=\kappa$ for some regular $\kappa$ such that there are $\kappa$ weakly inaccessibles below (that are limit of weakly inaccessibles that themselves are limit of weakly inaccessibles...), it implies that there is an extension of Lebesgue measure defined on all sets of reals, and more. And it is consistent as well that the continuum satisfies various strengthenings of this notion.