on the continuum hypothesis: only finitely many cardinals between $\aleph_0$ and $2^{\aleph_0}$

Question

Is it at least known that there are only finitely many cardinals between $\aleph_0$ and $2^{\aleph_0}$?

Answer 1

As the wikipedia explains, it is consistent with ZFC that $2^{\aleph_0} = \aleph_\alpha$ as long as some technical condition (the fact that $\alpha$ has uncountable cofinality) is satisfied. In particular, it is consistent that $\alpha$ could be absolutely enormous (the ordinals of uncountable cofinality aren't bounded), for example $\alpha = \omega_1$, the first uncountable ordinal, is a possibility.

Then there would be as many "intermediary" cardinals between $\aleph_0$ and $2^{\aleph_0}$ as there are ordinals $0 < \beta < \alpha$, of which there is in general infinitely many (for instance in the $\alpha = \omega_1$ example).

In short, the answer to your question is no.

Answer 2

The cardinality of a set of all these cardinals may be arbitrary (but still need to be lesser than or equal to continuum for obvious reasons). It is even possible than this cardinality is equal to continuum (by forcing $2^ {\aleph _0} =\kappa$, where $\kappa=\aleph { \kappa }$ is the fixed point of aleph hierarchy).