I recently edited an answer of mine on math.SE which discussed the implication of the two assertions:

- $AH(0)$ which is $2^{\aleph_0}=\aleph_1$, and
- $CH$ which says that if $A\subseteq 2^{\omega}$ and $\aleph_0<|A|$ then $|A|=2^{\aleph_0}$.

We know they are indeed equivalent under the axiom of choice (and actually much less). It is also trivial to see that $AH(0)\Rightarrow CH$. However the converse is not true, indeed in Solovay's model (or in models of AD) there are no $\aleph_1$ many reals, but $CH$ holds since every uncountable set of reals has a perfect subset.

While revising my answer I tried to find a reference whether or not in the Feferman-Levy model, in which the continuum is a countable union of countable sets, satisfies the continuum hypothesis (we already know that it does not satisfy $AH(0)$).

To my surprise the answer is negative. There exists a set whose cardinality is strictly between the continuum and $\omega$, the construction is described in A. Miller's paper [1] in which he remarks that in the Feferman-Levy the constructed set cannot be put in bijection with the continuum.

I was wondering whether or not this is *always* true in models in which the continuum is a countable union of countable sets, or is this just one of the peculiarities of the Feferman-Levy model.

Questions:

Let $V$ be a model of $ZF$ in which $2^{\omega}$ can be written as the countable union of countable sets. Does $CH$ fail in $V$?

Suppose that $V$ is a model of $\omega_1\nleq2^\omega$ and $CH$, does this imply that $\omega_1$ is regular (which means inaccessible in $L$)?

**Bibliography:**

- Miller, A.
**A Dedekind Finite Borel Set.***Arch. Math. Logic*50 (2011), no. 1-2, 1--17.