# The Continuum Hypothesis and Countable Unions

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  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:

1. 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 $$V$$?

2. 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:

1. Miller, A. A Dedekind Finite Borel Set. Arch. Math. Logic 50 (2011), no. 1-2, 1--17.
• I have added a second question which seems to be equivalent of the first. It may be an easier formulation... – Asaf Karagila Aug 1 '12 at 23:01
• The recent trend of completely arbitrary downvotes seems really strange. – Asaf Karagila Jul 9 '13 at 4:54
• So "Borel sets are analytic" is not a theorem of ZF? Interesting. – Paul Larson Dec 26 '15 at 23:38
• Paul, depends on how you define "Borel" and "analytic". – Asaf Karagila Dec 27 '15 at 5:38
• That's interesting too. – Paul Larson Dec 27 '15 at 10:11

The answer for the second question is no. Truss proved in  that if we repeat Solovay's construction from a limit cardinal $\kappa$, we obtain a model in which the following properties:

1. Countable unions of countable sets of real numbers are countable;
2. Every well-orderable set of real numbers is countable;
3. Every uncountable set of reals has a perfect subset;
4. DC holds iff $\omega_1$ is regular iff $\kappa$ is inaccessible in the ground model;
5. Every set of real numbers is Borel.

This shows that it is possible to have $CH+\aleph_1\nleq2^{\aleph_0}+\operatorname{cf}(\omega_1)=\omega$. However it does not answer the original (first) question.

Bibliography:

1. Truss, John, Models of set theory containing many perfect sets. Ann. Math. Logic 7 (1974), 197–219.
• Interesting. Does Truss use an inaccessible? Item 2 implies that $\omega_1$ is inaccessible to reals, so I suppose he must use one somewhere. Or am I missing something? – François G. Dorais Apr 8 '16 at 13:35
• Yes, it is inaccessible go reals, which only means it is a limit cardinal in L. But in Truss' models he can have $\omega_1$ singular also. – Asaf Karagila Apr 8 '16 at 14:14
• Ah! That makes sense and 4 is even more interesting now. – François G. Dorais Apr 8 '16 at 16:54
• I find (1) and (4) in conjunction to be the interesting part. :) – Asaf Karagila Apr 8 '16 at 17:42