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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:

  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.
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  • $\begingroup$ I have added a second question which seems to be equivalent of the first. It may be an easier formulation... $\endgroup$
    – Asaf Karagila
    Aug 1, 2012 at 23:01
  • $\begingroup$ The recent trend of completely arbitrary downvotes seems really strange. $\endgroup$
    – Asaf Karagila
    Jul 9, 2013 at 4:54
  • $\begingroup$ So "Borel sets are analytic" is not a theorem of ZF? Interesting. $\endgroup$ Dec 26, 2015 at 23:38
  • $\begingroup$ Paul, depends on how you define "Borel" and "analytic". $\endgroup$
    – Asaf Karagila
    Dec 27, 2015 at 5:38
  • 2
    $\begingroup$ That's interesting too. $\endgroup$ Dec 27, 2015 at 10:11

1 Answer 1

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The answer for the second question is no. Truss proved in [1] 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.
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  • $\begingroup$ 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? $\endgroup$ Apr 8, 2016 at 13:35
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Apr 8, 2016 at 14:14
  • $\begingroup$ Ah! That makes sense and 4 is even more interesting now. $\endgroup$ Apr 8, 2016 at 16:54
  • $\begingroup$ I find (1) and (4) in conjunction to be the interesting part. :) $\endgroup$
    – Asaf Karagila
    Apr 8, 2016 at 17:42

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