# Does ZF bound countable unions of countable sets?

ZF proves that whenever a countable union of countable sets can be well ordered then its cardinality is at most $\aleph_1$. But what if it cannot be well ordered? The Feferman-Levy model shows the continuum can be a countable union of countable sets. Is there a ZF proof that a countable union of countable sets must be smaller than or equal to that in size?

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I think that should be $\aleph_0$ in the first sentence. – Michael Greinecker Oct 15 '12 at 7:21
Michael, no. It should be $\aleph_1$. If $\aleph_1$ is singular then it is the countable union of countable ordinals; but $\aleph_2$ (even if singular) can never be the countable union of countable ordinals. – Asaf Karagila Oct 15 '12 at 7:47

In Jech The Axiom of Choice, problem 14 on chapter 5 states:

Let $M$ be a transitive model of ZFC, there exists $M\subseteq N$ with the same cardinals as $M$ [read: initial ordinals] and the following statement is true in $N$:

For every $\alpha$ there exists a set $X$ such that $X$ is a countable union of countable sets, and $\mathcal P(X)$ can be partitioned into $\aleph_\alpha$ nonempty sets.

Note that $\mathcal P(X)$ can only be mapped onto set many ordinals, so this ensures us that there is indeed a proper class of cardinalities which can be written as countable unions of countable sets.

Jech adds and points out that this $N$ is not an inner model of any transitive model of ZFC, as that would imply $2^{\aleph_0}$ can be partitioned into any number of sets. The reference given is to Morris from 1970:

D. B. Morris, A model of ZF which cannot be extended to a model of ZFC without adding ordinals, Notices Am. Math. Soc. 17, 577.

I haven't sat down to verify all the details, but it should probably work:

Using an Easton symmetry-like class forcing, we may obtain model add for every regular $\kappa$ a countable collection of countable subsets of $2^\kappa$ whose union is not countable, let us denote this union $A_\kappa$. Genericity arguments should give us that for $\kappa\neq\kappa'$, $|A_\kappa|$ is incomparable with $|A_{\kappa'}|$ - namely there are no injections between those two sets.

This shows that there is a proper class of different sets which can be represented as countable unions of countable sets. So no upper-bound is possible.

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I am working on something similar in nature, so I will have the answer whether or not this argument is correct some time in the future. I will post an update when I'm done. – Asaf Karagila Oct 15 '12 at 8:01
– Ricky Demer Oct 15 '12 at 8:59
Ricky, and? Smaller cardinality need not be countable. – Asaf Karagila Oct 15 '12 at 9:10
Oh yeah, I misread the question. $\:$ – Ricky Demer Oct 15 '12 at 9:22