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?



Does ZF bound countable unions of countable sets?
In Jech *The Axiom of Choice*, problem 14 on chapter 5 states:



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> 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 $N$:
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> 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.
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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.
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> 
> 




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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.


