MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 8 down vote accepted

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.

share|cite|improve this answer
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
1… $\:$ – 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.