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

Hi, i know that it is consistent with ZF without choice that the reals are the countable union of countable sets. Is there any good reference to read a proof? Thanks

share|cite|improve this question
Is there a reason that you say "the countable union of countable sets" instead of just "countable"? – Aaron Tikuisis Jun 26 '12 at 20:53
Aaroni Tikuisis: "Is there a reason to say `the countable union of countable sets' instead of just 'countable'?". Yes. The reals are not countable. – Steven Landsburg Jun 26 '12 at 21:00
To elaborate: the result "countable union of countable sets is countable" requires some amount of the axiom of choice, which is not provable in ZF. – Noah Schweber Jun 26 '12 at 21:13
By contrast, ZF does prove that $\mathbb{R}$ is not countable. – Noah Schweber Jun 26 '12 at 21:14

T. Jech, The Axiom of Choice. This particular proof appears in Chapter 10.

Essentially, the forcing goes through collapsing all the $\aleph_n$ (for finite $n$) to be countable, so in the full generic extension $\aleph_\omega$ of the ground model is countable too, but if we take permutations based on conditions based on finitely many collapses, then $\aleph_\omega$ of the ground model is not collapsed, and thus it becomes $\aleph_1$.

It is not difficult to show that if the ground model satisfied GCH then every real number in this symmetric extension came from a collapse of some $\aleph_n$, and those are countable. So we have that the real numbers are a countable union of countable sets.

share|cite|improve this answer
If you are more comfortable with partial orders than with Boolean algebras, you may want to read the presentation of this result in Ioanna Dimitriou's master thesis, available at – Andrés E. Caicedo Jun 26 '12 at 23:12

There are many references in Andres Caicedo's answer here: In particular he refers to Jech's book "The axiom of choice" for a proof of this result.

share|cite|improve this answer
It's theorem 10.6 in Jech's book, pp. 142 – godelian Jun 26 '12 at 21:21
Paul, you asked for references. Jech's book is very good. If you want references other than this you should say that. – Asaf Karagila Jun 26 '12 at 22:54

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.