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

  • 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
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    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
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    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
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    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.

There are many references in Andres Caicedo's answer here: https://math.stackexchange.com/questions/16246/ In particular he refers to Jech's book "The axiom of choice" for a proof of this result.

  • It's theorem 10.6 in Jech's book, pp. 142 – godelian Jun 26 '12 at 21:21
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    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

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