# ZF + the reals are the countable union of countable sets consistent

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
• 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

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.