TITLE: ZF +  the reals are the countable union of countable sets consistent
QUESTION [9 upvotes]: 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

REPLY [13 votes]: 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.