One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have cardinality $\aleph_n$, where $n$ is any non-zero ordinal (not necessarily finite). I have a sneaking suspicion that my solution is actually invalid, but I can't find any reference which invalidates my conclusion.
I have read that it is provable in ZF that there are no cardinals $\kappa$ such that $\aleph_0 < \kappa < \aleph_1$, but I believe the conclusion of my proof does not preclude the possibility that the cardinality is incomparable to $\aleph_1$ or some such.
I think the weakest point in my solution is where I claim that the supremum of a countable set of countable ordinals is again countable. This is true, of course, but it sounds uncomfortably close to the claim "a countable union of countable sets is countable", which is well-known to be unprovable in ZF. Can anybody confirm that the ordinal version is provable in ZF though? If not, I think I can weaken the claim to "the supremum of any set of countable ordinals is at most $\omega_1$", and this establishes the weaker result that the cardinality of a countable union of countable sets is not $\aleph_n$ for any ordinal $n \ge 2$.