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  • $\begingroup$ Asaf, I think you are mistaken. Here is a proof in ZF that if ${\mathbb R}$ is a countable union of countable sets (as in the Feferman-Levy model) then every well-orderable subset of ${\mathbb R}$ is countable: If $\omega_1$ injects into ${\mathbb R}$, then so does the set ${\omega_1}^\omega$, because $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. But then, by Schröder-Bernstein, ${\omega_1}^\omega$ has the same size as ${\mathbb R}$. (Cont.) $\endgroup$
    – Andrés E. Caicedo
    Commented Oct 20, 2011 at 1:53
  • $\begingroup$ (II). Now, here is the nice bit, essentially following the argument for König's lemma: If $X$ is a countable union of countable sets, then no map from $X$ to $\omega_1^\omega$ is onto. For if $f$ is a map and $X=\bigcup_n X_n$ with each $X_n$ countable, then each $f[X_n]$ is countable, so $T_n=\omega_1\setminus\{f(x)(n)\mid x\in X_n\}$ is non-empty. Let $\Phi:\omega\to\omega_1$ be the map that at $n$ picks the minimum of $T_n$. Then $\Phi$ is not in the range of $f$. $\endgroup$
    – Andrés E. Caicedo
    Commented Oct 20, 2011 at 1:58
  • $\begingroup$ The fact that $\omega_1$ doesn't embed into $\mathbb{R}$ in the Feferman-Levy model is apparently due to Cohen - math.wisc.edu/~miller/res/two-pt.pdf $\endgroup$
    – François G. Dorais
    Commented Oct 20, 2011 at 2:55
  • $\begingroup$ @Andres: Thanks a lot, as I wrote to Joel in a comment, it seems that this was a dream that convinced me that in the Feferman-Levy $2^\omega=\omega_1$, as I can't find that reference anywhere. @Francois: Thanks for the reference, it seems interesting and I'll give it a read. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 20, 2011 at 8:36