## Is it consistent relative to ZF that $\frak c = \aleph_\omega$?

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In ZFC we know that the continuum cannot have cofinality $\omega$.

However, in the Feferman-Levy model we have that $\frak c=\aleph_1$, and that $\operatorname{cf}(\omega_1)=\omega$. In fact in the Feferman-Levy model, $\aleph_\omega^L=\aleph_1^V$.

Is it consistent with ZF that $\frak c=\aleph_\omega$? Does that mean that the only restriction in ZF on the cardinality of the continuum is $\aleph_0<\frak c$?

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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.)  – Andres Caicedo Oct 20 2011 at 1:53
(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 $f$. – Andres Caicedo Oct 20 2011 at 1:58
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 – François G. Dorais♦ Oct 20 2011 at 2:55
@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. – Asaf Karagila Oct 20 2011 at 8:36

## 1 Answer

The answer is no. The continuum cannot be $\aleph_\omega$, and this can be proved in ZF, that is, without using the axiom of choice. To see this, suppose towards contradiction that $P(\omega)$ is equinumerous with $\aleph_\omega$. Since $P(\omega)$ is equinumerous with $P(\omega)^\omega$, and this does not require AC, it follows that there is a bijection $f:\aleph_\omega\cong (\aleph_\omega)^\omega$. Let $g(n)$ be the first ordinal $\alpha\lt\aleph_\omega$ that is not among $f(\beta)(n)$ for any $\beta\lt\aleph_n$. Since there are fewer than $\aleph_\omega$ many such $\beta$, it follows that there are fewer than $\aleph_\omega$ many such $f(\beta)(n)$, and so such an $\alpha$ exists. Thus, $g:\omega\to \aleph_\omega$. But notice that for any particular $\alpha\lt\aleph_\omega$, we have $\alpha\lt\aleph_n$ for some $n$ and consequently $g(n)\neq f(\alpha)(n)$, and thus $g\neq f(\alpha)$. Thus, $f$ was not surjective to $(\aleph_\omega)^\omega$ after all, a contradiction.

This is just a standard proof of Konig's theorem (that $\aleph_\omega^\omega\gt\aleph_\omega$), and the point is that it doesn't use AC.

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This argument is essentially the same as the argument that Andres gave in the comments above, but it seems to answer the whole question, and not just the issue about the reals being a countable union of countable sets. The proof shows more generally that the continuum, if well-orderable, cannot have countable cofinality, even in ZF. – Joel David Hamkins Oct 20 2011 at 3:41
Thanks a lot, Joel. I swear I read somewhere that in the Feferman-Levy model $2^\omega=\omega_1$. I can't find it, and the proofs you and Andres gave convince me that it was probably in a dream. – Asaf Karagila Oct 20 2011 at 8:33

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Just for the record, the standard proof of the more general Konig's theorem ($\kappa^{cof(\kappa)}>\kappa$ for $\kappa$ any aleph) doesn´t use AC (it is essentially the same proof that Joel already gave). – Ramiro de la Vega Oct 20 2011 at 13:00
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