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

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 $\aleph_0<\frak c$?

• 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.) – Andrés E. Caicedo Oct 20 '11 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 $f$. – Andrés E. Caicedo Oct 20 '11 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 '11 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 '11 at 8:36

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.
• 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 '11 at 8:33
• 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 '11 at 13:00