Is it consistent relative to ZF that $\frak c = \aleph_\omega$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:22:31Z http://mathoverflow.net/feeds/question/78627 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78627/is-it-consistent-relative-to-zf-that-frak-c-aleph-omega Is it consistent relative to ZF that $\frak c = \aleph_\omega$? Asaf Karagila 2011-10-19T23:52:22Z 2011-10-20T03:34:02Z <p>In ZFC we know that the continuum cannot have cofinality $\omega$. </p> <p>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, $\aleph_\omega^L=\aleph_1^V$. </p> <p>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 $\aleph_0&lt;\frak c$?</p> http://mathoverflow.net/questions/78627/is-it-consistent-relative-to-zf-that-frak-c-aleph-omega/78640#78640 Answer by Joel David Hamkins for Is it consistent relative to ZF that $\frak c = \aleph_\omega$? Joel David Hamkins 2011-10-20T03:28:45Z 2011-10-20T03:34:02Z <p>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.</p> <p>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.</p>