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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 $P(\omega)^\omega$ P(\omega)$is equinumerous with$P(\omega)$, P(\omega)^\omega$, 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 $g(n)=f(\alpha)(n)$, 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$), and the point is that it doesn't use AC. 1 The answer is no. The continuum cannot be$\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)^\omega$is equinumerous with$P(\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)=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.