FALSE: $2^{\aleph_0} = \aleph_\omega$

FALSE. Assume the Axiom of Choice (The Axiom of Choice). The continuum has cardinality $\aleph_\omega$:

$2^{\aleph_0} = \aleph_\omega$

(Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations, The successor cardinal $\kappa^{+}$, the alephs $\aleph_\alpha$, the beths $\beth_\alpha$, successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$).

The claim is plausible because ZFC really does leave the value of $2^{\aleph_0}$ open over a wide range of alephs, and $\aleph_\omega$ is a natural-looking candidate: it is above $\aleph_0$, as Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$, and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$ requires, and it is not a successor, so no obvious counting argument seems to touch it. Nevertheless ZFC refutes it outright, and the refutation is short.

What is being used, and what is not. The refutation uses only Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$, and $\operatorname{cf}(2^{\kappa}) > \kappa$; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$, itself a consequence of König's theorem: assuming the Axiom of Choice, if $\kappa_i < \lambda_i$ for every $i \in I$ then $\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i$, together with the ZF computation of $\operatorname{cf}(\aleph_\omega)$. No independence result is used anywhere: this is a theorem of ZFC, not a statement about what ZFC fails to decide, and it would be equally true in any model of ZFC.

Which values remain possible is a different question, and is not settled here. The cofinality constraint excludes candidate values whose cofinality is $\aleph_0$; it does not identify the value of $2^{\aleph_0}$, and the choice ledger at the end of the main page records what is and is not decided.

The parallel false claim about $\aleph_1$ is of a different kind. "$2^{\aleph_0} = \aleph_1$" is not refutable here at all: it is the continuum hypothesis, and neither it nor its negation follows from anything on this page or on the pages this one rests on, as What each result on this page costs in choice, and where the continuum escapes what ZFC can decide records. That difference is exactly why the present statement is on the page and the other is not: a false statement in this library carries a refutation, and a refutation must be a proof.