TITLE: Can the cardinality of continuum exceed all aleph numbers in ZF?
QUESTION [13 upvotes]: More precisely, is either of the following two statements consistent with ZF:

$2^{\aleph_0}\geq\aleph_{\alpha}$ for every ordinal number $\alpha$,
$2^{\aleph_0}\leq\aleph_{\alpha}\implies 2^{\aleph_0}=\aleph_{\alpha}$ for every ordinal number $\alpha$?

I'm asking mainly out of curiosity.

REPLY [5 votes]: I am reading your statement "$2^{\aleph_0} \geq \aleph_\alpha$" to say that there is an injection from $\omega_\alpha$ into $\mathbb{R}$.  (And similarly for "$2^{\aleph_0} \leq \aleph_\alpha$".)
(1) is not consistent, since given any set $X$, there are only set-many well-orderings of subsets of $X$.  (Without Choice, define $WO(X) = \{ R : R\text{ is a well-ordering of a subset of }X \}$.  This is clearly a set by Power Set, Pairing and Separation).  Using Replacement the family of order-types of elements of $WO(X)$ is also a set.  From this it follows that for every set $X$ there must be an $\alpha$ such that "$\aleph_\alpha \leq | X |$" does not hold.
(2) would only be consistent vacuously: where there is no well-ordering of the reals and so the statement $2^{\aleph_0} \leq \aleph_\alpha$ never holds.  (If there is an injection from $\mathbb{R}$ into $\omega_\alpha$, then there is also an injection from $\mathbb{R}$ into $\omega_{\alpha+1}$, and clearly $\aleph_\alpha \neq \aleph_{\alpha+1}$.)