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Aug 8, 2013 at 15:38 comment added Asaf Karagila @Andreas: True, we can make some statements about what sort of relations the cardinal of the continuum has with $\aleph$'s, but we can't really say "Ah. It's equal to this and that" without "this and that" being somewhat of a trivial statement (e.g. $\Bbb R$ or $\mathcal P(\Bbb N)$, or something similar).
Aug 8, 2013 at 15:18 comment added Andreas Blass "That is all we can say" might be a bit of an overstatement; we can make some comparisons between the continuum and well-ordered cardinals. For example, we can say (and you probably have said elsewhere) that $\mathbb C$ can be mapped onto $\omega_1$ (a theorem of ZF) and that in Solovay's model $\omega_1$ cannot be mapped one-to-one into $\mathbb C$.
Aug 8, 2013 at 12:07 comment added Asaf Karagila $\aleph_1$ is by definition in $\sf ZF$ equals to $\omega_1$. In every model where $\Bbb C$ cannot be well-ordered, its cardinality is not an ordinal. That's all we can say.
Aug 8, 2013 at 12:06 comment added user38200 And also, is $\aleph_1$ the same thing as $\omega_1$ in these models?
Aug 8, 2013 at 12:03 comment added user38200 And the same thing for Shelah's model, right?
Aug 8, 2013 at 12:01 comment added Asaf Karagila In the Solovay model, the cardinality of $\Bbb C$ cannot be described by an ordinal. To see more about the continuum hypothesis itself, math.stackexchange.com/questions/404807/…
Aug 8, 2013 at 11:58 vote accept user38200
Aug 8, 2013 at 11:57 comment added user38200 I read in Jech's axiom of choice that the generalized continuum hypothesis implies the axiom of choice. What for the continuum hypothesis itself?
Aug 8, 2013 at 11:51 comment added user38200 Thank you. But what is the cardinality of $\mathbb{C}$ in Solovay's model then?
Aug 8, 2013 at 11:47 history answered Asaf Karagila CC BY-SA 3.0