From what I understand, it's independent of ZFC whether $2^{\aleph_0} > \aleph_\alpha$ for any nonzero ordinal $\alpha$. But there are larger cardinalities than the $\aleph$ numbers, especially if we start including large cardinal axioms. So my question is twofold.

  1. Can ZFC alone prove the existence of a cardinal $\kappa$ such that $2^{\aleph_0} \le \kappa$?
  2. Do any of the large cardinal axioms together with ZFC imply the existence of a $\kappa$ with $2^{\aleph_0} \le \kappa$?
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    $\begingroup$ Under $\mathsf{ZFC}$ every infinite cardinal is an $\aleph_\alpha$ for some ordinal $\alpha$. $\endgroup$ – Brian M. Scott Jan 14 '17 at 20:35
  • $\begingroup$ @BrianM.Scott Oh. Well I guess that answers the first question. Follow-up: Can ZFC prove there exists $\alpha$ such that $2^{\aleph_0} = \aleph_\alpha$, even if it can't prove which one? $\endgroup$ – eyeballfrog Jan 14 '17 at 20:36
  • $\begingroup$ @eyeballfrog: Yes: ZFC proves that $2^{\aleph_0}$ is an infinite cardinal, and ZFC proves that every infinite cardinal is $\aleph_\alpha$ for some $\alpha$. $\endgroup$ – hmakholm left over Monica Jan 14 '17 at 20:41
  • $\begingroup$ @eyeballfrog That's a consequence of what he said. Under ZFC, every set is in bijection with an $\aleph_\alpha$. $\endgroup$ – Noah Schweber Jan 14 '17 at 20:41

The axiom of choice is equivalent to stating that every infinite cardinal is an $\aleph$ number. We can talk about non-$\aleph$ cardinals in the absence of choice, but it's still true there is a proper class of $\aleph$ cardinals, regardless of assuming choice or not.

The continuum, therefore, is certainly an $\aleph$ number under $\sf ZFC$, and since we cannot prove that $2^{\aleph_0}=\aleph_1$ or not, there is no ordinal that $\sf ZFC$ proves is the $\aleph_\alpha$ of the continuum. What $\sf ZFC$ does prove is that if $2^{\aleph_0}=\aleph_\alpha$, then $\alpha>0$ and $\operatorname{cf}(\alpha)>\aleph_0$; and we know that this is the only provable restriction on the size of the continuum from the works of Cohen and Solovay.

To recap: assuming $\sf ZFC$, the continuum is an $\aleph$ cardinal.

Large cardinal axioms do postulate the existence of "large cardinals", but this is a technical term—without a well-defined meaning—and not a term for a cardinal larger than all the $\aleph$'s. It is just usually the case that if $\kappa$ is a large cardinal, then $\kappa=\aleph_\kappa$ (not always, though!) so it is sort of hard to describe $\kappa$ using the $\aleph$ numbers.

Moreover, due to the work of Levy and Solovay, we also know that most large cardinal axioms are compatible with both the continuum hypothesis and its negation, so large cardinals do not determine the value of the continuum either.

  • $\begingroup$ Do I remember correctly that $\kappa=\aleph_\kappa$ can even happen below the continuum? $\endgroup$ – hmakholm left over Monica Jan 14 '17 at 20:50
  • $\begingroup$ @HenningMakholm Sure - start with a ground model $V$, and consider $\kappa\in V$ with $V\models \kappa=\aleph_\kappa$; now add $\kappa^+$-many Cohen reals. This forcing is c.c.c., so preserves cardinals, so $\kappa=\aleph_\kappa$ continues to hold in $V[G]$; but $V[G]$ also satisfies $2^{\aleph_0}>\kappa$. Essentially, the point is that since the statement "$\kappa=\aleph_\kappa$" doesn't involve exponentiation, if it can happen at all then it can happen below $2^{\aleph_0}$. $\endgroup$ – Noah Schweber Jan 14 '17 at 20:53
  • $\begingroup$ @Henning: Yes, the $\alpha\mapsto\aleph_\alpha$ is continuous, so it has a closed and unbounded class of fixed points. Using what Noah suggested we can even arrange that $2^{\aleph_0}$ is a fixed point, simply requiring that the cofinality of the chosen number of generic reals is uncountable. $\endgroup$ – Asaf Karagila Jan 14 '17 at 20:57
  • $\begingroup$ I'd be happy to hear about the downvote! $\endgroup$ – Asaf Karagila Jan 15 '17 at 11:09

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