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The GCH axiom basically says that for all infinite cardinal numbers $\kappa$, the number of cardinals lying strictly between $\kappa$ and $2^\kappa$ is as small as possible. Namely, there are none.

Is there an axiom which claims the opposite, in other words that the number of cardinal numbers lying strictly between $\kappa$ and $2^\kappa$ is as large (in some sense) as possible?

Edit. For example - and I don't know if this is a silly suggestions, I know very little set theory - is the following axiom for infinite cardinals $\kappa$ consistent with ZFC? And if so, is it interesting? $$|\{\mbox{cardinals } \nu \mid \kappa<\nu<2^\kappa\}|=2^\kappa$$

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Note that there is no "largest possible distance" between even $\aleph_0$ and $\mathfrak{c} = 2^{\aleph_0}$. It is an old result that as long as $\aleph_\alpha$ has uncountable cofinality, then it is relatively consistent with $\mathsf{ZFC}$ that $\mathfrak{c} = \aleph_\alpha$. As every infinite successor cardinal has uncountable cofinality, this implies that there is no bound on the number of cardinals strictly between $\aleph_0$ and $\mathfrak{c}$.

Easton's Theorem goes even further, and says that except for certain basic restrictions, the function $\aleph_\alpha \mapsto 2^{\aleph_\alpha} = \aleph_{G(\alpha)}$ restricted to the regular cardinals can be pretty much arbitrary. (As Andrés Caicedo notes in his comment below, under the assumption of certain large cardinal hypotheses, the arbitrariness is further restricted. As a basic example, the least (infinite) cardinal at which $\sf{GCH}$ fails cannot be measurable.)

The answers to these questions might also be of interest:

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  • $\begingroup$ Could there be other possible notions of 'distance between $\kappa$ and $2^\kappa$' that don't run into this problem? $\endgroup$ – goblin Jun 17 '13 at 6:40
  • $\begingroup$ @user18921: At the moment I cannot think of one that wouldn't fall under Easton's result. I'll think about this some more, however. $\endgroup$ – user642796 Jun 17 '13 at 6:50
  • $\begingroup$ Well, there are some restrictions. If $\kappa$ is a mildly large cardinal (measurable, for example), then the conditions in Easton's theorem are by far not all $2^\kappa$ needs to satisfy. Also, Easton's result does not touch singular cardinals, and the story there is much more complicated. $\endgroup$ – Andrés E. Caicedo Jun 17 '13 at 6:52
  • $\begingroup$ @ArthurFischer, thanks. This may be a silly suggestion - I don't know a whole lot about set theory - but what about the axiom $|\{\mathrm{cardinals}\; \nu \mid \kappa < \nu < 2^\kappa\}|=2^\kappa.$ $\endgroup$ – goblin Jun 17 '13 at 6:53
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    $\begingroup$ @user18921: It's not only consistent at the lowest level, but it is consistent at the lowest level (relative to a large cardinal hypothesis). $\endgroup$ – user642796 Jun 17 '13 at 7:34
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Let me [partially] address your edit, as the original question was well addressed by Arthur in his answer.

We start with a model of $\sf ZFC+GCH$. Consider the function defined for regular $\kappa$ as $F(\kappa)=\min\{\lambda\mid\lambda=\aleph_\lambda\land\operatorname{cf}(\lambda)>\kappa\}$, then this function satisfies the requirements of Easton's theorem. Therefore there exists a model of set theory such that for every regular $\kappa$ it holds:$$2^\kappa=F(\kappa)=\aleph_{F(\kappa)}=\aleph_{2^\kappa}.$$

It's not hard to see that the gap between $\kappa$ and $2^\kappa$ contains $2^\kappa$ cardinals, and in fact the gap itself is unbounded.

On the axiom of choice side of events, Truss showed that if there exists $\alpha$ such that for every $X$, $\alpha$ cannot be embedded into the cardinals between $X$ and $2^X$, then the axiom of choice holds. That is, a bounded gap between a set and its power set is a strengthening of the axiom of choice, much like $\sf GCH$ is.

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  • $\begingroup$ "It's not hard to see that the gap between $\kappa$ and $2^\kappa$ contains $2^\kappa$ cardinals, and in fact the gap itself is unbounded." This holds only when $\kappa$ is a regular cardinal of the model, right? $\endgroup$ – goblin Apr 3 '14 at 9:31
  • $\begingroup$ Yes, although if you try and calculate what happens to $2^{\aleph_\omega}$ you will see it gets fairly large as well. $\endgroup$ – Asaf Karagila Apr 3 '14 at 12:44

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