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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? $$|\{\mbox{cardinals } \nu \mid \kappa<\nu<2^\kappa\}|=2^\kappa$$

Asaf Karagila
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goblin GONE
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2 Answers2

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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:

Community
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user642796
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  • Could there be other possible notions of 'distance between $\kappa$ and $2^\kappa$' that don't run into this problem? – goblin GONE Jun 17 '13 at 06:40
  • @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. – user642796 Jun 17 '13 at 06:50
  • 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. – Andrés E. Caicedo Jun 17 '13 at 06:52
  • @ArthurFischer, thanks. This may be a silly suggestion - I don't know a whole lot about set theory - but what about the axiom $|{\mathrm{cardinals}; \nu \mid \kappa < \nu < 2^\kappa}|=2^\kappa.$ – goblin GONE Jun 17 '13 at 06:53
  • @user18921 While it's not a standardized axiom by any means - there's very little that follows from it! - that's certainly possible at the lowest level at least; it's consistent that $2^\omega$ is weakly inaccessible, in which case its cardinality $\kappa$ is a fixed-point of the aleph 'function' : $\kappa=\aleph_\kappa$, and so there are $\kappa$ cardinals less than $\kappa$. – Steven Stadnicki Jun 17 '13 at 07:03
  • @AndresCaicedo: Thanks. I was trying to hold the line between absolute correctness and understandability to the uninitiated, and perhaps I swayed too much on the "understandability" side of thing. I hope it's somewhat better now. – user642796 Jun 17 '13 at 07:03
  • @StevenStadnicki, is it only consistent at the lowest level? This is not at all obvious to me... – goblin GONE Jun 17 '13 at 07:18
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    @user18921: It's not only consistent at the lowest level, but it is consistent at the lowest level (relative to a large cardinal hypothesis). – user642796 Jun 17 '13 at 07:34
  • @ArthurFischer, thank you. – goblin GONE Jun 17 '13 at 07:39
  • @AndrésE.Caicedo: Hmm, why is it relatively consistent with ZFC that $c = \aleph_{c^+}$ where $c = 2^{\aleph_0}$, as Arthur's first paragraph seems to imply, since $c^+$ has uncountable cofinality? Don't we have by basic cardinal properties that $\aleph_{c^+} ≥ c^+ > c$? – user21820 Jun 17 '19 at 07:08
  • @user21820 Yes, that's not possible. But I don't see him saying that. What one can ensure is that, in some extension $V[G]$ of the universe $V$, $\mathfrak c^{V[G]}=\aleph_{({\mathfrak c}^+)^V}$. That is, as long as some mild conditions on $V$ and $\alpha$ are satisfied, in some rather tame extension $V[G]$ we can have $\mathfrak c=\aleph_\alpha$, regardless of whether, in $V$, $\alpha$ used to be $\mathfrak c^+$ or whatever. And by "rather tame" we mean that no cofinalities or cardinalities have changed (so, e.g., if $\alpha$ was ${\mathfrak c}^+$, then $\alpha$ is still a cardinal). – Andrés E. Caicedo Jun 17 '19 at 09:34
  • @AndrésE.Caicedo: Thanks for your clarification, but then I think the first paragraph of this post is indeed misleading, especially since it finishes with "there is no bound on the number of cardinals strictly between $\aleph_0$ and $\frak{c}$", when in fact there is, namely $\frak{c}$ itself. – user21820 Jun 17 '19 at 14:06
  • @user21820 Yes, it is a somewhat imprecise shorthand. It expresses the fact that for any ordinal $\alpha$, in some extension with the same cardinals, there are more than $\alpha$ cardinals below $\mathfrak c$. I don't know if it is misleading, though; more like informal jargon: we know what it really means, and the extra precision may end up being confusing as well. I agree it is not ideal, and I'm glad I could clarify. – Andrés E. Caicedo Jun 17 '19 at 14:26
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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.

Asaf Karagila
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  • "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? – goblin GONE Apr 03 '14 at 09:31
  • Yes, although if you try and calculate what happens to $2^{\aleph_\omega}$ you will see it gets fairly large as well. – Asaf Karagila Apr 03 '14 at 12:44