Define $N_n$ as $n$ th natural number: $N_0=0, N_1=1, N_2=2, ...$.

What happens after exponentiation?

We have the following equation: $2^{N_n}=N_{2^{n}}$.

(Which says: For all finite cardinal $n$ we have: $2^{n~\text{th finite cardinal}}=2^{n}~\text{th finite cardinal}$).

What this means?

The gap between $N_n$ and $2^{N_n}$ is rapidly increasing in exponential speed.

Now look at the $\text{GCH}$. It says that the gap between an infinite cardinal $\kappa$ and $2^{\kappa}$ is just a constant number $1$ in cardinals. Even in models for total failure of $\text{GCH}$ we usually have a finite gap between $\kappa$ and $2^{\kappa}$. Now if we look at the infinite cardinals as generalization of natural numbers it seems we should restate continuum hypothesis with more acceleration for the function $\kappa \mapsto 2^{\kappa}$ in order to uniform the behavior of exponentiation function in finite and infinite cardinals. Note to the following statement:

For all cardinal $\kappa$ we have $2^{\kappa~\text{th infinite cardinal}}=2^{\kappa}~\text{th infinite cardinal}$.

This is a direct generalization of the equation $\forall n\in \omega~~~~~2^{N_n}=N_{2^n}$ to the following form:

Natural Continuum Hypothesis (NCH): $~~~\forall \kappa\in Card~~~~~2^{\aleph_{\kappa}}=\aleph_{2^{\kappa}}$

Unfortunately $\text{NCH}$ is contradictory by Konig's lemma because assuming $\text{NCH}$ we have: $\aleph_{\aleph_0}<cf(2^{\aleph_{\aleph_0}})=cf(\aleph_{\aleph_1})\leq \aleph_1$ a contradiction.

(Thanks to Ramiro and Emil for their advices.)

But the acceleration problem remains open. The main question here is this:

Can we have a rapidly increasing gap between carinals by exponentiation? In the other words:

Question 1: Assuming consistency of $\text{ZFC}$ (plus some large cardinal axiom or axiom of constructibility), is the following statement consistent with $\text{ZFC}$?

$\forall \kappa\in Card~~~\exists \lambda \in Card~;~~~~~\lambda\geq 2^{\kappa}~\wedge~2^{\aleph_{\kappa}}=\aleph_{\lambda}$

Definition 1: Let $\kappa$ be a (finite or infinite) cardinal. Define $\kappa$-based beth function as follows:

$\beth_{(\kappa)}:Ord\longrightarrow Card$


$\forall \alpha\in Ord~~~\beth_{(\kappa)}(\alpha +1):=2^{\beth_{(\kappa)}(\alpha)}$

$\forall \alpha\in LimitOrd~~~\beth_{(\kappa)}(\alpha):=\bigcup_{\beta \in \alpha}\beth_{(\kappa)}(\beta)$

Definition 2: Let $F:Card\longrightarrow Card$ be an increasing function and $\delta$ an ordinal. We say that $F$ has acceleration rank $\delta$ if $\delta=min\{\alpha\in Ord~|~\forall \kappa\in Card~~~\beth_{(\kappa)}(\alpha)\leq F(\kappa) < \beth_{(\kappa)}(\alpha+1)\}$.

For example the functions $\kappa\mapsto \kappa^{+}$, $\kappa\mapsto 2^{\kappa}$, $\kappa\mapsto 2^{2^{\kappa}}$ have acceleration ranks $0, 1, 2$ respectively.

Question 2: Is there any limitation for acceleration of the continuum function? Precisely is the following statement true?

"For any ordinal $\delta$ there is an increasing function $F:Card\longrightarrow Card$ with acceleration rank $\delta$ such that assuming consistency of $\text{ZFC}$ (and some large cardinal axiom) one can prove the consistency of $\text{ZFC}$ with the statement $\forall \kappa\in Card~~~2^{\aleph_{\kappa}}=\aleph_{F(\kappa)}$."

According to Emil's interesting comment I added his question here:

Question 3: Assuming consistency of $\text{ZFC}$ (and some large cardinal axiom or axiom of constructibility), is the following consistent with $\text{ZFC}$?

$\forall \kappa\in Card~~~~\exists \lambda\in Card~~~~~2^{\aleph_{\kappa}}=\aleph_{\lambda}$

Note that it is not trivial that one can have a "cardinal index" for $\aleph$ as value of $2^{\aleph_{\kappa}}$ everywhere. Perhaps we will need some non-cardinal ordinals as indexes to avoid inconsistency.

  • 6
    $\begingroup$ Wouldn't $NCH$ imply $2^{\aleph_\omega}=\aleph_{\omega_1}$? $\endgroup$ Nov 13, 2013 at 14:16
  • 1
    $\begingroup$ @RamirodelaVega: Yes. We have $\text{NCH}\longrightarrow \text{CH}\wedge\neg \text{GCH}$ so $2^{\aleph_{\aleph_{0}}}=\aleph_{2^{\aleph_{0}}}=\aleph_{\aleph_{1}}$. $\endgroup$
    – user42090
    Nov 13, 2013 at 14:24
  • 3
    $\begingroup$ So, doesn’t that imply that NCH is contradictory, as ZFC proves $\mathrm{cf}(2^\kappa)>\kappa$ for every infinite $\kappa$? $\endgroup$ Nov 13, 2013 at 14:35
  • $\begingroup$ Very good! I deleted my answer (showing that NCH has large cardinal lower bounds in strength) in light of Ramiro's observation that NCH is inconsistent. $\endgroup$ Nov 13, 2013 at 14:40
  • 4
    $\begingroup$ I can't really see the point made by your five introductory lines. Your function $N$ is the identity, so all it seems to me that it says is that "exponentiation behaves in an exponential fashion". $\endgroup$ Nov 13, 2013 at 15:31

4 Answers 4


In the following answer, by Foreman-woodin model, I mean the model constructed by them in the paper "The generalized continuum hypothesis can fail everywhere. Ann. of Math. (2) 133 (1991), no. 1, 1–35. "

Questions 1 and 3 have positive answer: In Foreman-Woodin model for the total failure of GCH the following hold:

1) For all infinite cardinal $\kappa, 2^{\kappa}$ is weakly inaccessible, and hence a fixed point of the $\aleph-$function,

2) If $\kappa \leq \lambda< 2^{\kappa},$ then $ 2^{\lambda}= 2^{\kappa}.$

In this model for all infinite cardinals $\kappa, 2^{\kappa}=\aleph_{ 2^{\kappa}}$ in particular for all fixed points $\kappa$ of the $\aleph-$function, $2^{\aleph_\kappa}=\aleph_{ 2^{\kappa}}$. Also note that in this model for all infinite cardinals $\kappa,$ if we let $\lambda=2^{\aleph_\kappa},$ then $\lambda \geq 2^\kappa,$ and $2^{\aleph_\kappa}=\aleph_\lambda.$ So both of questions 1 and 3 have a positive answer.

For your question 2, $\delta$ can be arbitrary large: Start with GCH+there exists a supercompact cardinal $\kappa$+ there are infinitely many inaccessibles above it. Now let $\delta$ be any ordinal $<\kappa.$ Force with Foreman-Woodin construction above $\delta$ (in the sense that let the first point of the Radin club added during their forcing construction be above $\delta$). In their final model (which is $V_\kappa$ of some extension of the ground model) for all infinite cardinals $\lambda <\kappa, 2^\lambda \geq \lambda^{+\delta}$. So if $F$ is defined in the ground model by $F(\kappa)=\kappa^{+\delta},$ then the acceleration rank of $F$ is $\delta$ (using GCH), and in the finial model for all infinite cardinals $\kappa, 2^\kappa \geq F(\kappa).$

Remark. I may note that we can not define the function $F$ in the ground model, such that it is the realization of power function in the extension, but we can find some inner model of the final extension in which $GCH$ holds and such a function $F$ is definable.

  • $\begingroup$ Very nice Mohammad. This phenomenon happens under assumption of $\text{NCH}$, i.e. $\text{NCH}$ implies for each fixed point of $\aleph$ function like $\kappa$, $2^{\kappa}$ is also a fixed point of $\aleph$ function. $\endgroup$
    – user42090
    Nov 13, 2013 at 19:04
  • $\begingroup$ Excellent answer Mohammad. Please introduce some references for Foreman-Woodin construction. But about question 2, why the acceleration rank of the function $\kappa\mapsto \kappa^{+\delta}$ is $\delta$? What about Emil's upper bound on possible acceleration rank of continuum function? $\endgroup$
    – user42090
    Nov 14, 2013 at 8:35
  • $\begingroup$ Under GCH, $\beth_\kappa(\alpha)=\kappa^{+\alpha}$ $\endgroup$ Nov 14, 2013 at 10:20
  • $\begingroup$ Ah! Of Course. So is Emil's answer incorrect? $\endgroup$
    – user42090
    Nov 14, 2013 at 10:29
  • $\begingroup$ The problem of definability once again! Thank you Mohammad. $\endgroup$
    – user42090
    Nov 14, 2013 at 12:42

Let me provide a large cardinal lower bound for the statement in question 1. Since $\kappa\lt\kappa+1\lt 2^\kappa$ for $\kappa>1$, the statement implies that there is a universal failure of the GCH above $\aleph_1$ at cardinals of the form $\aleph_\kappa$ for $\kappa$ a cardinal. In particular, this implies that the negation of the singular cardinals hypothesis SCH. Since $\neg\text{SCH}$ is known to have large cardinal strength, we have large cardinal lower bounds for the consistency strength of your statement.

Meanwhile, since the statement also implies violations of the SCH arbitrarily high in the cardinals, it follows by a result of Solovay that it is inconsistent with the existence of strongly compact cardinals.

  • $\begingroup$ Really interesting. Thanks. Please introduce some references for Solovay's result. $\endgroup$
    – user42090
    Nov 13, 2013 at 14:18
  • $\begingroup$ I adapted my answer for the revised question. $\endgroup$ Nov 13, 2013 at 15:07

NCH is inconsistent with ZFC:

First, under NCH, we have: $$ 2^{\aleph_{\aleph_{\aleph_0}}} = \aleph_{2^{\aleph_{\aleph_0}}} = \aleph_{\aleph_{2^{\aleph_0}}} = \aleph_{\aleph_{\aleph_1}} . $$

Under NCH, $\aleph_{\aleph_{\aleph_0}}$ is strong limit, since, for any $n \in \omega$: $$ 2^{\aleph_{\aleph_n}} = \aleph_{2^{\aleph_n}} = \aleph_{\aleph_{2^n}} < \aleph_{\aleph_{\aleph_0}} . $$

Now, consider the Theorem 7.3 in Chapter 14 (Cardinal Arithmetic - Abraham and Magidor) of the Handbook of Set Theory (page 1210):

Suppose that $\delta$ is a limit ordinal and ${|\delta|}^{\operatorname{cf}(\delta)} < \aleph_{\delta}$. Then $$ {\aleph_{\delta}}^{\operatorname{cf}(\delta)} < \aleph_{({|\delta|}^{+4})} . $$

Then we have, under NCH: $$ 2^{\aleph_{\aleph_{\aleph_0}}} = {(\aleph_{\aleph_{\aleph_0}})}^{\aleph_0} < \aleph_{{(\aleph_{\aleph_0})}^{+4}} = \aleph_{\aleph_{\omega + 4}} . $$

However, $$ \aleph_{\aleph_{\aleph_1}} > \aleph_{\aleph_{\omega + 4}} . $$

  • 3
    $\begingroup$ As has been pointed out by Ramiro de la Vega, and mentioned in the question itself, NCH is inconsistent with ZFC because of König’s lemma. No need to involve pcf theory. $\endgroup$ Nov 13, 2013 at 15:39
  • 1
    $\begingroup$ What an amazing answer by pcf theory! Thanks for your effort, Alberto. $\endgroup$
    – user42090
    Nov 13, 2013 at 16:23

As for question 2, if $\forall\kappa\,2^{\aleph_\kappa}=\aleph_{F(\kappa)}$, then $F$ does not have an acceleration rank.

If the rank were $\delta\ge2$, we could take $\kappa$ to be any fixpoint of the $\aleph$ function to obtain a contradiction: $$2^{\aleph_\kappa}=2^\kappa\le\aleph_{2^\kappa}=\aleph_{\beth_{(\kappa)}(1)}.$$ On the other hand, a finite acceleration rank is impossible: this would imply $F(n)<\omega$ for every $n<\omega$, i.e., $\aleph_\omega$ is strongly limit. In particular, $\omega\le F(\omega)\le\beth_{(\omega)}(\delta+1)=\beth_{\delta+1}<\aleph_\omega$, hence $\mathrm{cf}(2^{\aleph_\omega})=\mathrm{cf}(\aleph_{F(\omega)})=F(\omega)<\aleph_\omega$, contradicting König’s lemma.

  • $\begingroup$ Interesting answer Emil. Thank you very much. $\endgroup$
    – user42090
    Nov 13, 2013 at 15:54
  • $\begingroup$ Your answer imposes an upper bound on the possible growth speed of continuum function. It is really surprising. $\endgroup$
    – user42090
    Nov 13, 2013 at 16:19
  • $\begingroup$ Oh! What a strange result! It seems I should change some parameters in my intuition about continuum function. $\endgroup$
    – user42090
    Nov 13, 2013 at 17:49
  • $\begingroup$ Emil, I added some edits in the definition of acceleration rank in order to clarify the difference between growth rate of the functions $\kappa\mapsto \kappa^{+}$, $\kappa\mapsto 2^{\kappa}$, $\kappa\mapsto 2^{2^{\kappa}}$. Please add updates in your answer. Thanks. $\endgroup$
    – user42090
    Nov 13, 2013 at 19:37
  • 1
    $\begingroup$ F is defined in the ground model, and it is asked it to be realized as a power function in some extension. See my answer. $\endgroup$ Nov 14, 2013 at 4:19

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