The exponentiation operator inflicts a subtle information loss on the transfinite numerical equations, pretty similar to the case of $a^2=b^2 \nRightarrow a=b$ in real numbers. In fact, for the infinite cardinals we have $2^\kappa=2^{\lambda}\nRightarrow \kappa=\lambda$ because, for example, one may force the transfinite exponentiation function to behave like this: $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. Thus merely knowing the number of the subsets of a set doesn't give you enough information about the definite number of its members.

We abuse such an information gap to derive some weaker versions of the Generalized Continuum Hypothesis, $GCH$ (i.e. $\forall \kappa~~~2^{\kappa}\leq \kappa^+$), as well as stronger versions of its total failure, $FCH$ (i.e. $\forall \kappa~~~2^{\kappa}>\kappa^+$). Our questions are mainly about the tension between these two families of statements.

Definition 1. For any $m, n\in \omega$ define:

  • $GCH_{n}^{m}:~~\forall \kappa\in Card~~~~\beth_{n+1}(\kappa)\leq \beth_{n}(\kappa^{+(m+1)})$

  • $FCH_{n}^{m}:~~\forall \kappa\in Card~~~~\beth_{n+1}(\kappa)> \beth_{n}(\kappa^{+(m+1)})$

Remark 1. In order to clarify the notation, note that $GCH_{0}^{0}$ is simply $GCH$ and $GCH_{1}^{0}$ is the statement $2^{2^{\kappa}}\leq 2^{\kappa^+}$, namely $2$ to the power of the two sides of the $GCH$ inequality. $GCH_{0}^{1}$ is $2^{\kappa}\leq \kappa^{+2}$, that is $GCH$ with a wider gap allowed, while $GCH_{1}^{1}$ indicates $2^{2^{\kappa}}\leq 2^{\kappa^{+2}}$, a weakening of the latter statement by power function. Also, for each $m, n\in \omega$, $FCH_{n}^{m}$ is the total failure of $GCH_{n}^{m}$ and the following diagrams hold.

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Now consider the limit of the statements in these diagrams, namely $‎\bigwedge‎_{m,n\in \omega} FCH_{n}^{m}$ and $\bigvee_{m,n\in\omega} GCH_{n}^{m}$. The first natural question that arises is about the consistency of $‎\bigwedge‎_{m,n\in \omega} FCH_{n}^{m}$ as a very strong total failure of $GCH$ (which is known to have some large cardinal strength).

Question 1. Is $ZFC+\bigwedge‎_{m,n\in \omega} FCH_{n}^{m}$ consistent?

Note that in any model of $\bigwedge‎_{m,n\in \omega} FCH_{n}^{m}$ we must have $2^{\aleph_0}>\aleph_{\omega}$. Also, no fixed-gap model for the total failure of $GCH$ (in which $\forall\kappa~~~2^{\kappa}=\kappa^{+t}$ for some natural number $t\geq 2$) works. However, I suspect that some non-fixed gap classical models for the total failure of $GCH$ may work here.

Certainly, there are a lot of consistency tensions (in the standard set-theoretic terms) between the left and right sides of the above diagram as any element on the left badly violates its corresponding element on the right (and many of its connected nodes). However, some instances of reconciliation might exist as well.

Definition 2. For any index set $I\subseteq \omega\times\omega$, we say the family $\{FCH_{n}^{m}\}_{(m,n)\in I}$ of strong total failures of $GCH$ is compatible with the weak instances of $GCH$ if assuming appropriate large cardinal assumptions, $ZFC+\bigwedge_{(m,n)\in I}FCH_{n}^{m} +\bigvee_{m,n\in\omega} GCH_{n}^{m}$ is consistent, namely there are $m',n'\in\omega$ such that $ZFC+\bigwedge_{(m,n)\in I}FCH_{n}^{m}+GCH_{n'}^{m'}$ is consistent.

Remark 2. An example that justifies such an attitude is $FCH_{0}^{0}$ which could be made consistent with $GCH_{0}^{1}$ in a fixed gap $2$ model where $\forall\kappa~~~2^{\kappa}=\kappa^{++}$, while being totally inconsistent with $GCH_{0}^{0}$. More generally, it sounds plausible that assuming appropriate large cardinal assumptions, for any $m,n\in\omega$, there exist $m',n'\in\omega$ such that $ZFC+FCH_{n}^{m}+GCH_{n'}^{m'}$ is consistent (Please correct me if it is not true!).

The next question asks about the number of such compatible families of strong failures of $GCH$.

Question 2. What is the size of the following subset of the continuum? Is it uncountable?

$\mathcal{C}:=\{I\subseteq\omega\times \omega~|~\{FCH_{n}^{m}\}_{(m,n)\in I}~\text{is compatible with weak instances of}~GCH\}$

  • 1
    $\begingroup$ It is worth mentioning that not all arrows in the above diagram are strict. As Mohammad pointed out in a personal discussion with me, despite their weak-looking appearances, all $GCH_{n}^{0}$s are equivalent to $GCH$, so the arrows in the first column of the right diagram are actually two-sided. The argument relies on proving the fact that under $GCH_{n+1}^{0}$ the function $\kappa\mapsto 2^{\kappa}$ satisfies a variant of being strictly increasing which itself implies $GCH_{n}^{0}$. $\endgroup$ Jun 21, 2018 at 13:26

1 Answer 1


In the Foreman-Woodin model (mathscinet MR1087344), for all $\kappa$, $\beth_n(\kappa)$ is at least the $n^{th}$ weakly inaccessible above $\kappa$. Furthermore, this model satisfies the following:

(1) $\forall \kappa \forall n (2^\kappa > \kappa^{+n})$

(2) $\forall \kappa \forall n (2^\kappa = 2^{\kappa^{+n}})$

From these statements, it follows that $\text{FCH}^m_n$ holds for all $m,n<\omega$. To show this, first note that $\text{FCH}^m_0$ holds for all $m$ by (1). Now, (2) says that $\beth_1(\kappa^{+m}) = \beth_1(\kappa)$ for all $m$. Thus $\beth_n(\kappa^{+m}) = \beth_n(\kappa)$ for all $n \geq 1$. Thus $\beth_n(\kappa^{+m}) < \beth_{n+1}(\kappa)$ by Cantor's Theorem.

  • $\begingroup$ (+1) Thanks! That is actually the "classical model" that I already suspected might satisfy $\bigwedge_{m,n\in\omega} FCH_{n}^{m}$ but I couldn't remember. Any idea about the question 2? $\endgroup$ Jun 21, 2018 at 14:00

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