# Consistency of: “The continuum function is injective, and for all infinite cardinals $\kappa$ we have that $2^\kappa$ is weakly inaccessible.”

I asked here about "large powerset axioms" and to my delight, learned that such axioms are being taken seriously. I've been toying with them ever since. My favourite is: "The continuum function is injective, and for all infinite cardinals $\kappa$ we have that $2^\kappa$ is weakly inaccessible," since this is easy to understand yet implies the existence of a rich universe of cardinal arithmetic. But it occurs to me that before toying any further, I should really find out whether or not this axiom is consistent with ZFC.

So, does anyone know a contradiction from this axiom? And if not, are any large cardinal axioms known to imply its consistency? I could not find the answers in the linked articles, nor do I really know where to start looking.

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The Foreman-Woodin model satisfies $2^\kappa$ is weakly inaccessible for all $\kappa,$ but it is not injective. –  Mohammad Golshani May 8 at 14:59

If you make your requirement only for regular cardinals $\kappa$, then we can easily get an equiconsistency.

Theorem. The following theories are equiconsistent over ZFC:

1. There are unboundedly many inaccessible cardinals.
2. The continuum function $\kappa\mapsto 2^\kappa$ is injective and $2^\kappa$ is weakly inaccessible for any infinite regular cardinal $\kappa$.

Proof. Statement 2 implies that there are unboundedly many (strongly) inaccessible cardinals in $L$, and so statement 1 holds in $L$. Conversely, assume statement 1 holds. By forcing, we may assume without loss that the GCH also holds in $V$. Let $\kappa_\alpha$ be the $\alpha^{th}$ infinite regular cardinal and let $\delta_\alpha$ be the $\alpha^{th}$ inaccessible cardinal. Define $$E(\kappa_\alpha)=\delta_{\alpha+1}.$$ This function is increasing and has $\text{cof}(E(\kappa))>\kappa$ for every regular cardinal $\kappa$. Thus, it satisfies the hypotheses of Easton's theorem, and so there is a cardinal and cofinality-preserving forcing extension $V[G]$ in which $2^\kappa=E(\kappa)$ for every regular cardinal $\kappa$. Since the $\delta_\alpha$ all remain weakly inaccessible in the extension, we have that $2^\kappa$ is weakly inaccessible for every regular cardinal $\kappa$. Furthermore, the continuum function in $V[G]$ is strictly increasing on successor cardinals and hence is injective. QED

Meanwhile, if you really want $2^\kappa$ to be weakly inaccessible even for singular $\kappa$, then the hypothesis becomes inconsistent with the singular cardinals hypothesis, and hence inconsistent with the existence of supercompact or strongly compact cardinals.

Theorem. If $2^\kappa$ is weakly inaccessible for all infinite cardinals $\kappa$, then the SCH fails at every singular strong limit. In particular, this hypothesis is inconsistent with the existence of supercompact or strongly compact cardinals.

Proof. If $\kappa$ is a singular strong limit, then under the SCH, it must be that $2^\kappa=\kappa^+$, which is definitely not weakly inaccessible. QED

The failure of SCH already has a fairly high large cardinal strength, and so we get lower bounds for the consistency strength of your hypothesis (when applied to all cardinals including singular cardinals). Perhaps the excellent inner model theory experts we have here on MO can state more precise claims about this.

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Expanding on Joel's last paragraph: in 1991, building off a long chain of results, Moti Gitik showed (ac.els-cdn.com/016800729190016F/…; see also download.springer.com/static/pdf/968/…, page 247) that $ZFC+\neg SCH$ is equiconsistent with $ZFC+$"There is a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}$." –  Noah S May 8 at 17:54
(cont'd) See e.g. cantorsattic.info/Mitchell_rank for a definition of Mitchell order. Basically, Gitik's assumption is that there is a measurable cardinal which admits as big measures as possible. –  Noah S May 8 at 17:58
Thanks for the additional information, Noah! –  Joel David Hamkins May 8 at 20:07
I don't know inner model theory, but I think the statement, if consistent, needs much larger cardinals. I think we even need a proper class of cardinals $\kappa$ of Mitchell order an inaccessible above $\kappa$, and even more. –  Mohammad Golshani May 10 at 4:52