# Is there a “large powerset axiom” so extreme that it disproves the existence of strongly inaccessible cardinals?

If $\kappa$ is a strongly compact cardinal, then the singular cardinal hypothesis holds above $\kappa$. Hence the existence of large cardinals at the level of "strongly compact" or above is incompatible with even (apparently) mild large powerset axioms like "$2^\kappa$ always strictly exceeds $\kappa^+.$"

This raises the question:

Is there a "large powerset axiom" $\varphi$ so extreme that $\mathrm{ZFC}+\varphi$ disproves the existence of strongly inaccessible cardinals? Let us also require that $\mathrm{ZFC}+\varphi$ does not prove $\neg \mathrm{Con}(ZFC + \varphi).$

• I think the statement, "There are no inaccessibles" is a large powerset axiom already: it asserts that, for every uncountable regular $\kappa$, there is some $\lambda<\kappa$ such that $2^\lambda\ge\kappa$; and we can make this even stronger by adding "$\mu<\nu\implies 2^\mu<2^\nu$, to demand a $\lambda<\kappa$ with $2^\lambda>\kappa$. In fact, these two axioms together imply that below a regular, we can find cardinals $\lambda$ with $2^\lambda$ arbitrarily large below the $\kappa$th cardinal above $\kappa$. To me, this is a large powerset axiom. – Noah Schweber May 22 '14 at 13:58
• I think you mean to raise the question, rather than to beg it. begthequestion.info – Joel David Hamkins May 22 '14 at 14:04
• Is there a way to construe (something like) the Inner Model Hypothesis as a large powerset axiom? I don't see it off the top of my head, but maybe it's equivalent to something along those lines. – Noah Schweber May 22 '14 at 14:16
• "Beg the question" is one of those phrases that should be shown the door. An antique and confusing translation of a Latin phrase "petitio principii", itself a poor medieval translation from Greek to Latin, as argued by those pesky "descriptivist" linguists here: languagelog.ldc.upenn.edu/nll/?p=2290 Used correctly (to mean "assume the conclusion"), chances are it will be misunderstood. Used incorrectly around certain sophisticated people, it becomes a shibboleth. Best to avoid it altogether. – Todd Trimble Apr 5 '15 at 2:59
• Actually, following the comments section of the language log link I provided earlier, I discovered a great catch-all term for this and similar semantic shifts here: languagelog.ldc.upenn.edu/nll/?p=2290#comment-65857 Namely, the phrase is skunked: "sticking to the older sense confuses those unfamiliar with it, while using the newer sense annoys traditionalists who feel that it is wrong." – Todd Trimble Apr 5 '15 at 7:22

Foreman's maximality principle is as you have requested, though it is not yet known if it is consistent or not.

Foreman's maximality principle: Any non-trivial forcing notion either it adds a real or colapses some cardinals.

It follows from it that:

1) $GCH$ fails everywhere,

2) there are no inaccessible cardinals.

This principle is stated in the following paper:

Foreman, Magidor, Shelah, "$0^\sharp$ and some forcing principles", J. Symbolic Logic, 51 (1986) 39-46.

• That's really nice! – Noah Schweber May 22 '14 at 14:15
• So it's still open whether or not this is consistent with $\sf ZFC$? – Asaf Karagila May 22 '14 at 17:35
• Also, are there known consistent "limited" versions (e.g. "every proper forcing ..." or some other reasonable class of forcing notions)? – Asaf Karagila May 22 '14 at 17:54
• @AsafKaragila Martin's maximum. – Andrés E. Caicedo May 22 '14 at 18:10
• @AsafKaragila It is not quite like that. A concrete theorem, due to Stevo, is that PFA implies that if a partial order adds a subset of omega_1, then it either adds a real or collapses omega_2. – Andrés E. Caicedo May 22 '14 at 20:53

Let me add another example, which is more known.

Consider the following:

Tree property holds at all regular cardinals $\geq \aleph_2.$

If this statement is consistent is a well-know question of Magidor (1970$^{th}$), and is more famous than Foreman's maximality principle. There are some results supporting this principle. It also implies:

1) $GCH$ fails everywhere (if $2^\kappa=\kappa^+$, then there is a special $\kappa^{++}$-Aronszajn tree),

2) There are no inaccessible cardinals (if $\kappa$ is inaccessible, then there is a special $\kappa^+$-Aronszajn tree).