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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 $\neg \mathrm{Con}(ZFC + \varphi).$

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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. – Joel David Hamkins May 22 '14 at 14:04
"While descriptivists and other such laissez-faire linguists are content to allow the misconception to fall into the vernacular, it cannot be denied that logic and philosophy stand to lose an important conceptual label should the meaning of BTQ become diluted to the point that we must constantly distinguish between the traditional usage and the erroneous "modern" usage. This is why we fight." @JoelDavidHamkins, an interesting statement. I'm almost convinced. – goblin May 22 '14 at 14:11
"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: 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 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: 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 at 7:22

2 Answers 2

up vote 10 down vote accepted

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.

See also "Questions about $\aleph_1-$closed forcing notions".

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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 Caicedo May 22 '14 at 18:10
@Andres: For which class of forcing notions? – Asaf Karagila May 22 '14 at 20:43

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).

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