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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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    $\begingroup$ 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. $\endgroup$
    – Noah Schweber
    Commented May 22, 2014 at 13:58
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    $\begingroup$ I think you mean to raise the question, rather than to beg it. begthequestion.info $\endgroup$
    – Joel David Hamkins
    Commented May 22, 2014 at 14:04
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    $\begingroup$ 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. $\endgroup$
    – Noah Schweber
    Commented May 22, 2014 at 14:16
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    $\begingroup$ "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. $\endgroup$
    – Todd Trimble
    Commented Apr 5, 2015 at 2:59
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    $\begingroup$ 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." $\endgroup$
    – Todd Trimble
    Commented Apr 5, 2015 at 7:22

2 Answers 2

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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 adds a real or collapses 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 "https://mathoverflow.net/questions/64280/questions-about-aleph-1-closed-forcing-notions".

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  • $\begingroup$ That's really nice! $\endgroup$
    – Noah Schweber
    Commented May 22, 2014 at 14:15
  • $\begingroup$ So it's still open whether or not this is consistent with $\sf ZFC$? $\endgroup$
    – Asaf Karagila ♦
    Commented May 22, 2014 at 17:35
  • $\begingroup$ Also, are there known consistent "limited" versions (e.g. "every proper forcing ..." or some other reasonable class of forcing notions)? $\endgroup$
    – Asaf Karagila ♦
    Commented May 22, 2014 at 17:54
  • $\begingroup$ @AsafKaragila Martin's maximum. $\endgroup$
    – Andrés E. Caicedo
    Commented May 22, 2014 at 18:10
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    $\begingroup$ @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. $\endgroup$
    – Andrés E. Caicedo
    Commented May 22, 2014 at 20:53
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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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  • $\begingroup$ This statement or Foremans principle consistent with existence weakly inaccessible cardinals or even weaky Mahlos? $\endgroup$
    – Alex O.
    Commented Dec 14, 2020 at 13:54
  • $\begingroup$ @AlexO. Though it implies the total failure of GCH, we do not know how. For example we don't know if it implies the continuum is bounded by some cardinal or if the principle is consistent with arbitrary large values of the continuum. $\endgroup$
    – Mohammad Golshani
    Commented Dec 16, 2020 at 11:16

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