13
$\begingroup$

Exercises 6, 7, and 8 in section 10.4 of Hodges' big model theory textbook contain an outline of a proof of the consistency of the following statement

There exists a countable first-order theory $T$ such that $T$ has no saturated models.

Call this statement $\mathrm{NSat}_\omega$ and call the analogous statement with no cardinality restriction $\mathrm{NSat}$.

The proof outlined by Hodges uses a result of Woodin's that it is consistent with $\mathrm{ZFC}$ that for every infinite cardinal $\kappa$, $2^\kappa=\kappa^{++}$, but he mentions that this result assumes some large cardinals and in particular 'a supercompact is more than enough'.

Are large cardinals necessary for this result? There are different meanings of a question like this but a typical formulation would be something like:

Does $\mathrm{ZFC}$ + 'there exists a transitive model of $\mathrm{ZFC}+\mathrm{NSat}_\omega$' prove the existence of transitive models of any typical large cardinal axiom? What about $\mathrm{ZFC}+\mathrm{NSat}$?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
15
$\begingroup$

Unless I'm missing something, if $\vert T\vert+\aleph_0<\kappa$ and $\kappa^+=2^\kappa$, then we can build a saturated model of $T$ of cardinality $2^\kappa$. So:

  • If we want a countable theory with no saturated model, we need GCH to fail everywhere.

  • If we want some theory with no saturated model, we need GCH to hold at most boundedly often.

And these principles have rather high consistency strength. A failure of GCH at a singular limit cardinal is in particular a failure of the Singular Cardinal Hypothesis, the failure of which is known to be equiconsistent with the existence of a $\kappa$ such that $o(\kappa)=\kappa^{++}$. So this gives a lower bound to both principles.

A better analysis of the strength involved, at least for NSat (although I suspect they'll be identical), is this wonderful MSE answer by Andres Caicedo (in particular, my previous paragraph is just a rephrasing of his second-to-last paragraph). However, that's from a while ago, so more might be known now.

Improve this answer
$\endgroup$
6
  • $\begingroup$ That is larger than I was expecting. It seems it's actually equiconsistent with $2^\kappa = \kappa^{++}$. $\endgroup$
    – James E Hanson
    Commented May 13, 2019 at 23:25
  • $\begingroup$ I have a followup question about the precise nature of the consistency result. If $\mathrm{GCH}$ fails everywhere do you get an inner model with large cardinals containing all of $2^{\aleph_0}$? $\endgroup$
    – James E Hanson
    Commented May 13, 2019 at 23:40
  • $\begingroup$ @JamesHanson I would guess so, based on vague recollections of things Steel told me, but I'm not at all confident. You'd have to actually think about the process for building relevant inner models and seeing if you could modify it (if necessary) to accommodate all the reals; this might be easy but it's not something I know anything about. The process for building such inner models is not simple enough for me to just glance at and say anything with confidence; see e.g. Mitchell's original lower bound paper "On the singular cardinal hypothesis." $\endgroup$
    – Noah Schweber
    Commented May 13, 2019 at 23:54
  • $\begingroup$ What about for a single real? I want to be able to say in $\mathrm{ZFC}$ for a given countable theory $T$ either $T$ has a saturated model or there is an inner model $M$ containing $T$ in which $T$ has a saturated model. $\endgroup$
    – James E Hanson
    Commented May 13, 2019 at 23:59
  • $\begingroup$ @JamesHanson If $r$ is a real, $L[r]$ satisfies $\mathsf{GCH}$,so letting $r$ be a real coding $T$, you have the result you suggest. $\endgroup$
    – Andrés E. Caicedo
    Commented May 14, 2019 at 0:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .