6
$\begingroup$

The following question arose while trying to read Shelah's papers Models with second-order properties I-V. For simplicity, I'm assuming a much stronger theory here than Shelah: throughout, we work in $\mathsf{ZFC+V=L}$. I'm also happy to go higher, e.g. to $\mathsf{ZFC+V=L}+ \{\varphi: \mathsf{ZFC}$ + "There is a measurable cardinal" $\vdash\varphi^L\}$, if that would help.

Say that a first-order theory $T$ in a finite language is nice iff first-order logic + the quantifier over automorphisms of $T$-models $$\mathsf{Q}_T(\Phi; \psi(F))\equiv \mbox{"If $\Phi$ defines a model of $T$, then $\Phi$ has an automorphism $F$ satisfying $\psi$"}$$

is fully compact. Shelah proves that the theories of Boolean algebras and ordered fields are nice.

Is there any "snappy" model-theoretic property which implies niceness?

(OK fine, "inconsistent" is such a property, but I'm looking for less trivial examples.) Basically, it's not clear to me when to expect that a given theory is nice in this sense. In particular, it's not clear to me that niceness is a "tameness" property: if $T$ is more expressive than $S$ then models of $T$ likely have fewer automorphisms than models of $S$, so the theory of "automorphisms-of-$T$-models" might still be tamer than the theory of "automorphisms-of-$S$-models."

Almost certainly this is answered somewhere in Shelah's papers, but I am having trouble parsing them; they are extremely information-dense!

Improve this question
$\endgroup$
7
  • $\begingroup$ If $V=L$, isn't $\phi^L$ the same as $\phi^V$, which is just $\phi$? $\endgroup$
    – Christopher King
    Commented Jul 12 at 15:46
  • 2
    $\begingroup$ @ChristopherKing I think you're passing assumptions through curly braces in a way which is not allowed. E.g. take $\varphi$ to be "there is a measurable cardinal." Trivially we have $\mathsf{ZFC+Meas}\vdash\varphi$, but it is not true that $\mathsf{ZFC+Meas}\vdash\varphi^L$ (hopefully!). Basically, the large cardinal theory I have in mind is the "shadow in $L$" of a measurable cardinal: what consequences of the existence of a measurable are compatible with $\mathsf{V=L}$? $\endgroup$
    – Noah Schweber
    Commented Jul 12 at 19:41
  • $\begingroup$ Here's another example. Let $T=\mathsf{ZFC+V=L+\{\varphi: ZFC+2Meas\vdash}\varphi^L\}$, where $\mathsf{2Meas}$ = "There are at least two measurable cardinals." The sentence "There is a transitive model of $\mathsf{ZFC+Meas}$" is in $T$ (via Shoenfield absoluteness) but $\mathsf{Meas}$ itself ("There is a measurable cardinal") is not in $T$. $\endgroup$
    – Noah Schweber
    Commented Jul 12 at 19:44
  • $\begingroup$ Oh right! (Technically adding V=L as an axiom is redundant than, since ZFC proves (V=L)^L.) $\endgroup$
    – Christopher King
    Commented Jul 12 at 20:08
  • 1
    $\begingroup$ @ChristopherKing Correct. :P $\endgroup$
    – Noah Schweber
    Commented Jul 12 at 20:09

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .