5
$\begingroup$

Let $T$ be a finite collection of axioms of $\mathrm{ZF}$, let $\sigma$ be a sentence in the language of $\mathrm{ZF}$ and consider the statement

$\tau$: “any transitive countable model of $T$ satisfies $\sigma$”.

Then $\mathrm{ZFC}\vdash\tau$ implies $\mathrm{ZFC}\vdash\sigma$, by the classical argument using Reflection, Downward Löwenheim–Skolem and Mostowki’s collapse to get a countable transitive model where finitely many sentences are absolute.

My question is: does $\mathrm{ZF}\vdash\tau$ imply $\mathrm{ZF}\vdash\sigma$? This may look trivial but, without choice, Downward Löwenheim–Skolem can’t be used as above. On the other hand, it could be argued that $\mathrm{ZF}\vdash$ “there is a proof of $\sigma$ from $T$”, but this does not mean that such a proof can be found in the meta-theory.

Thank you for your help with this (possibly trivial) matter.

Improve this question
$\endgroup$
13
  • 1
    $\begingroup$ I don’t know about the rest, but ZF certainly proves reflection in the form “if there is a proof of $\sigma$ in $T$, then $\sigma$”. This holds for any sequential theory that proves induction for all formulas in its language, see mathoverflow.net/a/87249 . $\endgroup$
    – Emil Jeřábek
    Commented Nov 14, 2019 at 15:06
  • 2
    $\begingroup$ @WillBrian Work inside ZFC, and assume $\neg\sigma$. By Jech’s remark, there exists a countable transitive model of $T+\neg\sigma$. But by assumption, all countable transitive models of $T$ satisfy $\sigma$, a contradiction. Thus, $\sigma$. (By the way, this actually shows the stronger statement $\mathrm{ZFC}\vdash\tau\to\sigma$.) $\endgroup$
    – Emil Jeřábek
    Commented Nov 14, 2019 at 15:53
  • 1
    $\begingroup$ @WillBrian That’s not what the OP is claiming. The claim is absolutely clear: $\mathrm{ZFC}\vdash\tau$ implies $\mathrm{ZFC}\vdash\sigma$. I have shown that even $\mathrm{ZFC}\vdash(\tau\to\sigma)$. Your interpretation is $\mathrm{ZFC}\vdash(\tau\to\mathrm{Pr}_{\mathrm{ZFC}}(\sigma))$, which may be false. $\endgroup$
    – Emil Jeřábek
    Commented Nov 14, 2019 at 16:16
  • 3
    $\begingroup$ See mathoverflow.net/questions/269682/… $\endgroup$
    – Elliot Glazer
    Commented Nov 14, 2019 at 21:34
  • 3
    $\begingroup$ @ElliotGlazer The answer to the post you suggested indicates a way to answer my question affirmatively. Thank you. $\endgroup$
    – dragoon
    Commented Nov 14, 2019 at 23:34

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .