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In $\mathsf{ZFC}$, a somewhat cheating way to buy "transitive models" without cost in consistency is to add to the language a constant symbol $M$ and add to $\mathsf{ZFC}$ the axioms stating that $M$ is countable transitive, and for each axiom of $\mathsf{ZFC}$, add its relativization to $M$. This allows us to "pretend" that we have a transitive model of $\mathsf{ZFC}$ (the catch is that this is a theorem schema).

I wonder if adding a constant symbol is necessary. In other words, can there be some theory $T$ (extending $\mathsf{ZFC}$) in the language $\{\in\}$, such that T can prove there is a set $M$, and $T$ proves each of its own axioms relativized to $M$?

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    $\begingroup$ @HanulJeon I think this is what I'm referring to in my first paragraph. Or are you suggesting this can be done without expanding the language? $\endgroup$
    – ikrto
    Feb 14, 2021 at 19:45
  • $\begingroup$ I misread your question, so I removed the previous comment. $\endgroup$
    – Hanul Jeon
    Feb 15, 2021 at 12:32

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$\mathsf{ZFC}$ already has this property. Specifically:

  • The reflection theorem shows that every model of $\mathsf{ZFC}$ - even of $\mathsf{ZFC+\neg Con(ZFC)}$ - contains a model of $\mathsf{ZFC}$. (This model might not internally be a model of $\mathsf{ZFC}$, which is why this isn't an obvious absurdity.)

  • The fact that satisfiability is absolute to $L$ then lets us pick out a specific model, via the $L$-ordering.

The details are as follows:

Working in $\mathsf{ZFC}$, fix some standard enumeration of the $\mathsf{ZFC}$ axioms and let $T_0$ be the largest initial segment of that enumeration which is consistent. Classically of course we have (under the usual assumptions) that $T_0=\mathsf{ZFC}$; meanwhile, note that $\mathsf{ZFC}$ proves "$T_0$ is consistent" (trivially) as well as "$\varphi\in T_0$" for each $\varphi\in\mathsf{ZFC}$ (by the reflection theorem).

Now since $\mathsf{ZFC}$ proves that $T_0$ is consistent, we can - in $\mathsf{ZFC}$ - consider $M=$ the least constructible set model of $T_0$, where "least" refers to the standard ordering on $L$. This is provably in $\mathsf{ZFC}$ a model of $T_0$, and so for each $\mathsf{ZFC}$-axiom $\varphi$ we have $\mathsf{ZFC}\vdash M\models\varphi$ as desired.

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  • $\begingroup$ Thanks! I believe the first bullet refers to the theorem in Hamkins's answer here? Another question: since $\mathsf{ZFC}$ proves $T_0$ is consistent, we can pick out the least constructible set model of $T_0$. This is because $V$ and $L$ agree on the consistency of c.e. theories by Shoenfield's theorem, right? $\endgroup$
    – ikrto
    Feb 14, 2021 at 20:30
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    $\begingroup$ @ikrto To the first, yes. To the second, invoking Shoenfield is galactic overkill and c.e.-ness isn't really needed. The right argument is the following. First, since $L$ and $V$ have the same natural numbers, they agree on consistency of constructible theories $L$. This means first that $T_0=T_0^L$ (think about how we defined $T_0$) and second that $\mathsf{ZFC}\vdash \mathsf{Con}(T_0)\leftrightarrow\mathsf{Con}(T_0)^L$. Now use the fact that the completeness theorem holds in $L$, which is a consequence of $L$ satisfying a tiny tiny amount of $\mathsf{ZFC}$ (e.g. $\mathsf{KP}$ is enough). $\endgroup$ Feb 14, 2021 at 20:33
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    $\begingroup$ It's also worth noting that $T_0$ is defined in a $\Pi^0_1$, not $\Sigma^0_1$, way. Of course being an initial segment of a computable theory according to a computable ordering it is itself computable, but somehow it's "morally" not computable or even c.e. This isn't a point which matters here, but it's neat and I'm easily distracted by shiny objects. $\endgroup$ Feb 14, 2021 at 20:36
  • $\begingroup$ What about transitivity, though? $\endgroup$ Feb 14, 2021 at 22:55
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    $\begingroup$ @spaceisdarkgreen: Obviously if $M$ is an $\omega$-model, it must agree with its meta-theory on the axioms of ZFC, so taking the minimal transitive model of ZFC we have a universe of ZFC where there are no transitive models of ZFC, but since the theory is the same, having a set which satisfies "each of the axioms" would mean that it satisfies ZFC (internally and externally), therefore there are no transitive models there. $\endgroup$
    – Asaf Karagila
    Feb 15, 2021 at 2:00

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