# Can a theory prove (schematically) its axioms relativized to a set?

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 $$M$$?

• @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? Feb 14, 2021 at 19:45
• I misread your question, so I removed the previous comment. Feb 15, 2021 at 12:32

$$\mathsf{ZFC}$$ 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.

• 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? Feb 14, 2021 at 20:30
• @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). Feb 14, 2021 at 20:33
• 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. Feb 14, 2021 at 20:36
• What about transitivity, though? Feb 14, 2021 at 22:55
• @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. Feb 15, 2021 at 2:00