This is a followup to a previous question of mine. To summarize the result of that question, by a result of Kanovei and Shelah, $\mathsf{ZFC}$ is enough to show that there is a uniform procedure for taking a structure and finding a proper class sized elementary extension of it (with a definable truth predicate) which is 'class-saturated' in the sense that any type over a set of parameters is realized.
Ultimately my question is whether or not an analogous statement can be made for building such models uniformly from theories, rather than from structures. The result of my previous question is enough to reduce that to this question:
Question 1. Is there a formula $\varphi(x,y,z)$ in the language of set theory such that for any first-order language $\mathcal{L}$ and any complete $\mathcal{L}$-theory $T$, there is a unique $\mathcal{L}$-structure $\mathfrak{M}$ such that $\mathfrak{M}\models T$ and $\varphi(\mathcal{L},T,\mathfrak{M})$ holds.
What if we allow $\varphi$ to have parameters (i.e. in principle $\mathsf{ZFC}$ might not have a global choice function for models of complete theories, but it might prove that there always exists a parameter over which it has such a function)?
A few observations and thoughts.
- If we add in a well-ordering of $\mathcal{L}$ then we can easily do this by adding Henkin constants and completing the resulting theory in a mechanical way (which we can do since we can well-order the set of formulas in this language).
- There is such a formula if instead of requiring that $\varphi(\mathcal{L},T,\mathfrak{M})$ hold for a unique $\mathfrak{M}$ we instead require that it holds for a unique isomorphism type. Specifically, we can let $\varphi(\mathcal{L},T,\mathfrak{M})$ hold whenever $\mathfrak{M}$ is a special model of cardinality $\kappa$ with $\kappa \geq |\mathcal{L}|$ the smallest such that $T$ has a special model of cardinality $\kappa$.
- Using Scott's trick we can get that down to a set of models of the same isomorphism type (i.e. we require that $\mathfrak{M}$ have minimal foundational rank among members of its isomorphism type).
- If we allow $T$ to be incomplete, then this implies a global Boolean prime ideal theorem, i.e. that there is a uniform choice of prime ideal for Boolean algebras. I'll call this $\mathsf{GBPI}$.
- It is easy to see that if we have $\mathsf{GBPI}$ then we can get such a formula, even for $T$ incomplete, which would imply that for $T$ incomplete this is equivalent to $\mathsf{GBPI}$. All you need to do is add Henkin constants to the theory (which can be done in a completely uniform way) and then choose a completion of the Henkinized theory using $\mathsf{GBPI}$. The term model gives the required model of the theory. That said, I doubt that $\mathsf{ZFC}$ proves $\mathsf{GBPI}$, or even fairly weak consequences of it, such as global choice for pairs, for that matter. I also suspect that $\mathsf{GBPI}$ is weaker than global choice.
- Just like with global choice, despite the seemingly non-first-order nature of the statement $\mathsf{GBPI}$, it is expressible as a single first-order sentence thanks to the reflection principle: Let $\chi(x,y)$ be a formula that says '$(V_\alpha,\in) \models \psi(x,y)$ for the smallest $\alpha$ and smallest $\psi$ (in some fixed well-ordering) such that $\exists ! z ((V_\alpha,\in) \models \psi(x,z)$ and $z$ is a prime ideal for $x)$.' As long as there is a formula that uniformly selects prime ideals, this formula will as well. A similar trick will work for $\mathsf{GBPI}$ with a parameter (although $\psi$ will also need a parameter) and for a global choice function for models.
I restricted Question 1 to complete theories in the hope that it would make the question easier, but there's a chance it's equivalent.
Question 2. Is the existence of a global choice function for models of complete theories equivalent to $\mathsf{GBPI}$?
