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I have stumbled upon a question while solving this problem:

If all finite subsets of the set of axioms of a first-order theory $K$ have models, prove that $K$ has a model.

I questioned: approaching such a problem, could one construct a model by counting axioms of $K$ and considering subsequent models of each singleton $\{\mathscr B_i\}$? This seems plausible, since a union of any such sets is consistent. Generalizing, can we conjoin models of $\{\mathscr B_1,\dots,\mathscr B_n\}$ and $\{\mathscr C_1,\dots,\mathscr C_m\}$, if we know that their union is consistent, to get a model of its union? If we can, how can we put it in words, so it is rigorous and does not miss anything?

noballpointpen
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    This is the compactness theorem. The proof is a little bit trickier than what you had in mind. – Rob Arthan Aug 08 '23 at 00:39
  • The answer seems to be: more or less. See Noah Schweber's answer here: https://math.stackexchange.com/questions/4455078/is-there-a-purely-topological-proof-that-a-certain-topological-space-derived-fro/4456537#4456537 – Nagase Aug 08 '23 at 00:45
  • @Nagase Actually, the point of that answer was if anything the opposite. Indeed, there's a good sense in which the answer to the question is negative: we can whip up a computable sequence of sentences $\varphi_i$ with a uniformly computable sequence of models $\mathcal{M}_i\models\varphi_i$ (that is, each $\mathcal{M}_i$ is computable, and moreover there is a computable function assigning to each $i$ a program computing $\mathcal{M}_i$) but such that ${\varphi_i:i\in\mathbb{N}}$ has no computable model. – Noah Schweber Aug 08 '23 at 01:46
  • @NoahSchweber Heh, my original intention was to write "More or less (more 'less' and less 'more')", but I thought it could have just sound weird. "More": the ultraproduct construction is a way of combining structures; "less": the result is not exactly a combination, as your answer shows – Nagase Aug 08 '23 at 10:35
  • If this was a problem, did it come after Gödel’s completeness theorem? It follows from that. – Carsten S Aug 08 '23 at 11:37

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This question is (necessarily) somewhat vague, but there's a good sense in which the answer to the question is negative: we can whip up a computable sequence of sentences $\varphi_i$ with a uniformly computable sequence of models $\mathcal{M}_i\models\varphi_i$ (that is, each $\mathcal{M}_i$ is computable, and moreover there is a computable function assigning to each $i$ a program computing $\mathcal{M}_i$) but such that $\{\varphi_i:i\in\mathbb{N}\}$ has no computable model. This tells us that any such "fusing" method must be very complicated.

Here's one way to do this: in the language of arithmetic + a new constant symbol $c$, let $\varphi_{2i}$ be the $i$th axiom of Peano arithmetic and let $\varphi_{2i+1}$ be the sentence "$c>\underline{i}$" where $\underline{i}$ is the numeral standing for $i$. Letting $\mathcal{M}_i$ be $\mathbb{N}$ expanded to interpret $c$ as $i+1$ gives the uniformly computable models of the individual $\varphi_i$s, and Tennenbaum's theorem handles the rest of the claim.

Similarly, we can find sentences $\varphi,\psi$ each of which has a computable model such that $\varphi\wedge\psi$ is consistent but has no computable models; this is a good exercise if you're already somewhat familiar with Tennenbaum's theorem, and in particular the fact that it is applicable even to some finitely axiomatizable theories such as $\mathsf{I\Sigma_1}$.

On the other hand, if we strengthen the hypothesis from "$\mathcal{M}_i\models\varphi_i$" to "$\mathcal{M}_i\models\bigwedge_{j\le i}\varphi_j$" so that the "basic" models get closer and closer to satisfying the whole theory (FWIW this doesn't affect the computability-theoretic point above), there is an important "fusion" method which is guaranteed to work: ultraproducts. If $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$ then $$\prod_{i\in\mathbb{N}}\mathcal{M}_i/\mathcal{U}$$ is a model of the whole theory $\{\varphi_i:i\in\mathbb{N}\}$ by Los' Theorem; see e.g. the discussion here. However, both Los' Theorem and the existence of a nonprincipal ultrafilter in the first place rely on (a fragment of) the axiom of choice. While ultraproducts are extremely important in model theory, this fact is not in tension with the observation in the previous paragraph.

Noah Schweber
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