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Stone Duality characterizes Boolean algebras in terms of spaces. I regard this as being done by identifying an algebra with its ``space of models'', and feel like the barrier for a similar thing to be done is that FOL language does not have the power of characterizing models, but propositional logic can. (I am not sure if my thoughts are relevant).

But here:

https://personalpages.manchester.ac.uk/staff/Marcus.Tressl/papers/StoneDualityBooleanAlgebras.pdf

on page 14, the paper says Stone duality is something like completeness, which does admit a counterpart in FOL.

So what is the real reason for FOL to lack of a description like Stone duality?

Y.X.
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    "FOL language does not have the power of characterizing models, but propositional logic can" I'm not sure what you mean by that? – Noah Schweber Feb 04 '24 at 22:55
  • You are probably looking for something along the lines of this result https://arxiv.org/abs/1008.3145. There are some details about why coherent logic is being used that I do not understand, but apparently you can view any classical first-order theory as a coherent theory via Morleyization and this process preserves the category of models. – Alvaro Pintado Feb 04 '24 at 23:48
  • Sorry, I am not able to make it precise either. I sort of think about FOL theories cannot give the characterization of the models, for instance, there exist non-standard models of arithematics. Another thing that I think maybe relevant is that if we attempt to identify a FOL theory with its models, you get a class (at least one for each cardinality) instead of a set of models (which can be captured by an ultrafilter). – Y.X. Feb 05 '24 at 02:57
  • @Y.X. To be fair, by downward Lowenheim-Skolem we only need to think about a set of models. – Noah Schweber Feb 05 '24 at 03:06

1 Answers1

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This got longer than I thought; skip to the end if you just want a couple possibly-useful references!

There is, unsurprisingly, extensive work on algebraic and topological aspects/interpretations of first-order logic. But you are correct that it is much murkier than the propositional setting.

The most obvious issue is that semantics and syntax are much further separated in the case of $\mathsf{FOL}$. A "model" for propositional logic in a fixed language (= set of atomic propositions) $P=\{p_i:i\in I\}$ is just a valuation map $\nu:P\rightarrow\{\top,\perp\}$, and there is no meaningful difference between models and complete (= maximal consistent) theories. The situation for first-order logic is completely different: even if you restrict to "nice" structures (e.g. countable ones), there are non-isomorphic structures with identical theories. In fact, there is no canonical way to associate a single model up to isomorphism with a complete theory, in the sense that $\mathsf{ZF}$ alone cannot prove that for each (first-order) language $\Sigma$ there is a function sending each complete $\Sigma$-theory to a model of that theory.

The situation is even worse than this indicates, though. The above might suggest that we could "fix" things by (assuming Choice and) using a stronger logic than $\mathsf{FOL}$. But there really is a fundamental jump in complexity of the semantics involved which can't be gotten around. Forgetting logics and theories entirely for a second, each $P$-valuation is determined entirely by its restrictions to the individual $p_i\in P$, and the space of $P$-valuations is the $I$-fold product of these atomic valuations. By contrast, given a first-order language $\Sigma=\{R_j:j\in J\}$, there is no good way to combine a family $\{\mathcal{M}_j:j\in J\}$ with each $\mathcal{M}_j$ an $\{R_j\}$-structure into a single $\Sigma$-structure, appropriate forgetful functors notwithstanding. In the first-order setting, the different parts of the language can interact meaningfully with each other, and that's not true of the propositional setting. This is what leads the obvious attempt to topologically prove the compactness of first-order logic to fail.

The topological analyses of first-order logic and general predicate logics are significantly more complicated than their propositional counterparts; see e.g. type spaces (which are Stone spaces!). For that matter, so are the algebraic analyses (cf. cylindrical algebras or polyadic algebras). So we should expect any duality theorems around first-order logic to be similarly complicated. I'm not an expert here, but at a glance Makkai-style conceptual completeness theorems and papers like van Gool/Marques' On duality and model theory for polyadic spaces seem relevant. In particular, quoting from Makkai's paper: "Strong conceptual completeness is most familiar in (classical) propositional logic, where it takes the form of the Stone duality theorem. "

Noah Schweber
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  • Thanks so much! This brings me much more interesting things than I expected. A quick question: as for " even if you restrict to "nice" structures (e.g. countable ones), there are non-isomorphic structures with identical theories." I believe it is true, but may I please ask for an explicit example? – Y.X. Feb 06 '24 at 04:20
  • @Y.X. Sure. Take for example Presburger arithmetic. By the compactness theorem, it has a model not isomorphic to $(\mathbb{N};0,1,<,+)$ - think about adding a constant symbol $c$ together with axioms saying that $c>0$, $c>1$, $c>1+1$, etc. – Noah Schweber Feb 06 '24 at 04:35
  • Sorry I do not get it. What explicitly are the two different structures, and what is the same theory of them? – Y.X. Feb 06 '24 at 06:20
  • @Y.X. I used the compactness theorem to avoid having to explicitly build structures. For explicit examples, consider the linear orderings $\mathbb{Z}$ (ordered as usual) and $\mathbb{Z}\times {0,2}$ (ordered lexicographically; think of this as "$\mathbb{Z}+\mathbb{Z}$"). These linear orders are countable and non-isomorphic, but we can show (e.g. via Ehrenfeucht-Fraisse games) that they satisfy exactly the same first-order sentences. – Noah Schweber Feb 06 '24 at 17:49
  • Back to my original comment, Presburger arithmetic is complete so any pair of non-isomorphic countable models of it will give an example. Unfortunately, Presburger arithmetic is also a bit more complicated. One model is obvious, namely $\mathbb{N}$. For an example of a countable nonstandard (that is, $\not\cong\mathbb{N}$) model, consider the set $\mathscr{X}$ of polynomials with rational coefficients which (if nonzero) have positive leading coefficient and whose constant term is an integer. $0,1,+$ are defined on $\mathscr{X}$ as usual, and we set $p<q$ iff $q(x)-p(x)>0$ for large enough $x$. – Noah Schweber Feb 06 '24 at 17:57
  • The two non-isomorphic models of Presburger arithmetic seem very non-trivial to come up with! Thank you very much for making it explicit! – Y.X. Feb 08 '24 at 22:41