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Does there exist a first-order signature $L$ and two first-order $L$-theories $T$ and $T'$ such that $T$ and $T'$ both have arbitrarily large finite models, and the finite models of $T$ and $T'$ are the same, but $T$ and $T'$ differ on infinite models?

user107952
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5 Answers5

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In the language of rings, the theory of integral domains and the theory of fields have the same finite models (since every finite integral domain is a field), and have arbitrarily large finite models (since there are finite fields of all prime power orders), but they differ on infinite models (since there are integral domains which are not fields).

Alex Kruckman
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    Ah, that's a nice one. Similarly we could use Wedderburn's little theorem and replace the theory of integral domains with the theory of division rings. – Qiaochu Yuan Feb 05 '25 at 19:01
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    Another example comes from topology: $T_1$ spaces and discrete spaces. – freakish Feb 05 '25 at 19:19
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    @freakish That's right, in spirit. But how do you treat $T_1$ spaces as models of a first-order theory? – Alex Kruckman Feb 05 '25 at 19:20
  • @AlexKruckman ops, you are right. Topology is not definable in first order. – freakish Feb 05 '25 at 19:22
  • @AlexKruckman: Well, not quite exactly, but you can treat them as two-sorted structures. Then you will have models in which the second sort is not closed under unions (it's a basis closed under finite unions and intersections), but the separation axioms remain elementary and the basic principle works just the same, no? – tomasz Feb 08 '25 at 17:21
  • @tomasz Yes, I agree that works. – Alex Kruckman Feb 08 '25 at 19:25
  • We can distill this to an even simpler example, the theory of cancelative monoids and the theory of groups. Then we note that every finite integral domain $R$ satisfies that $R \setminus {0}$ is a cancelative multiplicative monoid, hence a group, so $R$ is a field. – Mark Saving May 06 '25 at 21:31
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We can take $T$ to be any first-order theory with

  1. arbitrarily large finite models
  2. such that there exists a sentence $\varphi$ which is satisfied by some infinite model but no finite model

then take $T' = T \cup \{ \neg \varphi \}$. In Alex Kruckman's example, $T$ is the first-order theory of integral domains $D$, and $\varphi$ is "$D$ has a nonzero non-invertible element."

Another algebraic example: $T$ is the first-order theory of groups $G$, and $\varphi$ is "$G$ has at least $3$ elements, and every non-identity element of $G$ is conjugate." It's a nice exercise to show that there don't exist any finite such groups, and a harder one to show that there exist infinite such groups.

More examples with groups can be produced using finitely presented groups which are not residually finite, e.g. the Higman group, which lets us take $\varphi$ to be:

$G$ contains $4$ elements $a, b, c, d$, not all trivial, satisfying $$a^{-1} ba = b^2, b^{-1} cb = c^2, c^{-1} dc = d^2, d^{-1} ad = d^2.$$

For a proof that this works (that no finite group satisfies $\varphi$ but some infinite group does) see e.g. this blog post by Terence Tao.

Qiaochu Yuan
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For an algebra-free example, take for $T$ the theory of total discrete orders with a first element, and for $T'$ additionally ask that there is a last element.

Denis
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    Or let $T$ be the theory of linear orders and let $T'$ be the theory of discrete linear orders (or the theory of discrete linear orders with a least element and a greatest element). – Alex Kruckman Feb 06 '25 at 14:43
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For an easily-verified example: Take a single propositional constant $P$, and take $T_1$ to be axiomatised by just “$P$”, and $T_2$ to be axiomatised by the sentences “if there exist $\leq n$ elements, then $P$” for each $n \in \mathbb{N}$.

This is essentially a minimal modification of the obvious trivial example, where $T_1$ is inconsistent and $T_2$ says “there are $\geq n$ elements” for each $n$, which you’ve excluded by your non-triviality condition “arbitrarily large finite morels” — replacing $\bot$ by $P$ makes it non-trivial while still exhibiting this phenomenon.

Peter LeFanu Lumsdaine
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A couple more somewhat natural examples:

  • the theories of surjective and of injective (unary) functions agree on finite models,
  • the (universal) theory of finite groups and the theory of groups agree on finite models (although strict containment is not so obvious in this case, but follows from the observation that a finitely presentable group satisfies the universal theory of finite groups if and only if it is residually finite, and not all f.p. groups are residually finite).
tomasz
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  • Nice, I didn't know the second example. Do you know an explicit identity that holds for all finite groups but not for all groups? – Alex Kruckman Feb 09 '25 at 03:19
  • Oh, actually I see that Qiaochu Yuan already gave this example (with an explicit sentence) in his answer. – Alex Kruckman Feb 09 '25 at 03:26
  • @AlexKruckman: Yes, basically, any f.p. not residually finite group gives you an explicit example (as explicit as the presentation and the witness to failure of residual finiteness). – tomasz Feb 10 '25 at 21:30
  • @AlexKruckman, can one say that in all finite groups every element has finite order, but this sentence is not true for infinite groups? – Maryam Ajorlou Mar 16 '25 at 07:06
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    @MaryamAjorlou No, because "$x$ has finite order" is not a first-order definable property. – Alex Kruckman Mar 16 '25 at 11:58
  • @AlexKruckman, Aha! Thank you. what about this sentence, in every finite group, for every element $g$, we have $g^{|G|} = e$? – Maryam Ajorlou Mar 16 '25 at 13:31
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    @MaryamAjorlou Go ahead and try to write that as a first-order sentence in the language of groups. How do you express $|G|$ or exponentiation $g^n$ when $n$ is not fixed? – Alex Kruckman Mar 16 '25 at 14:01
  • Yes, you are right. I didn't notice that we can't have a quantifier over a number like $|G|$ – Maryam Ajorlou Mar 16 '25 at 19:25