7

A simple application of the Compactness theorem shows that for any first order theory $T$, if it has aribtrarily large finite models, then it also has an infinite model.

This is an interesting result, even surprising, one might say. However, I couldn't come up with any interesting examples for $T$ for which this result is not trivial. For the theory of groups, or of fields, or graphs - it's very easy to construct an infinite model. So, Are there any interesting consequences from this result?

35T41
  • 3,377
  • "Very easy" is in the eye of the beholder. For example, would you say it's very easy to exhibit an infinite field of characteristic p>0? – Z. A. K. Oct 16 '20 at 17:06
  • 1
    @Z.A.K.: sure, take $\mathbb{F}_p(t)$. – Qiaochu Yuan Oct 16 '20 at 17:18
  • @QiaochuYuan: That's exactly what I mean. It's obvious to me, and it's easy to understand the direct construction, but it's not obvious in the same sense that an infinite group $\mathbb{Z}$ or field $\mathbb{Q}$ are: I'd expect all students in my logic course to have seen the former two, but not $\mathbb{F}_p(t)$. – Z. A. K. Oct 16 '20 at 17:30
  • @Z.A.K. Or with less effort, fix an appropriate family of field embeddings $f_{i,j}: \mathbb{F}{p^i}\rightarrow\mathbb{F}{p^j}$ and take the direct limit. – Noah Schweber Oct 16 '20 at 18:07
  • @Z.A.K.: The algebraic closure of any finite field. – Asaf Karagila Oct 16 '20 at 18:09
  • @AsafKaragila The fact that every field has an algebraic closure is certainly not obvious (as pointed out by Zev Chonoles in the linked answer :) ) – Z. A. K. Oct 16 '20 at 18:11
  • @Z.A.K.: If you know that there are arbitrarily large finite fields of characteristics $p$, you may as well know that there are algebraic closures. – Asaf Karagila Oct 16 '20 at 18:12
  • @AsafKaragila why would the students need to know that there are arbitrarily large finite fields of characteristic p? You can prove it to them that fact when you give the application of the compactness theorem (especially for odd p). [at this point it'd be best to move this to chat] – Z. A. K. Oct 16 '20 at 18:16
  • Well, the compactness theorem is proved from AC (axiom of choice) or PIT (Boolean algebra prime ideal theorem). Maybe you can use your fact about models to prove some things that are usually proved from AC or PIT ? – GEdgar Oct 16 '20 at 18:17
  • @GEdgar: The problem here is that normally when talking about finite models, you're going to be dealing with a finite language (or at the very least a well-orderable language), in which case you don't need the full strength of BPI to prove there is an infinite model, it just follows in ZF. E.g., there are finite linear orders, so there's an infinite one, but that's easy: $\Bbb N$. – Asaf Karagila Oct 16 '20 at 18:21
  • 1
    It’s the easiest way to prove the finite Ramsey theorem: prove the infinite Ramsey theorem, which is rather easy, and then use the compactness theorem to get a contradiction if the finite Ramsey theorem fails. – Brian M. Scott Oct 16 '20 at 18:40

2 Answers2

3

I mostly see this result used in the contrapositive: if you have a class of objects for which there are arbitrarily large finite objects but no infinite object, then your class of objects is not the class of models of a first-order theory. For example, there is no first-order theory whose models are exactly the finite groups, etc.

In turn, I mostly see this result used as follows: suppose you have some theory $T_1$ whose models are some objects you're interested in (groups, etc.) and you want to know whether some property of these objects is finitely axiomatizable by a finite set of axioms $T_2$. You can rule this out by considering the single sentence consisting of the conjunction of the negation of every sentence in $T_2$, which together with $T_1$ gives the objects not satisfying your your property. If there are arbitrarily large finite objects but no infinite object not satisfying your property, then $T_2$ can't exist.

Unfortunately the examples I can think of of this situation usually involve using more of the strength of the compactness theorem. For example, you can show that for fields, having characteristic $0$ is not finitely axiomatizable, because the negation of this condition (having positive characteristic) does not satisfy the compactness theorem: if you add sentences equivalent to "the characteristic is not $2$, not $3$, not $5$," etc. then every finite subset of those has a model (a field of larger positive characteristic) but the entire set of sentences can't have a model (within fields of positive characteristic).

Qiaochu Yuan
  • 419,620
2

Is there a group on which the maps

$x \mapsto x^2, x \mapsto x^3, x \mapsto x^4, \dots$

are injective precisely if they are surjective? For example, $\mathbb{Z}$ is not such a group, since the map $x \mapsto x + x$ is injective, but not surjective.

You can use your knowledge of group theory to construct such a group explicitly (easy to think of Abelian examples, so let's make it non-Abelian), or you can appeal to compactness. Consider the first-order theory over the language of group theory that has the axioms of group theory (again, you can make it non-Abelian), along with axioms asserting that each of the functions above is injective precisely if it is surjective.

On a finite set, every function is injective precisely if it is surjective, so every finite group (every finite non-Abelian group) is a model of this theory. Therefore, there must be an infinite group that is a model of this theory as well.

One can, of course, immediately start asking less trivial questions along the same lines (until directly constructing the groups gets difficult enough): this eventually leads to the theory of pseudofinite structures, structures that satisfy all sentences that every finite structure of the same kind satisfies (if you thought of $(\mathbb{Q},+)$ in the Abelian case: that is actually a pesudofinite group, but proving that is much harder).

Z. A. K.
  • 11,359