# Satisfiability of sentences (Compactness theorem)

If L is languege $L-$sentence $\phi$ is satisfiable in every finite $L-$structure, is it satisfible also in every infinite $L-$structure?

I would say that's true. In my opinion, from the Compactness theorem it holds that if $\phi$ is satisfiable in every finite $L-$structure, it has to be satisfiable in any infinite $L-$structure. However, I'm not sure why it has to be satisfiable in every.

I've tried to prove it like that: Let ve have $\neg \phi$ and let suppose it has some infinite model (that means there exist some $L-$structure in which $\neg \phi$ is satisfiable). Then, according to the Compactness theorem, there has to exist some finite subset of our $L-$structure in which $\neg \phi$ is satisfiable, which is a contradiction.

Is it possible to do it like that or am I wrong in some point? Because I'm not really sure about it, so I'm afraid I'm doing some mistake without knowing it...

• I do not know what is meant by satisfiable in every (finite) $L$-structure. In any specific $L$-structure, $\phi$ is either true or false. Jun 11, 2015 at 6:32
• Yes, that's how I've understood the question. It was really formulated like this, however, I supposed that $\phi$ is satisfiable in some $L-$structure if it's true in it. Jun 11, 2015 at 6:36
• Compactness tells us that, if $\phi$ is true in finite models of arbitrarily large cardinality, then $\{\phi\}\cup\{\text{there are at least }$n$\text{ distinct elements} \mid n\in\mathbb{N}\}$ is consistent, so there exists an infinite model in which $\phi$ is true. You are using a (false) partial converse, that if $\psi = \neg\phi$ is true in some infinite model, then it is true in some finite model. Lots of theories have infinite models but no finite models—algebraically closed fields, for example. Jun 11, 2015 at 7:14

There are sentences $\phi$ that are true in every finite $L$-structure but such that there are infinite $L$-structures in which $\phi$ is not true.
Here is one example. Let $L$ have a single binary predicate symbol $\lt$. Let $\alpha$ be the conjunction of all the sentences that together say that $\lt$ is a total order. Let $\beta$ be the sentence that says there is a smallest element under $\lt$, and let $\phi$ be the sentence $(\alpha\to\beta)$.
Then $\phi$ is true in every finite $L$-structure. However, there are infinite totally ordered sets that do not have a least element, so there are infinite $L$-structures in which $\phi$ is false.
Remark: I am puzzled by the phrase "satisfiable in every (finite) $L$-structure. Given any $L$-structure $\mathbb{M}$, any sentence is either true of false in $\mathbb{M}$.
• You are welcome. Compactness tells us in particular that if $\phi$ has models of every finite cardinality, or even just of arbitrarily large finite cardinality, then $\phi$ has an infinite model. But there may well be infinite structures in which $\phi$ is false. Jun 11, 2015 at 7:09
• That's the reason I wasn't sure about it. I thought that there is an equivalence, that if $\phi$ has an infinite model, then there must exists any finite subset of this model for $\phi$. So this implication doesn't hold? Jun 11, 2015 at 7:13
• It is not true that if $\mathbb{M}$ is an infinite model, then some finite substructure of $\mathbb{M}$ is a model. There may not even be a finite substructure, Jun 11, 2015 at 7:16