# compactness model theory question

Let $$\sigma$$ be a set of first-order formulas including the axioms of equality. Suppose that for every $$n\in\mathbb{N}, \sigma$$ has a satisfying model $$M_n$$ whose domain is finite and has at least $$n$$ distinct elements. Prove that the set $$\sigma$$ must have a model with infinite domain.

Edit: Here's my revised attempt.

By the compactness theorem, a $$\sigma$$ has a model iff every finite subset of $$\sigma$$ has a model. To show that $$\sigma$$ has a model with infinite domain, I need to add sentences to $$\sigma$$ to construct an infinite model, that satisfies $$\sigma$$ equipped with these sentences and thus $$\sigma$$, though I'm not sure how to find these sentences.

• A bit more context would be helpful. If you don't know the compactness theorem, then I don't think you will be able to give the proof. If you do know the compactness theorem, then you can augment $\sigma$ with sentences $\phi_i$ saying that the universe has at least $i$ elements and then use the given assumption and the compactness theorem to get an infinite model. Commented Jul 12, 2020 at 21:53
• @RobArthan I do know the compactness theorem.
– user763400
Commented Jul 12, 2020 at 21:53
• OK - go for it. If you can't see how to do it, update your question with what you tried and we can give you more help. Commented Jul 12, 2020 at 21:54
• @RobArthan thanks for the help, but unfortunately I still don't know how to solve this.
– user763400
Commented Jul 12, 2020 at 22:06

Define sentences $$\phi_i$$ as follows:
$$\phi_i = \exists x_1, x_2, \ldots, x_i. \bigwedge_{1 \le m < n \le i} x_m \neq x_n$$
I.e., (given the equality axioms), $$\phi_i$$ holds in a model $$M$$ iff $$M$$ has at least $$i$$ distinct elements. Let $$\sigma' = \sigma \cup \{\phi_1, \phi_2, \ldots\}$$. By assumption, any finite subset of $$\sigma'$$ has a model. Hence by the compactness theorem, $$\sigma'$$ has a model,$$M$$, say. But then the equality axioms together with each $$\phi_i$$ all hold in $$M$$, so $$M$$ must be infinite.
[Aside: the only equality axiom that is actually relevant here is reflexivity: $$\forall x. x = x$$.]