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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. – Rob Arthan Jul 12 '20 at 21:53
  • @RobArthan I do know the compactness theorem. –  Jul 12 '20 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. – Rob Arthan Jul 12 '20 at 21:54
  • @RobArthan thanks for the help, but unfortunately I still don't know how to solve this. –  Jul 12 '20 at 22:06

1 Answers1

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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: $\forall x. x = x$.]

Rob Arthan
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