In 'A Mathematical Introduction to Logic' by Enderton, page $147,$ the author stated the following theorem.

Theorem $26$A If a set $\Sigma$ of sentences has arbitrarily large finite models, then it has an infinite model.

The following is my question.

Question: I would like to know the meaning of arbitrarily large model.

I know that finite model means that its universe has finitely many elements. But I fail to understand arbitrarily large model, especially when we talk about arbitrarily large finite model.


It's not that an individual model of $\Sigma$ is 'arbitrarily large', it's that there is no upper bound on the possible size of a finite model of $\Sigma$. To spell this out, what it means for $\Sigma$ to have arbitrarily large finite models is that, for any natural number $n$, there is a finite model of $\Sigma$ of size $\ge n$.

By contrast, consider the set $\Sigma = \{ \forall x. \forall y. x=y \}$. Any model of $\Sigma$ must have size $0$ or $1$, and so this does not have arbitrarily large finite models.

  • $\begingroup$ Thanks. Your explanation is very clear. So your $\Sigma$ has finite but not arbitrarily large model, right? $\endgroup$ – Idonknow Mar 2 '18 at 15:29
  • $\begingroup$ Right, it has a finite model, but it does not have arbitrarily large finite models (and it certainly doesn't have an infinite model). $\endgroup$ – Clive Newstead Mar 2 '18 at 15:30

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