Compactness theorem

Question

Let $\Sigma$ be a set of theorems, such that for every $\varphi\in\Sigma$ exists an arbitrarily large (<--- edited) finite model $\mathcal{M}$, with $\mathcal{M}\models\varphi$.

Show: It exists an infinite modell $\mathcal{M}$ with $\mathcal{M}\models\varphi$ for every $\varphi\in\Sigma$.

Hello,

I want to proof this statement, and might need some help. I think the compactness theorem is needed.

My idea ist the following:

I want to extend $\Sigma$, such that it has an infinite carrier set $M$ and conclude that for the resulting model $\mathcal{M}=(M,\dotso )\quad\mathcal{M}\models\varphi$ by using the compactness theorem. The actuall proof seems to be trivial:

Let $\Sigma':=\Sigma\cup\{v_i\neq v_j: i\neq j\}$. By assumption every finite $\overline{\Sigma}\subset\Sigma$ is satisfiable. Therefor $\Sigma'$ is satisfiable by the compactness theorem and has obviously an infinite carrier set, since we added an infinite amount of variables.

My thoughts do not need that $\Sigma$ contains theorems (hence formulas without free variables) and not general formulas. Am I mistaken with my proof?

Thanks in advance for your comments.

Answer 1

Theorem: Let $\Sigma$ be a set of sentences such that every finite set of sentences from $\Sigma$ has arbitrarily large finite models. Then there is an infinite model of $\Sigma$.

Proof. Let $\pi_n$ be the assertion that our structure has at least $n$ distinct elements, and let $\Pi$ be the set of all the $\pi_n$. Since $\Pi$ has only infinite models, it is sufficient to show that $\Sigma \cup \Pi$ is consistent. By compactness, it is sufficient to show that $\Sigma \cup \Pi$ is finitely consistent.

Assume we are given a finite subset of $\Sigma \cup \Pi$. That is, we are given $\Sigma_0$ a finite subset of $\Sigma$ and $\Pi_0$ a finite subset of $\Pi$. Since $\Pi_0$ is finite, it can only include sentences $\pi_n$ up to some finite maximum $n_\max$. Therefore, to be a model of $\Pi_0$, it is sufficient to have size at least $n_\max$. By assumption, we can find $M \models \Sigma_0$ of size at least $n_\max$. Therefore $M \models \Sigma_0 \cup \Pi_0$. By previous statements, we have shown the desired result.

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Note that it is necessary to strengthen the hypothesis to talk about finite sets of sentences from $\Sigma$, rather than just individual sentences. Otherwise, as @James mentioned in a comment, we obtain a counterexample whenever we allow $\Sigma$ to contain two sentences that contradict each other. Also, that's how compactness works: you need to know that $\Sigma$ is finitely consistent, not just that every sentence of $\Sigma$ is consistent.

We also have to allow that that every finite set of sentences has arbitrarily large finite models: it doesn't work if you just require $\Sigma$ to be finitely consistent and that each individual sentence has arbitrarily large finite models. Let $\varphi$ be a sentence such that both $\varphi$ and $\neg \varphi$ have arbitrarily large finite models. Then let $\Sigma$ consist of the following two sentences:

• "either the structure has exactly one element or $\varphi$ holds"
• "either the structure has exactly one element or $\neg\varphi$ holds."

Then $\Sigma$ is consistent, its only models have exactly one element, and each sentence in $\Sigma$ has arbitrarily large finite models.