# Compactness and axiomatisability

Let $C$ be an axiomatisable class of structures for some given first-order signature, i.e. there is a set $T$ of sentences whose models are exactly the members of $C$.

Apparently it follows from the compactness theorem that: If $C$ contains arbitrarily large finite structures then it must contain an infinite structure.

Can someone explain this result? The compactness theorem tells me that $T$ has a model if each finite subset of $T$ has a model but I don't see how this relates to the size of structures in $C$.

• @Taroccoesbrocco: In your edits, you need to put "the" before "compactness theorem". It is usual to say something follows "from compactness" (no "the"), but this means it follows from the compactness theorem May 15, 2018 at 0:53
• @CarlMummert - Thank you! May 15, 2018 at 4:56

Hint: $\phi_3 \equiv \exists x\exists y\exists z(x \neq y) \land (x \neq z) \land (y \neq z)$ is a sentence that only holds in a structure with at least three elements. Using this pattern, you can design a sequence of sentences $\phi_3, \phi_4 \ldots$ such that $\phi_n$ only holds in a structure with at least $n$ elements. Then any finite subset of $T' = T \cup\{\phi_3, \phi_4, \ldots\}$ is consistent (because you are given that $C$ contains a model for $T$ that is also a model for $\phi_n$ for arbitrarily large $n$). Compactness gives you a model for $T'$ and that model is then an infinite model for $T$.
If you adjoin infinitely many constants $c_n$ to the language of $T$, and add sentences $c_i \ne c_j, i \ne j$ to $T$, obtaining a new theory $T'$, then every finite subset of $T'$ has a model: Given finite $T'_0\subseteq T'$, let $N$ be the number of distinct $c_n$ appearing in $T'_0$. There is $M\in C$ of size $\ge N$ which is a model of $T'_0\cap T$. Clearly we can assign the $c_n$ occurring in $T'_0$ to distinct elements of $M$, and expand the signature of $M$ to a model $M'$ that includes those elements as interpretations of the $c_n$.
By compactness, $T'$ has a model $M' = (U, ..., (a_n)_{n\in\Bbb N})$. Because $M'$ models all the sentences $c_i\ne c_j$, all the $a_n$ are distinct, so $U$ is infinite. But now the reduct $M = (U, ...)$ is an infinite model of $T$, so $M\in C$.