# Compactness Theorem for Non-elementary classes

Let $$\mathcal{L}$$ be a first-order language and let $$\mathcal{C}$$ be a class of $$\mathcal{L}$$-structures which is not necessarily an elementary class (i.e. there does not exist a theory $$T$$ such that $$\mathcal{C}=mod(T)$$).

Let $$\Sigma$$ be a set of $$\mathcal{L}$$-sentences such that for every finite subset $$\Sigma_0$$ of $$\Sigma$$ there is a member $$\mathcal{M}_{}\in\mathcal{C}$$ so that $$\mathcal{M}\models \Sigma_0$$. Then, can we conclude that there is a member $$\mathcal{N}\in\mathcal{C}$$ such that $$\mathcal{N}\models \Sigma$$?

No, we cannot. For example, take $$\mathcal{C}$$ to be the class of all finite structures in our language, let $$\varphi_n$$ be the sentence asserting that there exist at least $$n$$ elements in the universe - that is, $$\varphi_n$$ is the sentence $$\exists x_1,..., x_n(\bigwedge_{1\le i which is clearly first-order - and let $$\Sigma=\{\varphi_n:n\in\mathbb{N}\}$$. Every finite subset of $$\Sigma$$ is satisfied in all sufficiently large finite structures, but $$\Sigma$$ itself is only satisfied in infinite structures.
Note that we can do a similar trick with larger cardinalities, assuming a large enough language. For any infinite cardinal $$\kappa$$, let $$C_\kappa$$ be the language consisting of $$\kappa$$-many distinct constant symbols $$c_\eta$$ ($$\eta<\kappa$$), take $$\Sigma=\{\neg c_\eta=c_\theta: \eta<\theta<\kappa\}$$, and take $$\mathcal{C}$$ to be the class of all structures of cardinality $$<\kappa$$. Then every subset of $$\Sigma$$ with cardinality $$<\kappa$$ (not just every finite subset) is satisfied in some element of $$\mathcal{C}$$, but no element of $$\mathcal{C}$$ satisfies all of $$\Sigma$$.
• If $\mathcal{C}$ has sufficiently large members. Then can we still filnd an example? – Alice.H Aug 23 '19 at 22:34
• @Alice.H You can replicate the same idea "in a formula" - e.g. take our language to have a single unary predicate symbol $U$, and let $\mathcal{C}$ be the class of all structures $M$ with $U^M$ finite. – Noah Schweber Aug 23 '19 at 22:43