Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 27 / 4 / b / Solution

For nonempty first-order structures $(M_i{)}_{i\in I}$ and an ultrafilter $\mathcal{U}$ on $I$, define the ultraproduct by quotienting the product domain byFunctions are interpreted coordinatewise; a relation holds when its coordinate truth set belongs to the ultrafilter. Closure under finite intersections makes these interpretations independent of representatives. The Łoś theorem asserts, for every formula $\phi$,

Prove it by structural induction. A term evaluates coordinatewise, by induction on its construction, so atomic equality and relations are exactly their defining truth sets. Conjunction uses intersection; negation uses the ultrafilter property that exactly one of a set and its complement belongs to $\mathcal{U}$. These establish all Boolean connectives.

For the existential step put $A=\{i:M_i\models \exists x\psi (x,{\bar{a}}_i)\}$. If an ultraproduct witness $[b]$ exists, induction says $\{i:M_i\models \psi (b_i,{\bar{a}}_i)\}\in \mathcal{U}$, and this set is contained in $A$, so $A\in \mathcal{U}$. Conversely if $A\in \mathcal{U}$, choose a witness $b_i$ at every $i\in A$ and any domain element at the other coordinates. Induction makes $[b]$ a witness. This coordinatewise selection uses choice in the metatheory; nonemptiness of the factors is essential. Universal quantification follows from negation and existence, completing the proof.

For the requested compactness theorem, let $I=[T{]}^{<\omega }$. For each finite $F\subseteq T$, choose $M_F\models F$. For each sentence $\sigma \in T$, its cone $C_{\sigma }=\{F:\sigma \in F\}$ is nonempty, and intersections of finitely many cones contain the finite union of their requirements. These cones generate a proper filter. The ultrafilter lemma extends it to $\mathcal{U}$. The coordinate models satisfying $\sigma$ include $C_{\sigma }$, so the Łoś theorem makes their ultraproduct satisfy $\sigma$. Since this holds for each $\sigma \in T$, the ultraproduct is a model of $T$. No completeness theorem is needed in this second compactness proof.