Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 20 / 9

A model-theoretic ultralimit is the direct limit of a chain of successive ultrapowers along their diagonal elementary maps. It is not the topological notion of a limit along an ultrafilter. We first prove the precise relative embedding lemma needed to synchronize two such limits.

If $A\prec B$, then $B$ embeds elementarily into an ultrapower of $A$, with the embedding restricting on $A$ to its diagonal map. To prove this Frayne embedding lemma, use the elementary diagram of $B$ with the elements of $A$ treated as named parameters. Every finite fragment $\Delta$ of that diagram has a realization in $A$: replace its finitely many names from $B\setminus A$ by variables and existentially quantify their conjunction. It is true in $B$, and hence true in $A$ by elementarity. Choose these simultaneous realizations for each finite fragment, keeping every parameter from $A$ interpreted as itself; give unmentioned names an arbitrary default in the nonempty $A$.

On the index set of finite fragments, extend the filter of cones $\{\Delta :\Delta \supseteq \Gamma \}$ to an ultrafilter $U$. For each $b\in B$, let $g_b(\Delta )$ be its value in the chosen coordinate realization. Every diagram sentence is satisfied on its cone, so the Łoś theorem makes $b\mapsto [g_b]$ an elementary embedding $B\to A^I/U$. Inequalities in the diagram ensure injectivity. If $b\in A$, all its coordinates equal $b$, giving exactly the diagonal map.

There is also the initial version: if merely $A\equiv B$, then $A$ embeds elementarily into an ultrapower of $B$. A finite fragment of the elementary diagram of $A$, with all its names replaced by variables, yields an existential sentence true in $A$ and hence in $B$. The identical finite-fragment and cone construction therefore proves this version without already assuming an embedding.

Start with $A_0=A$. Use the initial version to place $A_0$ elementarily in an ultrapower $B_0$ of $B$. Relabel embedded copies so we can regard the elementary maps as inclusions. Apply the relative version to $A_0\prec B_0$ to obtain an ultrapower $A_1$ of $A_0$ containing $B_0$, and whose inclusion of $A_0$ is its diagonal map. Next apply it to $B_0\prec A_1$ to obtain an ultrapower $B_1$ of $B_0$ containing $A_1$, again fixing the earlier diagonal inclusion. Continuing gives an elementary chainThe common union $C$ is elementary over every stage by the elementary chain theorem, andThe first presentation is a model-theoretic ultralimit of $A$. The second, preceded by the initial diagonal embedding of $B$ into $B_0$, is a model-theoretic ultralimit of $B$. Each consecutive same-letter inclusion is precisely the corresponding ultrapower diagonal map, not just an unspecified elementary embedding. Thus the two limits are identified with the same structure $C$, proving