Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 135 4 Solution Created 2026-10-03 Updated 2026-10-05
One form of the Ehrenfeucht-Mostowski theorem says that, for every infinite first-order structure and every total order , an elementary extension contains distinct elements forming an order-indiscernible sequence. After a suitable Skolem expansion, their Skolem hull is a first-order model of , and every order automorphism of extends to a structure automorphism of that hull. If has a definable infinite linearly ordered subset , the generators may lie in and their order may agree with that definable order. Uniqueness of the extended structure automorphism is asserted in the chosen Skolem expansion, rather than for every structure automorphism of its reduct.
Here is an ultraproduct proof of the Ehrenfeucht-Mostowski theorem. First expand to with Skolem functions for all existential first-order formulas, iterating through the enlarged languages if necessary. This is done in the ordinary classical metatheory; the choice restriction in question 1 is not a restriction on this question. Fix a nonprincipal ultrafilter on .
We can obtain an infinite sequence of distinct elements in an elementary extension of by one preliminary ultrapower. For each choose a finite list of distinct elements of . Represent by the function taking the th entry when , with an arbitrary default otherwise. Distinctness holds on a cofinite set, so follows from Łoś theorem. In the definably ordered variant, first name any parameters defining , and choose each finite list increasingly inside instead; then and for . Infinitude of an ordered set supplies every finite increasing chain. This preparatory step uses no Ramsey theorem.
For a finite subset of , define the ultrafilter on by nested membership, in this order:where means . This is the ordered Fubini product of ultrafilters. Closure under intersections and the decision between a set and its complement hold at each nested level, proving it is an ultrafilter. For take the principal ultrafilter on the one-point product. PutIf , pull a function back along the projection . A test independent of an omitted coordinate is unchanged by its ultrafilter quantifier, so projection pushes to . The induced map is therefore an elementary embedding by Łoś theorem. These maps are coherent, giving a directed system indexed by finite subsets of .
Take its directed limit of elementary embeddings . One can verify elementarity directly: representatives of a finite tuple occur at a common stage; functions and atomic relations are interpreted there. In the existential step, a witness in the limit occurs with the parameters at some later common stage, and elementarity pulls the existence statement back. Thus every embeds elementarily into , which contains an elementary copy of .
For , let be the class of the coordinate function . Projection coherence makes this independent of . If , then for each fixed the set is cofinite. The nested test therefore gives . In the ordered variant, the same argument with gives and .
For every first-order formula of the expanded language and every increasing tuple , Łoś theorem givesThe right-hand side depends on the first-order formula and tuple length, not on the indices. This proves order indiscernibility in the expanded language. Let be the Skolem hull of the generators in . The Tarski-Vaught test gives , and its reduct is a first-order model of .
An order automorphism of acts byIndiscernibility makes this well defined: combine the finite supports of two term expressions into one increasing tuple, and apply indiscernibility to their logical equality. Applying it to relation first-order formulas proves preservation of all relations; the inverse is induced by . Every element of the hull is such a term, so the extension is unique among structure automorphisms preserving the Skolem expansion. MoreoverThis proves the theorem, including its ordered version and structure automorphism conclusion, using ultrapowers and a directed limit throughout.