Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 25 11 i Solution Created 2026-10-03 Updated 2026-10-14
We prove the Ehrenfeucht-Mostowski theorem by ultraproducts, with no application of Ramsey theorem. Expand the infinite structure by Skolem functions, choose distinct elements , and choose a nonprincipal ultrafilter on . Let be the desired total index order and put .
For every finite ordered subset , define an ultrafilter on by the ordered Fubini product of ultrafilters: a set belongs to exactly whenHere means , with the quantifiers nested in the displayed order. The resulting collection is an ultrafilter, by induction using the ultrafilter laws for negation and conjunction. For , the inverse image of under coordinate projection belongs to exactly when : quantifiers on unused coordinates leave a truth value unchanged.
Consequently there is a well-defined ultrafilter on the Boolean algebra of finite-coordinate cylinders in , declaring a cylinder large by this test. Coherence proves finite intersection closure and ensures that the empty cylinder is not large. Extend its generated filter to an ultrafilter on all subsets of . Form the ultrapower , and for let be the class of .
If , the coordinate equality test belongs to no corresponding two-coordinate product ultrafilter: for every value of the outer coordinate, the inner equality set is a singleton, excluded by nonprincipality. Hence the are distinct. For any first-order formula and any , the Łoś theorem givesThe right side is independent of the actual increasing index tuple, so the form an order-indiscernible sequence. Constant functions embed elementarily into . Take the Skolem hull of the generators. The Tarski-Vaught test gives an elementary substructure, and transporting terms along an index-order automorphism gives a well-defined automorphism of the hull, with inverse obtained by transporting along the inverse index automorphism. Equality of term representations is preserved by indiscernibility. This is the indiscernible and automorphism conclusion of Ehrenfeucht–Mostowski, obtained entirely from ultrafilters and Łoś's theorem. The ordered Fubini products need not be symmetric; order indiscernibility is exactly what their coherence supplies.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 27 1 Solution Created 2026-10-03 Updated 2026-10-14
The Ehrenfeucht-Mostowski theorem says that an infinite first-order structure , after Skolem expansion, has an elementarily equivalent model generated as a Skolem hull by distinct order-indiscernible sequence for any prescribed linear order . Every order automorphism of extends uniquely to an automorphism preserving this Skolem expansion. The reduct still models , but uniqueness of an automorphism in the reduct alone is not asserted. Iterate adjoining witness functions through all successively enlarged languages, so every existential formula in the final language has its own Skolem function. Write for the resulting expanded structure.
For the Ramsey theorem proof, add constants and require that they are distinct and that every two increasing tuples of the same length agree on every formula of the expanded language. Include . A finite fragment mentions finitely many formulas and finitely many indices, say of them. Choose distinct . For each arity occurring in those formulas, color the increasing tuples of natural numbers by their finite truth vectors in . Repeated applications of the infinite Ramsey theorem give an infinite subset homogeneous for all these finitely many colorings. Interpret the constants by its first elements, in their prescribed order. All the fragment's indiscernibility requirements hold. Thus compactness theorem produces a model of the whole scheme.
Let be the Skolem hull of the . Whenever an existential formula with parameters in holds in , its chosen Skolem witness belongs to . The Tarski-Vaught test, proved by the existential step of induction on formulas, therefore makes . For an order automorphism , defineTo compare different term presentations, collect all their indices into one increasing tuple. Indiscernibility transfers the equality of the terms to the transported tuple, so this map is well defined. The same argument transfers every relation; it preserves all functions by construction, and the map for is its inverse. Since every hull element is a term in the generators, uniqueness follows.
For a genuinely ultraproduct proof, keep the distinct and choose a nonprincipal ultrafilter on . For each finite ordered subset , takewhere is the ordered Fubini product of ultrafilters. Explicitly, a subset is large if its indicator is true after successively applying “for -almost every ” in the increasing coordinate order. Projection to any sublist has exactly that sublist's product ultrafilter. Hence for , pulling functions back along gives an elementary embedding by the Łoś theorem. These embeddings commute. Form their directed limit of elementary embeddings; each finite tuple lies in one stage, so induction on formulas makes the stage embeddings elementary.
At a stage containing , let be the class of the coordinate function . If , equality of these coordinate functions is small: for a fixed first coordinate there is only one possible equal second coordinate, and a singleton is not in . Thus . For any increasing , the truth of is the same nested ultrafilter test applied to , independent of the actual indices. This is precisely indiscernibility in the expanded language. Its Skolem hull, and the term-transport construction above, finish the second proof. Both constructions give the required model and the extended index automorphisms.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 25 6 Solution Created 2026-10-03 Updated 2026-10-14
We prove the Ehrenfeucht-Mostowski theorem by ordered products of ultrafilters, without using Ramsey's theorem. Choose a Skolem expansion of the infinite structure and distinct elements . Fix a nonprincipal ultrafilter on .
For a finite increasing subset of the desired index order , define the ordered Fubini product of ultrafilters on byThe notation means that the set of satisfying belongs to ; the leftmost quantifier is outermost. Upward closure and intersection follow successively at each coordinate, and applying complement at each step proves the ultrafilter dichotomy. Products are ordered; no false claim of symmetry is needed.
Put , with . If , pullback of a function by coordinate projection inducesFor a cylinder condition involving only coordinates in , deleting the unused ultrafilter quantifiers makes its membership in exactly its membership in . The Łoś theorem therefore makes an elementary embedding, not merely a homomorphism. The projection identities give .
Take the directed limit of this system: elements from two stages are identified if their images agree at a common larger finite stage, and formulas are evaluated at such a stage. Elementary coherence makes this well-defined. It also makes each stage elementary in the limit: a formula uses finitely many parameters, and an existential witness at a later stage can be pulled back at the level of existential truth because the transition maps are elementary. Denote the limit by .
For each , let be the limit image of the coordinate function . These elements are distinct. For , after fixing , equality holds for at most one , a set not in ; hence equality fails in the two-coordinate ultrapower and therefore in the limit.
For any increasing , Łoś evaluatesThe right-hand side depends on the formula and the number of increasing coordinates, not their particular indices. Thus is order-indiscernible for every formula of the expanded language simultaneously; no preliminary homogeneous thinning is used. If elements of were named before expanding, the same computation gives indiscernibility over those names and the initial stage embeds the original structure into the limit.
The Skolem hull of these generators is elementary by the witness property. Transporting the generators by any order automorphism of transports every term, with well-definedness and relation preservation supplied by the formula equivalence just proved. The inverse transport supplies an inverse automorphism. The ordered-ultrapower construction gives the full Ehrenfeucht–Mostowski model and its order automorphisms without Ramsey's theorem.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 20 8 Solution Created 2026-10-03 Updated 2026-10-14
Let be a full Skolem expansion of an infinite structure , choose distinct for , and extend the cofinite filter to a nonprincipal ultrafilter on . For a finite subset of the desired index order , define the ordered Fubini product of ultrafilters on bywhere means . These nested tests decide complements and preserve intersections by induction on , so is indeed an ultrafilter. For the empty set use the one-point index set and its principal ultrafilter.
Put . If , projection of coordinates inducesA truth test independent of a coordinate is unchanged by its dummy ultrafilter quantifier. Consequently pullback preserves exactly the -large subsets, proving that the map is well defined and injective. The Łoś theorem, applied to every formula rather than just to equality, proves that it is an elementary embedding. Projections compose, so these elementary maps form a coherent directed system.
Take its directed limit of elementary embeddings . Explicitly identify two stage elements when their images agree at a common larger stage, and interpret functions and relations there. For an existential formula true in the limit, its parameters and a witness occur at a common stage; elementarity then pulls truth back to the stage of the original parameters. This verifies that all stage maps into the limit are elementary.
Let be the image in this limit of the coordinate class . These elements are distinct: in , equality would require , whose inner-coordinate section is a singleton and hence not in the nonprincipal . For , the truth of any expanded-language formula on is exactlyThis expression depends on the formula and the tuple length, not on the chosen increasing indices. Thus is an order-indiscernible sequence. The empty-coordinate stage embeds elementarily into .
Their Skolem hull is elementary by the witness-function argument, and an order automorphism of extends to it by transporting the generators in every term. Equality and relation preservation follow from the displayed uniform truth test, and the inverse order automorphism gives the inverse map. Therefore this proves the Ehrenfeucht-Mostowski theorem through ultrapowers and their directed limit, with no Ramsey-theoretic step.
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.
Expand an infinite first-order structure by Skolem functions, choose distinct elements , and fix a nonprincipal ultrafilter on . For each finite subset of the required index order, form the ultrapower over using the ordered Fubini product of ultrafilters. Pullback along coordinate projections gives coherent elementary maps by the Łoś theorem. In their directed limit of elementary embeddings, the classes of the coordinate functions are distinct and form an order-indiscernible sequence; every first-order formula on increasing coordinates has the same nested ultrafilter test. Their Skolem hull gives an Ehrenfeucht-Mostowski model. An index-order automorphism sends a term in the generators to the same term in the transported generators; indiscernibility makes this well defined. Uniqueness is in the specified Skolem expansion, not necessarily in its reduct. If generators must respect a definable infinite order, first produce an increasing sequence by an ultrapower of finite increasing chains.