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One form of the Ehrenfeucht-Mostowski theorem says that, for every infinite first-order structure $M$ and every total order $I$, an elementary extension contains distinct elements $(a_i{)}_{i\in I}$ forming an order-indiscernible sequence. After a suitable Skolem expansion, their Skolem hull is a first-order model of $\mathrm{Th}(M)$, and every order automorphism of $I$ extends to a structure automorphism of that hull. If $M$ has a definable infinite linearly ordered subset $P$, the generators may lie in $P$ 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 $M$ to $M^{+}$ 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 $\mathcal{U}$ on $\mathbb{N}$.

We can obtain an infinite sequence of distinct elements $b_0,b_1,\dots$ in an elementary extension $M_0$ of $M^{+}$ by one preliminary ultrapower. For each $n\ge 1$ choose a finite list of $n$ distinct elements of $M^{+}$. Represent $b_j$ by the function taking the $j$th entry when $n>j$, 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 $P$, and choose each finite list increasingly inside $P$ instead; then $M_0\models P(b_j)$ and $b_j<b_k$ for $j<k$. Infinitude of an ordered set supplies every finite increasing chain. This preparatory step uses no Ramsey theorem.

For a finite subset $F=\{i_1<\cdots <i_r\}$ of $I$, define the ultrafilter ${\mathcal{U}}_F$ on ${\mathbb{N}}^F$ by nested membership, in this order:where $(\mathcal{U}n)Q(n)$ means $\{n:Q(n)\}\in \mathcal{U}$. 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 $F=\varnothing$ take the principal ultrafilter on the one-point product. PutIf $F\subseteq G$, pull a function back along the projection ${\mathbb{N}}^G\to {\mathbb{N}}^F$. A test independent of an omitted coordinate is unchanged by its ultrafilter quantifier, so projection pushes ${\mathcal{U}}_G$ to ${\mathcal{U}}_F$. The induced map $N_F\to N_G$ is therefore an elementary embedding by Łoś theorem. These maps are coherent, giving a directed system indexed by finite subsets of $I$.

Take its directed limit of elementary embeddings $N$. 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 $N_F$ embeds elementarily into $N$, which contains an elementary copy of $M^{+}$.

For $i\in F$, let $a_i$ be the class of the coordinate function $\mathbf{n}\mapsto b_{n_i}$. Projection coherence makes this independent of $F$. If $i<j$, then for each fixed $n_i$ the set $\{n_j:n_j\ne n_i\}$ is cofinite. The nested test therefore gives $a_i\ne a_j$. In the ordered variant, the same argument with $n_j>n_i$ gives $P(a_i)$ and $a_i<a_j$.

For every first-order formula $\phi$ of the expanded language and every increasing tuple $i_1<\cdots <i_r$, 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 $H$ be the Skolem hull of the generators in $N$. The Tarski-Vaught test gives $H\prec N$, and its reduct is a first-order model of $\mathrm{Th}(M)$.

An order automorphism $\pi$ of $I$ 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 ${\pi }^{-1}$. 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.

The Ehrenfeucht-Mostowski theorem says that if a first-order theory $T$ has an infinite model, then for every total order $I$ there is a model $M\models T$ generated as the Skolem hull of distinct order indiscernibles $(a_i{)}_{i\in I}$, and every order automorphism of $I$ extends to an automorphism of $M$.

Given an infinite cardinal $\kappa$, let $I=\kappa \times \mathbb{Q}$ with the lexicographic order, viewed as $\kappa$ consecutive copies of the rational order. In each copy independently choose either the identity or a fixed nonidentity order automorphism of $\mathbb{Q}$. These choices give $2^{\kappa }$ distinct order automorphisms of $I$.

Apply the theorem to this order. Distinct order automorphisms act differently on the distinct generators $a_i$, so their extensions give an injection into the automorphism group of a first-order structure $\mathrm{Aut}(M)$. Therefore $|\mathrm{Aut}(M)|\ge 2^{\kappa }$, proving that $T$ has models with arbitrarily large automorphism groups.

The inductive definition of ordinal addition isfor a limit ordinal $\lambda$.

The synthetic definition takes the order type of the disjoint union of a copy of $\alpha$ followed by a copy of $\beta$: every point of the first copy precedes every point of the second, and each copy retains its original total order.

Fix $\alpha$ and use transfinite induction on $\beta$. The synthetic sum with the empty order is $\alpha$. Appending a greatest element produces the successor ordinal rule. At a limit ordinal $\lambda$, the second copy is the union of its initial segments of types $\beta <\lambda$, so the whole order has type
${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the inductive recursion; uniqueness in transfinite recursion proves the definitions equivalent.

This class is not axiomatisable. It contains an infinite model, for example the total order $\mathbb{Q}$. If a theory $T$ in any fixed expanded language axiomatized it, choose a cardinalThe Upward Lowenheim-Skolem theorem would give a model of $T$ of cardinality at least $\kappa$. Its underlying order cannot be isomorphic to a subset of $\mathbb{R}$, because every such subset has cardinality at most $|\mathbb{R}|$. This contradiction proves the claim.

A well-order is a total order in which every nonempty subset has a least element.