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Here an ultralimit in the model-theoretic sense is a model-theoretic ultralimit: start with a first-order structure, take successive ultrapowers along their diagonal elementary embeddings, and take the directed limit of elementary embeddings of that chain. The elementary chain theorem, proved by induction on formulas after identifying the embeddings with inclusions, makes every stage elementary in the limit. This definition concerns structures, rather than the topological ultrafilter limit.

We first prove the needed Frayne embedding lemma. If $C\equiv D$, every finite subset of the elementary diagram of a structure $D$ can be realized in $C$: existentially quantify its finitely many named elements, and use elementary equivalence. Index by finite diagram fragments, choose such realizations, and extend the filter of cones requiring each diagram formula to an ultrafilter. For each $d\in D$, let $f_d$ give its chosen coordinate realizations, filling unspecified coordinates arbitrarily. The Łoś theorem shows that $d\mapsto [f_d]$ is an elementary embedding of $D$ into an ultrapower of $C$. The formulas $c_d\ne c_e$ ensure injectivity.

There is a relative version: if $C⪯D$, keep all named elements from $C$ constant in every coordinate. Each finite diagram fragment is realizable in $C$ over those parameters by elementarity, so the resulting embedding extends the diagonal embedding of $C$ into its ultrapower.

Now use alternating ultrapowers give isomorphic ultralimits. First embed $B$ into an ultrapower $A_1$ of $A$. Apply the relative lemma to $B⪯A_1$, embedding $A_1$ into an ultrapower $B_1$ of $B$ over its diagonal copy. Next embed $B_1$ into an ultrapower $A_2$ of $A_1$ over $A_1$, and continue. After coherent identifications this givesEach $A_{n+1}$ is an ultrapower of $A_n$ along its diagonal embedding, and each $B_{n+1}$ is an ultrapower of $B_n$ likewise. The two subsequences are cofinal in the same chain of elementary substructures, so their limits are the same union. Hence $\boxed{A\equiv B⟹A_{\infty }\cong B_{\infty }}$. No isomorphism theorem for single ultrapowers has been assumed.

The Ehrenfeucht-Mostowski theorem asserts that an infinite first-order structure has, for any prescribed total order $I$, an elementary extension containing distinct order-indiscernibles $(c_i{)}_{i\in I}$. After a Skolem expansion, the Skolem hull of those generators is a model of the original complete theory. Every order automorphism of $I$ extends to a unique automorphism of this specified Skolem hull in the expanded language.

Choose a Skolem expansion $M^{\ast }$ of the original infinite structure: for each existential formula, select a witness function, including nullary functions when needed. Add constants $c_i$ for $i\in I$ and the elementary diagram of a structure of $M^{\ast }$, using separate names for its elements. Require $c_i\ne c_j$ for distinct indices, and for each formula $\phi$ in the Skolem language requirewhenever both index tuples are strictly increasing. These are the order-indiscernible sequence requirements.

Every finite fragment is satisfiable in $M^{\ast }$. Choose a countably infinite sequence of distinct elements of $M^{\ast }$. The fragment mentions only finitely many formulas and arities. Color increasing $k$-tuples of that sequence by the finite vector of truth values of its relevant $k$-ary formulas. The Ramsey theorem supplies an infinite homogeneous subsequence; repeat for the finitely many arities. Assign the finitely many named $c_i$ to distinct elements of the final homogeneous subsequence in increasing index order. The fixed finite part of the elementary diagram of a structure already holds in $M^{\ast }$. By the compactness theorem proved in Question 5, the full set of requirements has a model $N^{\ast }$, giving the required elementary extension and order-indiscernible sequence.

Let $H$ comprise all values of Skolem-language terms in finitely many $c_i$. It is closed under the witness functions. If $N^{\ast }\models \exists x\psi (x,\bar{h})$ for parameters in $H$, the specified witness function supplies a witness in $H$. Induction on formulas therefore gives $H⪯N^{\ast }$; its reduct is a model of $\mathrm{Th}(M)$. For infinite $I$, its cardinality is at most $\max (|I|,|L^{\ast }|,{\aleph }_0)$ and at least $|I|$.

For an order automorphism $\alpha :I\to I$, defineRewrite any finite list of indices in increasing order, allowing a term to repeat or omit variables. The order-indiscernible sequence property transfers every equality between terms and every relation among their values, making this definition well defined and elementary. Applying ${\alpha }^{-1}$ gives its inverse. All of $H$ is generated by the $c_i$ and the named Skolem functions, so an automorphism extending $\alpha$ must have precisely this action. Thus $\boxed{\mathrm{Aut}(I)\hookrightarrow \mathrm{Aut}(H^{\ast })}$, with uniqueness in the Skolem expansion. This uniqueness need not hold in the unexpanded reduct.

An ultrapower $A^I/\mathcal{U}$ comes with its diagonal elementary embedding $a\mapsto [i\mapsto a]$, by the Łoś theorem. A model-theoretic ultralimit is the direct limit of successive ultrapowers along these diagonal embeddings, with elementary unions at limit stages. It is not a topological ultrafilter limit. We prove the needed embedding lemma with parameters and then build the isomorphism.

If $A\equiv B$, name every element $b\in B$ by a constant $c_b$ and let $D(B)$ be its elementary diagram of a structure: all sentences true in this fully named structure. For each finite $F\subseteq D(B)$, the conjunction mentions finitely many constants. Its existential closure in the original language is true in $B$, hence in $A$. Choose an interpretation of those constants in $A$ making $F$ true, and give unmentioned constants arbitrary values. Index these interpretations by $I=[D(B){]}^{<\omega }$. Extend the filter of finite-diagram cones to an ultrafilter $\mathcal{U}$.

Map each $b$ to the class of its coordinate interpretations. Every diagram formula holds on the cone containing it, so the Łoś theorem says this map preserves every formula, including negated equalities between distinct named elements. It is therefore an elementary embeddingThis is the Frayne embedding lemma. Its relative version is crucial: if $A\prec C$, interpret all constants naming elements of $A$ constantly by those same elements. Every finite fragment of the diagram of $C$ with these fixed parameters is realized in $A$, since its existential closure with parameters from $A$ is true in $C$ and elementarity transfers it to $A$. The same construction embeds $C$ into an ultrapower of $A$ while extending exactly the diagonal embedding of $A$.

Now start with $A\equiv B$. Embed $B$ into an ultrapower $A_1$ of $A$. Apply the relative lemma to $B\prec A_1$ to embed $A_1$ into an ultrapower $B_1$ of $B$, extending the diagonal map on $B$. Apply it to $A_1\prec B_1$ to embed $B_1$ into an ultrapower $A_2$ of $A_1$, extending the diagonal map on $A_1$. Continue alternately. Replace structures by isomorphic copies to turn these elementary embeddings into inclusions; the commuting identities supplied by the relative lemma giveIn this common chain the map $A_n\to A_{n+1}$ is its ultrapower diagonal, and so is $B_n\to B_{n+1}$. The initial $A\to A_1$ is also diagonal. The union of the $A$-stages and the union of the $B$-stages are cofinal unions of the same chain, hence are the same structure after these coherent identifications. The elementary chain theorem follows by induction on formulas, with every finite tuple and every existential witness lying in some sufficiently late stage. It shows this union is elementary over all its stages. Thus the two constructed ultralimits are isomorphic.

Conversely every successive ultrapower and elementary union preserves the original complete theory. If ultralimits of $A$ and $B$ are isomorphic, they satisfy the same sentences, and their elementary embeddings of the original structures give $A\equiv B$. ThereforeThis proves Frayne theorem by an explicit alternating construction; it does not assume the stronger theorem about isomorphic single ultrapowers or any cardinal-arithmetic hypothesis.

The compactness theorem states that a set of first-order sentences has a model if every finite subset has a model. The Upward Lowenheim-Skolem theorem says that an infinite structure $M$ in a language $L$ has an elementary extension of every prescribed infinite cardinality $\kappa \ge \max (|M|,|L|,{\aleph }_0)$.

Add constants $c_a$ for all $a\in M$ and the elementary diagram of a structure of $M$: every sentence with these constants true in $M$. Add $\kappa$ further constants $d_i$ and sentences $d_i\ne d_j$ for $i\ne j$. Any finite subset is satisfiable in $M$, by choosing the finitely many new constants distinctly, since $M$ is infinite. Compactness gives a model containing an elementary copy of $M$ and at least $\kappa$ elements.

For the exact cardinality, start with that copy and all the $d_i$, a set of cardinality $\kappa$. For every existential formula with parameters from this set that holds in the large model, choose one witness. Also close under the language's functions. Repeat countably many times. At each stage there are at most $\kappa$ formulas and finite parameter tuples, so the union still has cardinality $\kappa$. Every existential formula true in the large model with parameters in the union has a witness in the union; an induction on formulas proves that this substructure is elementary. This establishes the exact-size statement without assuming an unproved downward theorem.

A dense total order with $a<b$ is not a well-order: the nonempty set $\{x:a<x\}$ has no least element, since density supplies a point strictly between $a$ and any proposed least element.

In the strict-order language, the required axioms areThese axiomatize dense total orders, allowing endpoints. If more than one point is required, add $\exists x\exists y(x<y)$. For a non-strict poset language replace $x<y$ by $x\le y\wedge x\ne y$.

Expand the language by constants for every element of $M$ and a new tuple $c$. Let $\mathrm{ElDiag}(M)$ be the elementary diagram of a structure associated with $M$. Every finite subset ofis realized in the expansion of $M$, because the finitely many formulas from $p$ are simultaneously satisfiable in $M$. The compactness theorem gives a model $N^{\prime }$ of the whole set. The interpretations of the named constants give an elementary embedding $M\to N^{\prime }$, and the interpretation of $c$ realizes $p$. Identifying $M$ with its image produces an elementary extension $M⪯N$ realizing $p$. This is the finitely satisfiable type is realized in an elementary extension argument.