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A model-theoretic ultralimit is the elementary direct limit of a chain of successive ultrapowers along their diagonal embeddings. We will build two such chains with isomorphic limits. This is not a topological ultrafilter limit and does not require the stronger claim that one ultrapower of each structure is already isomorphic.

First prove the Frayne embedding lemma, including the relative form needed to make the diagrams commute. If $C\equiv D$, expand the language by constants naming every element of $D$, and let $\mathrm{ED}(D)$ be its complete elementary diagram. Index by finite subsets $s$ of this diagram. The finitely many constants in $s$ can be interpreted in $C$ so that $s$ holds: existentially quantifying them gives a sentence true in $D$ and hence in $C$. Choose such interpretations, and give unused constants an arbitrary fixed value in the nonempty domain of $C$.

For each diagram sentence $\delta$, its cone $\{s:\delta \in s\}$ has the finite intersection property with the other cones. Extend the generated filter to an ultrafilter $U$. For $d\in D$, let $f_d(s)$ be its chosen interpretation at coordinate $s$. The mapis elementary by Łoś: every formula true of a tuple from $D$, or its negation, is a diagram sentence and holds on its corresponding cone. In particular distinct named elements stay distinct.

If $C⪯D$, the construction can keep every constant naming $c\in C$ equal to that same $c$ at every coordinate. For a finite diagram fragment, existentially quantify just the names outside $C$; the resulting formula with parameters in $C$ is true in $D$, hence true in $C$. Thus the finite realizations still exist with all those old names fixed. ConsequentlyThis relative extension of the diagonal map is the essential strengthening.

Now let $A_0\equiv B_0$. The nonrelative lemma embeds $B_0$ into an ultrapower $A_1$ of $A_0$. Identify the image with a copy of $B_0$ inside $A_1$. Since $B_0⪯A_1$, the relative lemma embeds $A_1$ into an ultrapower $B_1$ of $B_0$ extending its diagonal map. Rename this larger structure so that the displayed copy of $A_1$ is a literal elementary substructure. Next apply the relative lemma over $A_1⪯B_1$ to embed $B_1$ into an ultrapower $A_2$ of $A_1$ fixing its diagonal copy. Continue alternately. We obtain a coherent elementary chainwith $A_0⪯A_1$ as well. Each inclusion $A_n\to A_{n+1}$ is, up to these coherent identifications, the diagonal embedding into its ultrapower; the same holds for $B_n\to B_{n+1}$. It is the relative clause, not merely the existence of arbitrary elementary embeddings, which guarantees this.

By the elementary chain theorem, the unions of the two subsequences are elementary limits. They are the same structure: each $A_n$ lies in a subsequent $B_n$, and each $B_n$ lies in $A_{n+1}$. Hence the alternating ultrapowers give isomorphic ultralimits construction provesConversely each ultrapower and each elementary direct limit has the same first-order theory as its starting structure. Isomorphic ultralimits therefore imply $A_0\equiv B_0$ as well. Thus the conclusion exactly characterizes elementary equivalence at the level of ultralimits.

Let $M^{\ast }$ be a full Skolem expansion of an infinite structure $M$, choose distinct $b_n\in M$ for $n\in \mathbb{N}$, and extend the cofinite filter to a nonprincipal ultrafilter $U$ on $\mathbb{N}$. For a finite subset $F=\{i_1<\cdots <i_r\}$ of the desired index order $I$, define the ordered Fubini product of ultrafilters $U_F$ on ${\mathbb{N}}^F$ bywhere $UnP(n)$ means $\{n:P(n)\}\in U$. These nested tests decide complements and preserve intersections by induction on $r$, so $U_F$ is indeed an ultrafilter. For the empty set use the one-point index set and its principal ultrafilter.

Put $M_F=(M^{\ast }{)}^{{\mathbb{N}}^F}/U_F$. If $F\subseteq G$, projection of coordinates inducesA truth test independent of a coordinate is unchanged by its dummy ultrafilter quantifier. Consequently pullback preserves exactly the $U_F$-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 $N^{\ast }$. 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 $a_i$ be the image in this limit of the coordinate class $[n\mapsto b_n]\in M_{\{i\}}$. These elements are distinct: in $U_{\{i,j\}}$, equality would require $n_i=n_j$, whose inner-coordinate section is a singleton and hence not in the nonprincipal $U$. For $i_1<\cdots <i_r$, the truth of any expanded-language formula on $(a_{i_1},\dots ,a_{i_r})$ is exactlyThis expression depends on the formula and the tuple length, not on the chosen increasing indices. Thus $(a_i{)}_{i\in I}$ is an order-indiscernible sequence. The empty-coordinate stage embeds $M^{\ast }$ elementarily into $N^{\ast }$.

Their Skolem hull is elementary by the witness-function argument, and an order automorphism of $I$ 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.

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

First use the fact that a small set is absent from a complete nonprincipal ultrafilter. Indeed, if $A\subseteq \kappa$ and $|A|<\kappa$, then every $\kappa \setminus \{\alpha \}$, $\alpha \in A$, belongs to $U$ by nonprincipality. Kappa-completeness gives $\kappa \setminus A\in U$, so $A\notin U$. In particular every final segment $\{i:i>\alpha \}$ belongs to $U$.

Suppose for contradiction that the ultrapower has at most $\kappa$ elements. List representatives $f_{\alpha }:\kappa \to M$ for all its classes, indexed by $\alpha <\kappa$; repetitions are allowed. At coordinate $i<\kappa$, fewer than $\kappa$ values occur among $f_{\alpha }(i)$ with $\alpha <i$. Since $|M|\ge \kappa$, chooseThis diagonal argument for ultrapower cardinality uses the axiom of choice. For each fixed $\alpha$, the functions $g$ and $f_{\alpha }$ disagree throughout the final segment $i>\alpha$, which belongs to $U$. Their equality set therefore cannot belong to $U$, and $[g]\ne [f_{\alpha }]$.

This contradicts the assumed enumeration of the ultrapower. HenceThe proof uses only that each ordinal $i<\kappa$ has cardinality below $\kappa$; it does not require a separate assumption of regularity.

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 relation that $X$ is finitely representable in $Y$ means that every finite-dimensional subspace $E\subseteq X$ has, for every $\epsilon >0$, a linear copy in $Y$ of distortion below $1+\epsilon$. A superreflexive Banach space is one for which every Banach space finitely representable in it is a reflexive Banach space.

Suppose first that $X$ is reflexive and $(x_i)\subseteq B_X$. By weak compactness, the sequence has a weak cluster point $x\in B_X$. The point $x$ belongs to the weak closure of every tail convex hull, and the weak and norm closures of a convex set agree by the Hahn-Banach separation theorem. Choose a convex combinationwithin $\theta /2$ of $x$, enlarging $n$ beyond all indices used. Then choosewithin $\theta /2$ of $x$. It follows that $\Vert y-z\Vert <\theta$.

Conversely, the James reflexivity criterion in convex-block form says that a nonreflexive Banach space has a number $\theta >0$ and a sequence $(x_i)\subseteq B_X$ such thatfor every $n$. One obtains this form from the usual bidual separation proof by applying the principle of local reflexivity to each finite-dimensional stage. Such a sequence contradicts the asserted property. This proves the convex-block separation criterion for reflexivity.

Consider now the uniform finite version. If it fails for some $\theta >0$, choose for every $N$ a sequence $(x_i^{(N)}{)}_{i=1}^N\subseteq B_X$ for which every cut has convex-hull distance at least $\theta$. In a free ultrapower $X_{\mathcal{U}}$, letfilling the finitely many missing coordinates arbitrarily. Every initial-tail pair of convex hulls of $({\xi }_i)$ remains at distance at least $\theta$, so the first criterion makes $X_{\mathcal{U}}$ nonreflexive. Since an ultrapower is finitely representable in $X$, this contradicts superreflexivity.

Conversely, if $X$ is not superreflexive, choose a nonreflexive space $Y$ finitely representable in $X$. The first criterion supplies a sequence in $B_Y$ whose convex blocks are separated by some $\theta >0$. For each $N$, transfer the span of its first $N$ vectors to $X$ with distortion arbitrarily close to one and normalize. The resulting $N$-term sequence violates the uniform condition, with separation at least, say, $\theta /2$. This proves the uniform finite convex-block criterion for superreflexivity.

A purely metric equivalent is the diamond-graph characterization of superreflexivity:For sufficiency, argue contrapositively. If $X$ is not superreflexive, the preceding uniform criterion supplies arbitrarily long unit-ball sequences whose convex hulls stay a fixed distance apart across every cut. At each replacement step in the diamond graph, map the two new branches to convex combinations on opposite sides of the corresponding cut. The upper bound follows from convexity and the lower bound from the fixed separation. The standard recursive diamond construction therefore embeds every $D_m$ into $X$ with one distortion constant independent of $m$. Thus divergence of the diamond distortions forces superreflexivity.

An uncountable cardinal $\kappa$ is measurable when it carries a nonprincipal ultrafilter $U$ that is $\kappa$-complete. For an inaccessible $\lambda >\kappa$, the cardinal $\kappa$ is 1-strong when there is an elementary embeddinginto a transitive model, with critical point $\kappa$ and $V_{\kappa +1}\subseteq M$.

The fundamental theorem on measurable cardinals constructs from $U$ the well-founded ultrapower and its ultrapower embedding $j:V_{\lambda }\to M$, whose critical point is $\kappa$. The embedding fixes $V_{\kappa }$. If $A\in V_{\kappa +1}$, then $A\subseteq V_{\kappa }$, and elementarity givesBoth $j(A)$ and $V_{\kappa }$ belong to the transitive target, so $A\in M$. Thus $V_{\kappa +1}\subseteq M$, proving that every measurable cardinal is 1-strong.