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A Boolean algebra homomorphism $B\to \mathcal{P}(A)$ is equivalent to a function assigning to every $a\in A$ an ultrafilter on $B$: test membership of $a$ in every image subset. This gives a left adjoint valued in the opposite category of sets. Its unit sends an algebra element $b$ to the set of ultrafilters containing $b$. On an infinite power-set algebra this map is not surjective, because the subset consisting of the principal ultrafilters cannot arise this way in the presence of a nonprincipal ultrafilter.

A subset $E$ of a set $X$ is cofinite in $X$ when its complement $X\setminus E$ is a finite set. For infinite $X$, the cofinite subsets form the cofinite filter. A nonprincipal ultrafilter on $\mathbb{N}$ contains every cofinite subset: it contains no singleton, hence no finite set, and decides between each set and its complement.

In a countable language, an ultraproduct over a nonprincipal ultrafilter on $\omega$ realizes every countable finitely satisfiable type, including types over countably many parameters. Enumerate the type ${\phi }_1,{\phi }_2,\dots$, and let $A_n$ be the ultrafilter-large set of coordinates where the first $n$ first-order formulas have a simultaneous witness. Put $B_n=\{i:i\ge n\}\cap {\bigcap }_{j\le n}{A}_j$. At coordinate $i$, choose a witness for the largest $n\le i$ with $i\in B_n$, or an arbitrary element when there is none. For each fixed $n$, the resulting function satisfies the first $n$ first-order formulas on $B_n$. The Łoś theorem then realizes the entire type. Countability of the language and parameter set makes the enumeration possible.

An ultraproduct over a nonprincipal ultrafilter on $\omega$ realizes each countable finitely satisfiable type. For the first $n$ formulas, their coordinate simultaneous-witness set $E_n$ belongs to the ultrafilter by the Łoś theorem. Replace it by $E_n\cap [n,\infty )$ and choose at coordinate $i$ a witness to as many initial formulas as that coordinate permits. This finite number tends to infinity modulo the ultrafilter, so the resulting product class realizes every formula. More generally countably incomplete ultrafilters admit the same descending-set diagonalization. This statement does not assert saturation at every larger cardinal.

If $U$ is a kappa-complete nonprincipal ultrafilter on $\kappa$ and $|M|\ge \kappa$, then $|M^{\kappa }/U|>\kappa$. Otherwise enumerate representatives $f_{\alpha }$ for its classes and choose $g(i)\notin \{f_{\alpha }(i):\alpha <i\}$. Every tail of $\kappa$ belongs to $U$, so $g$ differs modulo $U$ from every representative. The small set is absent from a complete nonprincipal ultrafilter lemma supplies the tail property.

For a prime number $p$, a nonprincipal ultrafilter on $\mathbb{N}$ givesThe Łoś theorem makes $K$ a field of characteristic $p$. For every natural number $m$, almost all factors have at least $m$ elements, so $K$ is infinite.

For a free filter $F$, a subset $S$ belongs to $F$ if and only if every ultrafilter extending $F$ contains $S$. If $S\notin F$, every $A\in F$ meets $I\setminus S$ in an infinite set; otherwise a cofinite restriction of $A$ would be contained in $S$. Adjoining $I\setminus S$ therefore generates a proper free filter. The ultrafilter lemma extends it to a nonprincipal ultrafilter witnessing failure of membership.

A cardinal is omega-measurable if it carries a countably complete ultrafilter that is a nonprincipal ultrafilter. Countable completeness rules out countable underlying sets. Such a measure is weaker than the cardinal's own full completeness requirement, but the least omega-measurable cardinal is measurable.

We take $\mathbb{N}=\{1,2,\dots \}$. An ultrafilter $p$ on $\mathbb{N}$ is a proper filter on a set which decides every subset: it contains $\mathbb{N}$ but not $\varnothing$, is closed under finite intersections and passage to supersets, and contains exactly one of $A$ and $\mathbb{N}\setminus A$ for every $A\subseteq \mathbb{N}$. A principal ultrafilter has the form ${\delta }_n=\{A:n\in A\}$; all other ultrafilters are nonprincipal ultrafilters.

To prove existence, start with the cofinite filter and order its proper filter extensions by inclusion. The union of a chain is again a proper filter on a set: the finitely many sets involved in an intersection all belong to one member of the chain, and no member contains the empty set. Thus Zorn's lemma supplies a maximal proper filter $p$. If $A\notin p$, adjoining $A$ must make the generated filter improper; equivalently, some $B\in p$ has $A\cap B=\varnothing$. Then $B\subseteq \mathbb{N}\setminus A$, so $\mathbb{N}\setminus A\in p$. This proves that $p$ is an ultrafilter. It contains the complement of every finite set and hence no finite set. In particular, it contains no singleton and is nonprincipal.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters with basic open setsWe have $\hat{A}\cap \hat{B}=\widehat{A\cap B}$ and $\beta \mathbb{N}\setminus \hat{A}=\widehat{\mathbb{N}\setminus A}$, so these basic sets define a topology and are clopen sets. If $p\ne q$, choose $A\in p\setminus q$. The disjoint basic neighbourhoods $\hat{A}$ and $\widehat{\mathbb{N}\setminus A}$ separate them. Thus the space is Hausdorff.

For compactness, it suffices to show every basic open cover has a finite subcover, since an arbitrary open cover can be refined by basic open sets. Suppose $\{\widehat{A_i}:i\in I\}$ has no finite subcover. The complements $\mathbb{N}\setminus A_i$ have the finite intersection property: if a finite intersection were empty, the corresponding finite family of $\widehat{A_i}$ would cover every ultrafilter. Their generated proper filter on a set extends to an ultrafilter, by the same Zorn's lemma argument. This ultrafilter contains every $\mathbb{N}\setminus A_i$, and therefore belongs to none of the alleged cover members, a contradiction. We have proved that $\beta \mathbb{N}$ is a compact Hausdorff space. The map $n\mapsto {\delta }_n$ embeds the discrete natural numbers densely: any nonempty $\hat{A}$ contains ${\delta }_n$ for some $n\in A$.

For addition on the Stone-Čech compactification of the natural numbers, defineand putThe map $D_q$ preserves intersections and complements, because translation does and $q$ is an ultrafilter. It also sends $\mathbb{N}$ to $\mathbb{N}$ and the empty set to itself. Consequently the displayed rule defines an ultrafilter $p+q$. On principal points it satisfies ${\delta }_n+{\delta }_m={\delta }_{n+m}$, so it extends ordinary addition.

Write $({\forall }^{p}n)\phi (n)$ to mean that $\{n:\phi (n)\}\in p$. The ultrafilter addition rule is then an iterated quantifier, with its order fixed. Expansion givesOrdinary addition is an associative operation, and the quantifiers are in the same order on both lines. Thus $\boxed{(p+q)+r=p+(q+r)}$. This argument does not interchange ultrafilter quantifiers and does not assert that this extended addition is commutative.

For fixed $q$, the translation $R_q:p\mapsto p+q$ hasa basic clopen set. Therefore addition is continuous in its left variable. This is the left-continuity convention used here: right translations are continuous, giving a left-topological semigroup. Continuity in both variables jointly is not part of the assertion.

Hindman's theorem says that every finite colouring of the positive integers has an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. Assume, as permitted, that the addition just defined has an idempotent ultrafilter $p$, so $p+p=p$. This ultrafilter is nonprincipal: a principal ${\delta }_n$ would satisfy ${\delta }_n+{\delta }_n={\delta }_{2n}\ne {\delta }_n$ for positive $n$. Hence every member of $p$ is infinite and every cofinite set belongs to $p$.

Exactly one cell $A$ of the finite colour partition belongs to $p$. Define its star set bySince $A\in p+p$, we have $D_p(A)\in p$ and therefore $A^{\ast }\in p$. Moreover, for each $n\in A^{\ast }$, the set $A-n$ belongs to $p$. Idempotence gives $D_p(A-n)\in p$, and translation of the defining formula showsThis proves the idempotent-ultrafilter star-set lemma rather than merely invoking it.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_k$ have been chosen with every nonempty finite sum in $A^{\ast }$. Choose $x_{k+1}$ fromThis is a finite intersection of members of the ultrafilter $p$, and so it is nonempty. The new sums are $x_{k+1}$ and $s+x_{k+1}$; all lie in $A^{\ast }$ by the choice. Mathematical induction yields an increasing infinite sequence withThus the assumed idempotent gives the full Hindman theorem, with distinct increasing generators and all their nonempty finite sums of one colour.

The saturated model condition in degree $\kappa$ means that a first-order model $M$ realizes every one-variable complete type over a parameter set $A\subseteq M$ of size less than $\kappa$. Equivalently every finitely satisfiable such complete type is realized, since it can be extended to a complete type. Without a specified degree, saturated usually means $|M|$-saturated. Countable saturation means ${\aleph }_1$-saturation, including complete types over countably many parameters.

Here is a general saturated elementary extension theorem, with proof: for every infinite regular cardinal $\kappa$, every first-order structure $M$ has a $\kappa$-saturated elementary extension. At a stage $M_{\alpha }$, list all complete types over its subsets of size less than $\kappa$. There is a set of them, since their formulas lie in a set-sized language. Add a witness constant for each complete type to the elementary diagram of $M_{\alpha }$, together with every formula in that complete type evaluated at its witness constant. A finite part mentions finitely many complete types and finitely many formulas from each; each finite conjunction is realizable in $M_{\alpha }$, and the witnesses for different complete types are unconstrained relative to each other. Thus the finite part is satisfiable. The compactness theorem produces an elementary extension $M_{\alpha +1}$ realizing all those complete types. At limits take elementary unions, justified by the elementary chain theorem.

Iterate for $\kappa$ stages and put $N={\bigcup }_{\alpha <\kappa }{M}_{\alpha }$. If $A\subseteq N$ has size below $\kappa$, regularity bounds the stages at which its elements enter, so $A\subseteq M_{\alpha }$ for some $\alpha <\kappa$. A complete type over $A$ consistent with $N$ is finitely satisfiable in $M_{\alpha }$: each finite conjunction is an existential formula with parameters from $M_{\alpha }$, and elementarity transfers its truth from $N$. The complete type is therefore realized in $M_{\alpha +1}$. This provesFor a singular desired degree, apply the theorem at a larger regular cardinal. If $M$ is infinite and an uncountable regular $\kappa$ satisfies ${\kappa }^{<\kappa }=\kappa$, with $|L|<\kappa$ and $|M|\le \kappa$, the construction can keep every stage of size $\kappa$. There are then at most $\kappa$ parameter sets and complete types, and downward Löwenheim-Skolem gives that size after each extension; adding $\kappa$ distinct elements initially ensures final size exactly $\kappa$. The resulting model is saturated in the unqualified sense. The cardinal-arithmetic hypothesis belongs to this size refinement, not to the general existence theorem.

An explicit ultraproduct version is useful too. In a countable language let $U$ be a nonprincipal ultrafilter on $\omega$ and $N={\prod }_{i<\omega }{M}_i/U$, where each factor is nonempty. Enumerate a countable finitely satisfiable complete type, including its chosen parameter representatives, as ${\phi }_1(x),{\phi }_2(x),\dots$. By the Łoś theorem, the set $A_n$ of coordinates where the first $n$ formulas have a simultaneous witness belongs to $U$. PutThese are decreasing members of $U$. At coordinate $i$, let $k(i)$ be the greatest $n\le i$ with $i\in B_n$, if there is one. Choose a witness to the first $k(i)$ formulas there, and an arbitrary element when there is none. For fixed $n$, every coordinate in $B_n$ has $k(i)\ge n$, so the chosen function satisfies ${\phi }_n$ on a $U$-large set. Its ultraproduct class realizes the whole complete type. A countable language over a countable parameter set has only countably many formulas, so this proves countable saturation of a nonprincipal ultraproduct over omega.

For NFU, we give the rank-indiscernible construction of an NFU model explicitly. NFU has a unary predicate $S$ for sets, atoms have no members, extensionality applies to sets, and comprehension applies to every stratified formula. A stratification assigns integer types to variables so that equality uses equal types and $x\in y$ requires $\mathrm{type}(y)=\mathrm{type}(x)+1$. A unary set predicate adds no type difference.

Start with the actual set structure $B=(V_{\omega +\omega },\in )$. This structure satisfies extensionality and every pure-membership separation instance: the subset defined in a set $a$ of rank below $\omega +\omega$ still has rank below $\omega +\omega$. We do not assert that $B$ satisfies replacement or all of ZFC. Expand it by Skolem functions. In it use the sequence of rank objects $V_{\omega +n+6}$, $n<\omega$. For two such objects $d<e$ in sequence order, we havewith quantifiers in $B$. The second statement holds because every subset of $V_{\alpha }$ belongs to $V_{\alpha +1}$, and the later rank is at least $\alpha +1$.

Introduce constants $d_i$, $i\in \mathbb{Z}$, and require them to be order indiscernibles in the Skolem language, satisfying these two assertions whenever $i<j$. Also require $V_{\omega +5}\subseteq d_i$ for every $i$; this fixed rank object is first-order definable in $B$. Every finite part of these requirements together with $\mathrm{Th}(B^{\ast })$ is satisfiable: color finite increasing tuples of the original rank sequence by the truth values of the finitely many formulas mentioned, and use Ramsey's theorem to get a homogeneous subsequence. The rank assertions hold for every increasing tuple there. The compactness theorem gives a model of the full theory. In its Skolem hull $N$ generated by the $d_i$, the integer shift extends to an automorphism $j$ with $j(d_i)=d_{i+1}$, by the term-transport proof of the Ehrenfeucht-Mostowski theorem. Thus $N$ still satisfies extensionality and all pure-membership separation instances, andThis is a set-sized first-order model; no satisfaction predicate for a proper-class universe is being assumed.

Externally let $D_i=\{x\in N:N\models x\in d_i\}$. The automorphism bijects $D_i$ onto $D_{i+1}$. Interpret a typed structure with sort $i$ equal to $D_i$. At sort $i+1$, declare $y$ a set exactly when $N\models y\subseteq d_i$; otherwise it is an atom. Define adjacent-sort membership byThe guard is necessary: a typed atom might have some ordinary $N$-members in $D_i$, and those must not become its typed members. Set-extensionality follows from extensionality in $N$, because the ordinary members of a typed set all lie in $D_i$. For any typed formula on sort $i$, its quantifiers can be restricted to the corresponding $d_l$, its set predicates replaced by subset assertions, and its memberships replaced by the guarded relation. This is an ordinary pure-membership formula of $N$ with finitely many parameters. Separation gives its extension $X\subseteq d_i$, and the displayed closure assertion gives $X\in d_{i+1}$. It is therefore a typed set witnessing comprehension. The automorphism shifts these sorted domains and preserves the set flags and guarded memberships.

Collapse the types onto $D_0$. DefineObjects not satisfying $S$ have no $E$-members. If two sets have the same $E$-members, their images under $j$ are subsets of $d_0$ with the same ordinary members, so are equal by extensionality of $N$; injectivity of $j$ gives equality of the original objects.

To verify every stratified comprehension instance, assign a variable $v$ its stratification type $t(v)$ and translate its value $a\in D_0$ to $j^{t(v)}(a)\in D_{t(v)}$. Equality is preserved. An atomic $E(x,y)$ translates to the guarded adjacent-sort membership because $t(y)=t(x)+1$; the unary predicate $S(y)$ translates to $j^{t(y)}(y)\subseteq d_{t(y)-1}$. Induction on formulas, using the domain bijections for quantifiers, preserves truth. If the free variable $x$ has type $t$, typed comprehension produces its extension $X\subseteq d_t$, with $X\in d_{t+1}$. Put $b=j^{-(t+1)}(X)\in D_0$. Then $S(b)$ holds, and for every $a\in D_0$,This is exactly the required NFU set. No formula containing the external automorphism $j$ has been used in a separation or comprehension scheme; $j$ only transports values after the pure formula is formed.

The construction also permits the usual infinity requirement. The ordinary $\omega$ and its successor graph $f$ are definable in $N$, hence fixed by $j$, and lie below $V_{\omega +5}$, so belong to all the required domains. Under $E$, the members of $\omega$ are still its ordinary $N$-members. The new Kuratowski pair satisfiesSince $j(f)=f$, membership of this pair in $f$ under $E$ is equivalent to $⟨x,y{⟩}_N\in f$. Thus $f$ is an internal injection of $\omega$ into itself omitting zero, giving a Dedekind-infinite set.

We have built a model of NFU, with infinity if it is included in the formulation, in ordinary ZFC metatheory:In particular $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{NFU})$. This proof does not assume $\mathrm{Con}(\mathrm{ZFC})$ as an additional internal axiom: its starting structure is the set $V_{\omega +\omega }$, not a model of ZFC.

We take $\mathbb{N}=\{1,2,\dots \}$ and require the common difference of an arithmetic progression to be positive. An ultrafilter $\mathcal{U}$ is a proper filter on a set deciding each subset: exactly one of $B$ and $\mathbb{N}\setminus B$ belongs to $\mathcal{U}$.

For the fixed length $m$, let ${\mathcal{B}}_m$ be the family of sets containing no m-term arithmetic progression. Any acceptable ultrafilter must contain every complement $\mathbb{N}\setminus B$ with $B\in {\mathcal{B}}_m$. These mandatory sets have the finite intersection property. Otherwise finitely many $B_1,\dots ,B_r\in {\mathcal{B}}_m$ would cover $\mathbb{N}$. Colour an integer by the first $i$ for which it belongs to $B_i$. The Van der Waerden theorem gives a monochromatic m-term arithmetic progression, contained in the corresponding $B_i$, a contradiction.

Generate a proper filter from those complements and extend it by the ultrafilter lemma. If some $B\in \mathcal{U}$ were in ${\mathcal{B}}_m$, its complement would also belong to $\mathcal{U}$, forcing the empty set into the filter. ThusNotice that sets avoiding one fixed length need not be closed under finite unions; the argument uses the finite intersection property, not an unsupported ideal claim.

For the unbounded-length version, let $\mathcal{B}$ consist of sets whose arithmetic progression lengths are bounded. Each $B_i\in \mathcal{B}$ fails to contain a progression of some length $m_i$, and hence of every greater length. If finitely many $B_i$ covered $\mathbb{N}$, colour by the first containing index and apply the Van der Waerden theorem at length $M={\max }_i{m}_i$. Again a contradiction results. The complements of all members of $\mathcal{B}$ therefore have the finite intersection property. Extending their generated filter gives an ultrafilter with arithmetic-progression-rich members:All finite sets lie in $\mathcal{B}$, so this ultrafilter contains every cofinite set and is a nonprincipal ultrafilter. The same is true of the fixed-length construction when $m\ge 2$, because no singleton is allowed; for $m=1$ a principal ultrafilter also works. These constructions illustrate ultrafilter selection from a partition-rich family.

The infinite-length answer is no. Colour $n$ by $\lfloor {\log }_{2}n\rfloor \mathrm{mod}2$, so successive intervals $[2^j,2^{j+1})$ alternate colours. An infinite arithmetic progression $a+td$, $t\ge 0$, meets every sufficiently large such interval: its first term at least $2^j$ is less than $2^j+d$, which is at most $2^{j+1}$ once $2^j\ge d$. It therefore has terms of both colours. Neither colour class contains an infinite arithmetic progression, but every ultrafilter contains one of the two colour classes. This dyadic block obstruction to infinite arithmetic progressions rules out the proposed ultrafilter.

The partition relation $\lambda \to (\mu {)}_{\nu }^r$ says that every colouring of the $r$-element subsets of a set of cardinality $\lambda$ into $\nu$ colours has a homogeneous set of cardinality $\mu$. Define the beth numbers over an infinite cardinal $\kappa$ by ${\beth }_0(\kappa )=\kappa$ and ${\beth }_{n+1}(\kappa )=2^{{\beth }_n(\kappa )}$. The Erdős-Rado theorem for finite arities gives the precise boundThus any fixed finite arity and any fixed infinite number of colours admit arbitrarily large infinite monochromatic sets once the starting cardinal is sufficiently large. For pairs, the especially useful case is $(2^{\kappa }{)}^{+}\to ({\kappa }^{+}{)}_{\kappa }^2$. The successor in this bound matters: a binary first-difference colouring on a suitably ordered set of binary sequences supplies the familiar obstruction at $2^{\kappa }$. Also, a theorem for every fixed finite arity is not a theorem homogenizing all arities simultaneously.

The mechanism is to build an end-homogeneous routing tree: each new vertex is routed according to its colours to earlier vertices on its branch. A long branch is end-homogeneous, meaning the colour from a fixed earlier vertex is independent of the later vertex. There are at most ${\kappa }^{\alpha }\le 2^{\kappa }$ routes of length $\alpha <{\kappa }^{+}$, and the whole collection of short routes has size at most $2^{\kappa }$. Inserting $(2^{\kappa }{)}^{+}$ vertices therefore forces a branch of length ${\kappa }^{+}$. One colour occurs at ${\kappa }^{+}$ earlier vertices on that branch, giving the pair theorem. Iterating the higher-arity version of this routing construction explains the successive powers in the Erdős-Rado theorem for finite arities.

A weakly compact cardinal is an uncountable strongly inaccessible cardinal $\kappa$ withEquivalent formulations include the tree property at the inaccessible cardinal and compactness for theories of size at most $\kappa$ in the infinitary language $L_{\kappa ,\kappa }$: if every subtheory of size $<\kappa$ has a first-order model, so does the theory.

A measurable cardinal is an uncountable $\kappa$ carrying a nonprincipal ultrafilter closed under intersections of fewer than $\kappa$ members. Such an ultrafilter contains no set of size $<\kappa$, since all singletons have complements in it and it is kappa-complete. Consequently a cofinal sequence of length $<\kappa$ would partition $\kappa$ into too few small pieces, so $\kappa$ is regular cardinal. If $\lambda <\kappa$ and $\kappa$ distinct subsets of $\lambda$ existed, choose, for each coordinate of $\lambda$, the bit selected by the ultrafilter. Kappa-completeness would make all those bits simultaneously constant on a large set of indices; only one distinct subset could have that bit pattern, a contradiction. Hence $2^{\lambda }<\kappa$: $\kappa$ is also a strong limit cardinal.

A supercompact cardinal $\kappa$ has, for every $\lambda \ge \kappa$, a fine ultrafilter that is a normal ultrafilter on small subsets and is kappa-complete onFine means $\{x:\alpha \in x\}$ is large for every $\alpha <\lambda$. Normal means that a function $f$ with $f(x)\in x$ on a large set is constant on some large set. Equivalently, for every such $\lambda$ there is an elementary embedding $j:V\to M$ with critical point $\kappa$, $j(\kappa )>\lambda$, and $M$ closed under $\lambda$-sequences from the universe.

These notions form the implication chain supercompact $\Rightarrow$ measurable $\Rightarrow$ weakly compact. For the first implication, an embedding as above gives the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$. For the second, use a normal ultrafilter on a cardinal obtained by normalizing its measure. Indeed the ultrapower embedding gives the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$; for a regressive $f$, $j(f)(\kappa )<\kappa$ is a fixed ordinal, yielding the required constant large fibre. Given a pair-colouring, select for each $\alpha$ the measure-one colour on the tail above $\alpha$. One selected colour occurs on a measure-one set of $\alpha$'s. Normality makes the diagonal intersection of their selected tails measure one; any two members of that intersection have the selected colour. This gives the required partition relation, and the preceding regularity and strong-limit argument gives inaccessibility.

An ultraproduct converts coordinatewise finite information into a single first-order structure. Let $(M_i{)}_{i\in I}$ be nonempty structures in a common first-order language, and let $U$ be an ultrafilter on $I$. In ${\prod }_i{M}_i$, setThe ultraproduct ${\prod }_i{M}_i/U$ has these equivalence classes as its elements. Functions are interpreted coordinatewise; a relation holds of classes exactly when it holds on a $U$-large set of indices. Intersecting finitely many large agreement sets shows that these interpretations are independent of representatives.

The Łoś theorem, which explains the construction's usefulness, saysHere is the proof. Induction on terms establishes the coordinatewise interpretation of atomic formulas. Finite intersections handle conjunction, and the ultrafilter property handles negation, since precisely one of a set and its complement is large. For an existential quantification, a witness class gives coordinate witnesses on a large set. Conversely, when the coordinate existential truth set is large, select a witness in each of those structures and a default element elsewhere. Its class is a witness in the ultraproduct. This uses choice. Structural induction on formulas now proves the theorem, with universal quantification obtained by negation.

Compactness. If every finite subset of a first-order theory $T$ has a model, take $I$ to be the finite subsets $\Gamma \subseteq T$ and choose $M_{\Gamma }\models \Gamma$. The cones $\{\Gamma :\Delta \subseteq \Gamma \}$ for finite $\Delta$ have the finite intersection property. Extend their filter to an ultrafilter by the ultrafilter lemma. For each $\phi \in T$, its cone is large, so the Łoś theorem makes the ultraproduct satisfy $\phi$. This proves the compactness theorem, including theories of arbitrary set size.

Elementary extensions and nonstandard elements. An ultrapower uses a fixed structure at every coordinate. The diagonal map sending $a$ to the class of the constant function $a$ is an elementary embedding by the Łoś theorem. For a nonprincipal ultrafilter on $\omega$, the class of $i\mapsto i$ in an ultrapower of the natural numbers exceeds every standard integer, because each corresponding tail is large. In an ultrapower of the real ordered field, the reciprocal of this element is positive and smaller than every positive standard reciprocal integer: an infinitesimal. These examples implement ordinary finite statements in a larger structure without claiming that every external subset or second-order assertion transfers.

Saturation. Such an ultraproduct over $\omega$ is countably saturated. To see the actual diagonal argument, list a finitely satisfiable type ${\phi }_0(x),{\phi }_1(x),\dots$ with parameters in the ultraproduct and fix representatives for those parameters. For each $n$, the set $E_n$ of coordinates admitting a simultaneous witness to its first $n$ formulas belongs to $U$ by the Łoś theorem. Arrange that the $E_n$ decrease and put $D_n=E_n\cap \{i:i\ge n\}$. The $D_n$ decrease, belong to $U$, and have empty intersection. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n$ with $i\in D_n$, or choose a default element if there is no such $n$. For each fixed $n$, $m(i)\ge n$ on the large set $D_n$. Therefore the resulting class realizes every formula. This proves countable saturation of an ultraproduct over omega; it does not assert saturation at all larger cardinalities.

Algebraic limits. Take finite fields ${\mathbb{F}}_p$ indexed by the primes and a nonprincipal ultrafilter. Their ultraproduct is a field of characteristic zero: for every fixed positive integer $n$, only finitely many primes divide $n$, so $n\cdot 1\ne 0$ on a large set. Every first-order sentence true in all finite fields holds in this infinite field, and more generally so does any sentence true for all but finitely many of the factors. This is a way to create an infinite field retaining uniform first-order information from finite fields. It does not make the field algebraically closed: the first-order sentence that every element is a square already fails in infinitely many of the factors.

Large cardinals. A countably complete ultrafilter gives a well-founded membership ultrapower of the universe. If it had an infinite descending chain represented by $f_n$, each setwould be large. Countable completeness makes ${\bigcap }_n{A}_n$ nonempty, giving an actual infinite membership descent at any index in the intersection, contrary to foundation. The Mostowski collapse therefore gives a transitive target and an ultrapower embedding $j:V\to M$.

For a measure on a measurable cardinal $\kappa$, every function $\kappa \to \alpha$ with $\alpha <\kappa$ is constant on a large set: otherwise intersect the fewer than $\kappa$ complements of all its fibres. Consequently $j$ fixes the ordinals below $\kappa$. The class of the identity function is below $j(\kappa )$ and above every constant class below $\kappa$, since all tails are large. Thus $j(\kappa )>\kappa$ and its critical point is $\kappa$. Fine ultrafilters on $P_{\kappa }(\lambda )$ that are normal ultrafilters on small subsets and kappa-complete yield the stronger embeddings witnessing supercompact cardinals.

These applications share a single principle: an ultraproduct preserves precisely the first-order information seen by its ultrafilter. Principal ultrafilters merely select a factor; nonprincipal ones can produce new sizes, elements and saturation. Existence of the needed ultrafilters and selection of coordinate witnesses are set-theoretic assumptions, and changing the ultrafilter may change the resulting model.

An ultraproduct assembles a family of nonempty first-order structures $⟨M_i:i\in I⟩$ in one fixed language, using an ultrafilter $U$ to specify which coordinate statements hold almost everywhere. Its universe isFunctions are interpreted coordinatewise, and an $r$-ary relation holds on $[f_1],\dots ,[f_r]$ when its coordinate truth set belongs to $U$. Changing representatives changes such a truth set only off a $U$-large set, so these definitions are well defined. A principal ultrafilter simply recovers its distinguished factor; a nonprincipal ultrafilter can produce new elements and structures.

The fundamental result is the Łoś theorem:Prove it by structural induction on first-order formulas. Atomic formulas are the definition. Intersections handle conjunction, while exactly one of a set and its complement belongs to an ultrafilter, handling negation. For the existential step, if the coordinate existential truth set is in $U$, use choice to select witnesses there and arbitrary values elsewhere; their product function is an ultraproduct witness. Conversely a product witness makes the coordinate existential truth set contain a $U$-large set. Other connectives and universal quantification follow.

In an ultrapower, all factors are one structure $M$. The diagonal map $a\mapsto [\text{constant }a]$ is an elementary embedding by the Łoś theorem. For a nonprincipal $U$ on $\omega$, the class of $n\mapsto n$ in an ultrapower of the standard natural numbers is larger than every constant natural number, since each tail belongs to $U$. This gives a nonstandard model of Peano arithmetic. Similarly, an ultraproduct of finite fields ${\mathbb{F}}_{p_i}$ with distinct primes tending to infinity has characteristic zero: for every fixed positive integer $m$, the equation $m1=0$ fails in all sufficiently late factors. An infinite structure can therefore arise from finite factors.

Ultraproducts prove the compactness theorem. If each finite subset $\Delta$ of a theory $T$ has a model $M_{\Delta }$, index these models by $[T{]}^{<\omega }$. The cones $\{\Delta :\phi \in \Delta \}$ have the finite intersection property. Extend their filter to an ultrafilter using the ultrafilter lemma. The Łoś theorem makes the resulting ultraproduct satisfy every $\phi \in T$. This construction also gives elementary extensions realizing prescribed finitely satisfiable types.

A useful stronger property is countable saturation of an ultraproduct over omega. Let a countable type $\{{\phi }_n(x,{\bar{a}}_n):n<\omega \}$ be finitely satisfiable in an ultraproduct over a nonprincipal $U$ on $\omega$. Choose coordinate representatives of its parameters. Let $E_n$ be the coordinates where the first $n$ formulas have a simultaneous witness; then $E_n\in U$, the $E_n$ decrease, and we can replace $E_n$ by $E_n\cap [n,\infty )$. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n\le i$ with $i\in E_n$, taking an arbitrary value if there is none. For each fixed $n$, the set where $m(i)\ge n$ is $U$-large. Its class realizes the entire type. This proof uses countable incompleteness of $U$, not arbitrary completeness or automatic saturation at larger cardinals.

Finally, complete ultrafilters connect ultraproducts to large cardinals. A countably complete ultrafilter prevents an externally infinite descending membership chain in an ultrapower of a well-founded universe: intersect the countably many coordinate truth sets and obtain an impossible descending chain at one coordinate. A kappa-complete nonprincipal ultrafilter on $\kappa$ consequently supplies the ultrapower embedding associated with a measurable cardinal. Thus the same construction supports both elementary compactness arguments and large-cardinal embeddings, with the completeness hypothesis deciding which well-foundedness and closure properties survive.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a kappa-complete nonprincipal ultrafilter $U$ on $\kappa$. An elementary embedding $j:M\to N$ preserves all first-order formulas with parameters:For a measurable $\kappa$, the Łoś theorem gives an elementary map from the universe into its ultrapower. Countable completeness makes that ultrapower well founded, so the Mostowski collapse gives a transitive target $M$ and an ultrapower embedding $j:V\to M$. The least moved ordinal, its critical point of an elementary embedding, is $\kappa$: every function into an ordinal below $\kappa$ is constant on a $U$-large set by kappa-completeness, whereas the identity function on $\kappa$ represents an ordinal above every constant ordinal below $\kappa$ and below the image of $\kappa$. Conversely a suitably available embedding with critical point $\kappa$ defines the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$.

The identity $V\to V$ is elementary. In ZFC there is no nonidentity elementary class embedding $j:V\to V$, by the Kunen inconsistency theorem. Here the embedding is a class for which restrictions to sets and the critical sequence are available, as in the usual class-theoretic formulation. The axiom of choice is part of this conclusion; it is not a claimed contradiction for the choice-free Reinhardt-cardinal hypothesis.

Here is the stationary-set proof. A nonidentity $j$ has a least moved ordinal $\kappa$; if all ordinals were fixed, induction on rank would fix every set. Put ${\kappa }_0=\kappa$, ${\kappa }_{n+1}=j({\kappa }_n)$ and $\lambda ={\sup }_{n<\omega }{\kappa }_n$. Since $j$ fixes $\omega$ and shifts this sequence, $j(\lambda )=\lambda$. It therefore fixes the regular successor cardinal $\delta ={\lambda }^{+}$. The stationarity of ordinals of prescribed cofinality gives the stationary setUse the stationary partition at a successor cardinal to split $S$ into $\kappa$ disjoint stationary pieces $⟨S_{\alpha }:\alpha <\kappa ⟩$, merging pieces if necessary. Applying $j$ gives a partition $⟨T_{\alpha }:\alpha <j(\kappa )⟩$ of the same $S$ into stationary pieces. The additional piece $T_{\kappa }$ is stationary.

The closure pointsform a club set. For unboundedness, repeatedly close a bound under the values of $j$ below it and take the supremum of the resulting countable sequence; regularity of $\delta$ keeps it below $\delta$. Closure under limits is immediate. Choose $\beta \in C\cap T_{\kappa }$ and an increasing sequence $⟨{\beta }_n:n<\omega ⟩$ cofinal in $\beta$. Thenwhile every ordinal satisfies $j(\beta )\ge \beta$, so $j(\beta )=\beta$. There is an $\alpha <\kappa$ with $\beta \in S_{\alpha }$. Since $j(\alpha )=\alpha$, elementarity gives $\beta =j(\beta )\in j(S_{\alpha })=T_{\alpha }$, contrary to disjointness from $T_{\kappa }$.

For a Boolean algebra $B$, let $U(B)$ be its set of ultrafilters, equivalently the Boolean homomorphisms $B\to 2$. Regard $U(B)$ as an object of ${\mathbf{Set}}^{\mathrm{op}}$. A homomorphism $k:B\to B^{\prime }$ acts by inverse image of ultrafilters, a set map $U(B^{\prime })\to U(B)$, hence an arrow $U(B)\to U(B^{\prime })$ in the opposite category.

A Boolean homomorphism $h:B\to \mathcal{P}(A)$ gives the functionConversely, a function $v:A\to U(B)$ gives $h(b)=\{a:b\in v(a)\}$. The ultrafilter laws are exactly what makes this map preserve the Boolean operations, including complements and $0,1$. These constructions are inverse and natural, proving the Boolean ultrafilter-power-set adjunctionIts unit is ${\eta }_B(b)=\{u\in U(B):b\in u\}$.

Take $B=\mathcal{P}(\mathbb{N})$, an object in the image of the right adjoint. There is a nonprincipal ultrafilter on $\mathbb{N}$, by extending the cofinite filter in the usual choice setting. The subset $P\subset U(B)$ consisting of all principal ultrafilters is not ${\eta }_B(b)$ for any $b\subseteq \mathbb{N}$: containing the principal ultrafilter at every integer forces $b=\mathbb{N}$, but ${\eta }_B(\mathbb{N})=U(B)$ also contains the nonprincipal ultrafilters. Thus ${\eta }_B$ is not surjective and is not an isomorphism. By the necessary condition proved in the root solution, this adjunction is not idempotent.

We prove the Ehrenfeucht-Mostowski theorem by ultraproducts, with no application of Ramsey theorem. Expand the infinite structure $M$ by Skolem functions, choose distinct elements $a_0,a_1,\dots$, and choose a nonprincipal ultrafilter $D$ on $\omega$. Let $I$ be the desired total index order and put $J={\omega }^I$.

For every finite ordered subset $F=\{i_1<\cdots <i_n\}$, define an ultrafilter $D_F$ on ${\omega }^F$ by the ordered Fubini product of ultrafilters: a set $B$ belongs to $D_F$ exactly whenHere $(Dm)\psi (m)$ means $\{m:\psi (m)\}\in D$, 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 $F\subseteq H$, the inverse image of $B\subseteq {\omega }^F$ under coordinate projection belongs to $D_H$ exactly when $B\in D_F$: 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 $J$, 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 $U$ on all subsets of $J$. Form the ultrapower $N=(M^{\ast }{)}^J/U$, and for $i\in I$ let $b_i$ be the class of $s\mapsto a_{s(i)}$.

If $i\ne j$, 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 $b_i$ are distinct. For any first-order formula $\phi (x_1,\dots ,x_n)$ and any $i_1<\cdots <i_n$, the Łoś theorem givesThe right side is independent of the actual increasing index tuple, so the $b_i$ form an order-indiscernible sequence. Constant functions embed $M^{\ast }$ elementarily into $N$. 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.

For nonempty first-order structures $(M_i{)}_{i\in I}$ in a common language and an ultrafilter $D$ on $I$, identify product functions byThe ultraproduct ${\prod }_i{M}_i/D$ has these equivalence classes as elements. Functions are interpreted coordinatewise, and a relation holds of classes exactly when its coordinatewise truth set belongs to $D$. Closure under finite intersections makes these interpretations independent of representatives.

The Łoś theorem assertsProve it by induction on first-order formulas. Atomic first-order formulas hold by the definitions and induction on terms. Conjunction uses intersections, and negation uses the fact that an ultrafilter contains exactly one of a set and its complement. For an existential first-order formula, a product witness gives a coordinate witness on a $D$-large set. Conversely, on the $D$-large set where a coordinate witness exists, choose one in each factor, and choose arbitrary values elsewhere. Its class is a product witness by the induction hypothesis. This is the only witness-selection step; the surrounding argument is made in a choice metatheory. The special case with all factors equal is an ultrapower, and constant functions give an elementary embedding.

Here is a direct compactness theorem proof. Suppose every finite subset of a theory $T$ has a model. Index factors by the finite subsets $i\subseteq T$, choosing $M_i\models i$. For finite $F\subseteq T$, the cone $C_F=\{i:F\subseteq i\}$ is nonempty, and $C_F\cap C_H=C_{F\cup H}$. Extend the generated proper filter to an ultrafilter $D$. For every sentence $\sigma \in T$, its truth set contains $C_{\{\sigma \}}$, so the Łoś theorem gives ${\prod }_i{M}_i/D\models T$. Finite satisfiability therefore implies satisfiability, directly from the ultraproduct construction rather than from a syntactic completeness argument.

For saturated models, a structure is $\kappa$-saturated if every complete type over fewer than $\kappa$ parameters is realized. A useful concrete case is countable saturation of a nonprincipal ultraproduct over omega. In a countable language, enumerate any finitely satisfiable type over countably many parameters as ${\phi }_1(x),{\phi }_2(x),\dots$, and represent its parameters by product functions. By the Łoś theorem, the set $A_n$ of coordinates where the first $n$ first-order formulas have a simultaneous witness belongs to the nonprincipal ultrafilter $D$. PutThese are decreasing $D$-large sets. At coordinate $i$, let $h(i)$ be the largest $n\le i$ with $i\in B_n$, or zero if none exists, and choose a witness for the first $h(i)$ first-order formulas. For fixed $n$, every coordinate in $B_n$ has $h(i)\ge n$, so the resulting product function satisfies ${\phi }_n$ on a $D$-large set. Its class realizes the whole type. Thus such an ultraproduct is ${\aleph }_1$-saturated.

There is also an existence construction at every prescribed degree $\kappa$. Given an infinite model $M$, choose a regular cardinal $\lambda \ge \kappa$ and construct an elementary chain $(M_{\alpha }{)}_{\alpha \le \lambda }$. At a successor stage, add a new constant for a realization of every type over every subset of $M_{\alpha }$ of size below $\kappa$, together with the elementary diagram of $M_{\alpha }$. Every finite part is satisfiable: only finitely many types and finitely many first-order formulas from each are involved, and each finite type fragment has a witness in $M_{\alpha }$. Compactness therefore supplies a simultaneous elementary extension. At limits take unions. To justify elementarity of a union, any existential first-order formula with parameters in an earlier stage has a witness in that stage whenever it has one at a later stage, by elementarity; this is the Tarski-Vaught test.

Any fewer-than-$\kappa$ parameters of $M_{\lambda }$ occur together in some $M_{\alpha }$, by regularity of $\lambda$. A type over them which is finitely satisfiable in $M_{\lambda }$ is finitely satisfiable in $M_{\alpha }$, again by elementarity, and was realized in $M_{\alpha +1}$. Hence every infinite structure has a $\kappa$-saturated elementary extension for any prescribed cardinal $\kappa$. No bound on the size of this extension is asserted; full saturation at the model's own cardinality has additional cardinal-arithmetic issues.

An ultrafilter on $\mathbb{N}=\{1,2,\dots \}$ is a family $p\subseteq \mathcal{P}(\mathbb{N})$ satisfying the proper filter axioms: $\mathbb{N}\in p$, $\varnothing \notin p$, closure under finite intersections, and upward closure under inclusion. In addition, for every $A\subseteq \mathbb{N}$, exactly one of $A$ and $A^c$ belongs to $p$. Equivalently, an ultrafilter is a maximal proper filter. The principal ultrafilter at $n$ consists of all subsets containing $n$; an ultrafilter not of this form is a nonprincipal ultrafilter.

To prove existence, start with the cofinite filter and order the proper filters containing it by inclusion. The union of a chain is still a proper filter: finitely many of its members lie in one filter of the chain, and no constituent contains the empty set. Thus Zorn lemma supplies a maximal such filter $p$. If $A\notin p$, adjoining $A$ cannot generate a proper filter; otherwise maximality fails. Consequently some $B\in p$ has $B\cap A=\varnothing$, so $A^c\in p$ by upward closure. This proves the ultrafilter alternative. No finite $A$ can belong to $p$, because $A^c$ is cofinite and already belongs to $p$. In particular $p$ contains no singleton and is a nonprincipal ultrafilter. The same argument, starting from any proper filter, proves the ultrafilter lemma used below.

Define $\beta \mathbb{N}$ to be the set of all ultrafilters on $\mathbb{N}$, with basic open setsThey form a topological basis because $\hat{\mathbb{N}}=\beta \mathbb{N}$ and $\hat{A}\cap \hat{B}=\widehat{A\cap B}$. Moreover $(\hat{A}{)}^c=\widehat{A^c}$, so they are clopen sets. This is the compact Hausdorff topology on ultrafilters. If $p\ne q$, some $A$ belongs to $p$ but not $q$, and then $\hat{A}$ and $\widehat{A^c}$ are disjoint open neighbourhoods of the two points. Thus $\beta \mathbb{N}$ is Hausdorff.

For compactness, take an open cover and refine it by basic open sets $\widehat{A_{\lambda }}$. If no finite subfamily covers, then for every finite collection of indices the intersection $\bigcap A_{\lambda }^c$ is nonempty. Indeed, if that intersection were empty, the corresponding $A_{\lambda }$ would cover $\mathbb{N}$, and every ultrafilter would contain at least one of them by the finite ultrafilter alternative. The complements therefore have the finite intersection property and generate a proper filter. Extend it to an ultrafilter $q$ by the proved ultrafilter lemma. Then $A_{\lambda }^c\in q$ for every $\lambda$, so $q$ belongs to none of the covering $\widehat{A_{\lambda }}$, a contradiction. A finite basic subcover lies inside a finite subcover of the original cover. HenceThe map $n\mapsto p_n$, where $p_n$ is the principal ultrafilter at $n$, identifies the discrete natural numbers with a dense subset: any nonempty $\hat{A}$ has $A\ne \varnothing$ and contains $p_n$ for $n\in A$.

Hindman theorem states that every finite colouring of the positive integers admits an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. We now give the Idempotent-ultrafilter proof of Hindman's theorem. For $A\subseteq \mathbb{N}$, write $A-x=\{y\in \mathbb{N}:x+y\in A\}$. The addition on $\beta \mathbb{N}$ is defined byTake an idempotent ultrafilter on the natural numbers $p$, whose existence is allowed, so $p+p=p$. It is a nonprincipal ultrafilter: addition of principal points is $p_m+p_n=p_{m+n}$, whereas no positive integer satisfies $2m=m$.

Exactly one colour class $A$ belongs to $p$. DefineBy idempotence, $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma needed for the recursion is that $A^{\ast }-x\in p$ whenever $x\in A^{\ast }$. To prove it, put $D=A-x\in p$. Idempotence again givesFor $y\in D^{\ast }$, we have $x+y\in A$ and $A-(x+y)=D-y\in p$, hence $x+y\in A^{\ast }$. Thus $D^{\ast }\subseteq A^{\ast }-x$, proving the lemma by upward closure.

Choose $x_1\in A^{\ast }$. Suppose $F_n=FS(x_1,\dots ,x_n)\subseteq A^{\ast }$ has been constructed. Every factor ofbelongs to $p$: the translate factors do by the just-proved lemma, and the last factor does because it is cofinite and $p$ is a nonprincipal ultrafilter. The finite intersection is therefore nonempty. Choose $x_{n+1}$ in it. Then $x_{n+1}$ and every $s+x_{n+1}$ with $s\in F_n$ lie in $A^{\ast }$, so $F_{n+1}\subseteq A^{\ast }$. This constructs a strictly increasing sequence, indeed $x_{n+1}>{\sum }_{i\le n}{x}_i$, and proves Hindman theorem.

Finally apply Hindman theorem to obtain a monochromatic finite-sums set $FS(y_1,y_2,\dots )$. We construct a divisibility chain inside a finite-sums set using disjoint successive blocks of indices. Start with $x_1=y_1$. If $x_i=m$, take $2m$ unused terms after all indices used so far, and split them into two groups of $m$ consecutive terms. In each group its $m+1$ partial sums, including $0$, contain two in the same congruence class modulo $m$ by the pigeonhole principle. Their difference is a nonempty consecutive sum divisible by $m$, and positivity makes it at least $m$. Let $x_{i+1}$ be the sum of these two selected sums. Then $m\mid x_{i+1}$ and $x_{i+1}\ge 2m>m$.

The index set defining $x_{i+1}$ lies strictly after all previous index sets. Discard the other terms in these two groups before continuing. Because the chosen index sets are disjoint, every finite sum of distinct $x_i$ is a finite sum of distinct $y_j$, retaining the same colour. Therefore