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Let $E={\bigcup }_{j\ge 1}{B}_{2j}$. For every even index $2j$, $B_{2j}\subseteq E$ and $B_{2j}\in U_{2j}$ imply $E\in U_{2j}$. For every odd index $2j-1$, disjointness gives $B_{2j-1}\subseteq E^c$, so $E^c\in U_{2j-1}$ and $E\notin U_{2j-1}$. ConsequentlyThe limit ultrafilter $U$ contains either $E$ or $E^c$. Whichever it contains, the corresponding basic clopen set $\hat{E}$ or $\widehat{E^c}$ is a neighbourhood of $U$, but infinitely many sequence terms lie outside it. This contradicts convergence. Therefore pairwise distinct ultrafilters cannot form a convergent sequence in $\beta \mathbb{N}$, the key step in no nontrivial convergent sequences in the Stone-Čech compactification.

Use the isolating sets from the preceding part and make the finite disjointificationFor $j<n$, $A_j\notin U_n$, so $A_j^c\in U_n$. Thereforeby finite-intersection closure of the ultrafilter. If $m\ne n$, $B_n\subseteq A_n$ and $A_n\notin U_m$ imply $B_n\notin U_m$. Also $B_n$ avoids every earlier $A_j$, hence every earlier $B_j$. ThusOnly finite intersections were used; no countable-intersection closure of an ultrafilter is being assumed.

The ultrafilter space. An ultrafilter $U$ on $\mathbb{N}$ is a proper filter on a set: it is upward closed, closed under finite intersections, excludes $\varnothing$, and contains exactly one of $A,A^c$ for each $A\subseteq \mathbb{N}$. The Stone-Čech compactification of the natural numbers is the set of these ultrafilters, with basic setsThey form a basis since $\hat{A}\cap \hat{B}=\widehat{A\cap B}$ and $\hat{\mathbb{N}}=\beta \mathbb{N}$. Moreover $(\hat{A}{)}^c=\widehat{A^c}$, so each is a clopen set. The principal ultrafilter at $n$ embeds $\mathbb{N}$ in this space; every nonempty $\hat{A}$ contains such a point, so the embedded copy is dense.

Distinct ultrafilters $U,V$ have some $A\in U$ with $A\notin V$. Then $A^c\in V$, and $\hat{A},\widehat{A^c}$ are disjoint open sets separating them. Thus the space is Hausdorff.

For compactness, any open set cover can be refined to a cover by basic sets. If $\{\widehat{A_i}\}$ has no finite subcover, the sets $A_i^c$ have the finite intersection property: otherwise finitely many $A_i$ cover $\mathbb{N}$, and their basic sets cover every ultrafilter. Extend the proper filter on a set generated by these complements to an ultrafilter $U$. This extension follows from Zorn lemma: unions of chains of proper filters remain proper, and a maximal proper filter decides every $A$ versus $A^c$. Indeed, if adding $A$ made the filter improper, some existing filter member would be disjoint from $A$, forcing $A^c$ into the filter. This $U$ contains every $A_i^c$ and misses every $\widehat{A_i}$, a contradiction. HenceThis proves the compact Hausdorff topology on ultrafilters directly.

An idempotent. The addition on the Stone-Čech compactification of the natural numbers can be defined byAssociativity is supplied. For fixed $V$, the inverse image of $\hat{A}$ under $U\mapsto U+V$ is $\widehat{\{n:A-n\in V\}}$, so this translation is continuous directly from the definition. This makes a left-topological semigroup in the convention used here.

We prove the Ellis–Numakura lemma in this case. Among nonempty compact subsemigroups choose one, $K$, minimal by inclusion. It exists by Zorn lemma: a descending chain has nonempty intersection by compactness, and that intersection is again a compact subsemigroup. For $a\in K$, the set $K+a$ is a nonempty compact subsemigroup contained in $K$. Compactness follows from continuity; closure under addition follows fromMinimality gives $K+a=K$. Hence there is $b\in K$ with $b+a=a$. The setis nonempty, compact, and a subsemigroup: if $x,y\in L$, then $(x+y)+a=x+(y+a)=x+a=a$. Thus minimality gives $L=K$, and in particular $a+a=a$. ThereforeThis is an idempotent ultrafilter on the natural numbers.

Isolating the sequence terms. Now use the hypothetical convergent sequence of distinct ultrafilters. For each $n$, choose $C_n\in U_n$ with $C_n^c\in U$, possible since $U_n\ne U$. Convergence implies $C_n^c\in U_m$ for all sufficiently large $m$. Thus only finitely many indices $m\ne n$ have $C_n\in U_m$. Call this finite set $E_n$.

For each $m\in E_n$, choose $D_{n,m}\in U_n$ with $D_{n,m}\notin U_m$, using distinctness. SetFinite-intersection closure puts $A_n$ in $U_n$. For $m\ne n$, either $C_n\notin U_m$ or one of the chosen $D_{n,m}$ is absent from $U_m$; upward closure then prevents $A_n$ from belonging to $U_m$. Hence

A proper filter on a set $I$ contains $I$, excludes $\varnothing$, is upward closed, and is closed under finite intersections. Order its proper extensions by inclusion. The union of a nonempty chain is again a proper filter: finitely many members lie together in one chain member. Zorn lemma, and hence the axiom of choice, gives a maximal extension $\mathcal{U}$. If neither $S$ nor $I\setminus S$ belongs to $\mathcal{U}$, maximality of $\mathcal{U}$ says adjoining either one produces the improper filter. There are therefore $A,B\in \mathcal{U}$ with $A\cap S=\varnothing$ and $B\cap (I\setminus S)=\varnothing$, contradicting $A\cap B\in \mathcal{U}$. Thus $\mathcal{U}$ is an ultrafilter.

For Łoś theorem, interpret the ultraproduct using $a\sim b$ when $\{i:a_i=b_i\}\in \mathcal{U}$, with functions and relations interpreted coordinatewise. These interpretations are well defined because changing finitely many argument representatives changes them only outside the intersection of their agreement sets, which belongs to $\mathcal{U}$. For any first-order formula and sequences of parameters,Prove this by structural induction. Atomic formulas follow from the interpretation of terms. Conjunction uses finite intersections; negation uses the ultrafilter dichotomy between a set and its complement. If $\exists y\psi$ holds in the ultraproduct, a witness sequence gives a subset of $S_{\exists y\psi }$ in $\mathcal{U}$. Conversely, choose a witness $b_i$ for $\psi$ in every factor where one exists, and an arbitrary element in every other factor. This is the second explicit use of the axiom of choice; all factors have nonempty carriers. The inductive hypothesis applied to $\psi (\mathbf{a},\mathbf{b})$ supplies the ultraproduct witness. The other connectives and universal quantification follow from these cases.

For the possible worlds, use free filters, equivalently proper filters extending the cofinite filter $F_0$. Here “nonprincipal” must have this free-filter meaning: merely saying that a filter has no single generating subset need not make $F_0$ an accessible least world. The designated world is $F_0$, and all worlds have carrier $D={\prod }_i{M}_i$. Atoms, including equality, are forced byEquality may identify distinct sequences at a world; it is a persistent congruence, and at an ultrafilter world its quotient is the ordinary ultraproduct.

The required equivalence with all ultraproducts uses dense-extension filter semantics. Its recursive clauses areEvery extension in these clauses is a free proper filter. The dense clause for disjunction is essential; for propositional formulas, local disjunction would instead give a Kripke model for intuitionistic propositional logic and would invalidate the claimed equivalence.

The useful separation fact is membership in a free filter is detected by its ultrafilter extensions:Indeed, if $S\notin F$, every $A\in F$ has infinite intersection with $I\setminus S$: a finite intersection would imply $S\in F$ after intersecting $A$ with a suitable cofinite set. Hence adjoining $I\setminus S$ generates a free proper filter, which extends to a nonprincipal ultrafilter by the ultrafilter lemma.

Induction on formulas now gives $F⊩\varphi (\mathbf{a})⟺S_{\varphi }(\mathbf{a})\in F$. Negation and implication follow by adjoining the positive set on which their classical truth conditions fail. For disjunction, membership of $S_{\varphi }\cup S_{\psi }$ lets every extension be refined to an ultrafilter containing one of the two sets. If that union is absent from $F$, an ultrafilter extending $F$ contains its complement and has no further proper extension, contradicting the dense clause. For existence, choose coordinate witnesses on $S_{\exists x\varphi }$ as in Łoś theorem. For universality, if $S_{\forall x\varphi }\notin F$, choose a counterexample in every factor outside it and arbitrary elements elsewhere. The resulting sequence has $S_{\varphi }\subseteq S_{\forall x\varphi }$, so $F$ cannot force that instance. The converse follows from upward closure and persistence.

At the designated world, therefore,This semantics validates classical first-order logic, including law of excluded middle. For example, every free filter has an ultrafilter refinement deciding an atomic formula, so its excluded middle is forced even when neither disjunct is forced at the original world. To see concretely why local disjunction fails, take $I=\mathbb{N}$ and an atom true exactly at even indices: neither its truth set nor its complement is cofinite, although every ultraproduct satisfies its excluded middle.

Yes. Order the proper filters containing the cofinite filter by inclusion. The union of every chain is again a proper filter containing $\mathcal{F}$: any finite collection of its members already lies in one filter in the chain. Zorn lemma gives a maximal proper filter $\mathcal{U}$.

A maximal proper filter is an ultrafilter. Indeed, if $A\notin \mathcal{U}$, adjoining $A$ cannot generate another proper filter, so some $U\in \mathcal{U}$ has $U\cap A=\varnothing$. Thus $U\subseteq A^c$, and upward closure gives $A^c\in \mathcal{U}$. Finally, $\mathcal{U}$ cannot equal any principal ultrafilter ${\mathcal{U}}_n$, because it contains the cofinite set $\mathbb{N}\setminus \{n\}$ whereas ${\mathcal{U}}_n$ does not. Thus $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$.

The cofinite filter $\mathcal{F}$ is nonempty, because $\mathbb{N}\in \mathcal{F}$; finite unions of finite complements show closure under intersections, and taking a superset can only shrink the complement. The empty set is excluded because its complement is infinite. It is not an ultrafilter: the even numbers and the odd numbers both have infinite complements, so neither belongs to $\mathcal{F}$. Finally, $\mathbb{N}\setminus \{n\}$ is cofinite but does not belong to ${\mathcal{U}}_n$, proving $\mathcal{F}⊈{\mathcal{U}}_n$ for every $n$.

Assume condition ii. In a finite coloringan ultrafilter contains exactly one color class: at least one must belong to it because their union is $\mathbb{N}$, and two disjoint classes cannot both belong to a proper filter. The chosen $C_i$ contains a member of $\mathcal{S}$, which is therefore monochromatic. This proves condition i.

A filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes the empty set, is upward closed, and is closed under finite intersections. An ultrafilter is a proper filter that contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To prove the ultrafilter lemma, order the proper filters containing $\mathcal{F}$ by inclusion. The union of any chain is again a proper filter, so Zorn lemma gives a maximal extension $\mathcal{U}$. If neither $A$ nor $X\setminus A$ belonged to $\mathcal{U}$, adjoining either one would generate an improper filter. There would then be $U,V\in \mathcal{U}$ with $U\cap A=\varnothing$ and $V\cap (X\setminus A)=\varnothing$. But $U\cap V=\varnothing$, contradicting propriety. Hence $\mathcal{U}$ is an ultrafilter.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters on $\mathbb{N}$, with basic setsThe identitiesshow that these sets form a basis of clopen sets. Distinct ultrafilters disagree on some $A$; one lies in $\overline{A}$ and the other in the disjoint set $\overline{\mathbb{N}\setminus A}$. Thus $\beta \mathbb{N}$ is a Hausdorff space.

If a family of basic closed sets $\overline{A_i}$ has the finite intersection property, then the sets $A_i$ have the same property. They generate a proper filter, which extends to an ultrafilter lying in every $\overline{A_i}$. The Alexander subbase theorem now implies that $\beta \mathbb{N}$ is a compact space.

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on the measurable cardinal $\kappa$, and fix $\lambda <\kappa$. Suppose for a contradiction that $2^{\lambda }\ge \kappa$. Choose distinct subsets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$. For each $\xi <\lambda$, the ultrafilter property chooses exactly one ofas a member $H_{\xi }$ of $U$. Since $\lambda <\kappa$, $\kappa$-completeness gives $H={\bigcap }_{\xi <\lambda }{H}_{\xi }\in U$.

Any two indices in $H$ give subsets having the same membership decision at every $\xi <\lambda$, so they give the same $A_{\alpha }$. The chosen subsets were distinct, hence $|H|\le 1$. This contradicts the fact that a small set is absent from a complete nonprincipal ultrafilter. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, which is precisely the strong limit cardinal condition. This is the measurable cardinal is a strong limit cardinal argument.

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$ satisfyingIf $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ is an adjunction with unit $\eta$ and counit $\epsilon$, thendefines the monad induced by an adjunction. The two triangle identities give the unit laws, while naturality of $\epsilon$ gives associativity.

Let $\mathcal{D}$ now be a full subcategory of $[\mathcal{C},\mathcal{C}]$ that contains the identity endofunctor and is closed under composition, and suppose $T$ is terminal in $\mathcal{D}$. There is exactly one natural transformationand exactly oneBoth sides of either unit law are endomorphisms of the terminal object $T$, so they equal $1_T$; both sides of associativity are maps $T^3\to T$, so they are equal as well. This gives a monad, and terminality also makes both structure maps unique. This is the monad structure on a terminal endofunctor.

For a set $A$, let $\mathcal{U}(A)$ be the set of ultrafilters on $A$. A function $f:A\to A^{\prime }$ induces the pushforwardwhich makes $\mathcal{U}$ the ultrafilter functor. There is no ultrafilter on the empty set. Moreover, every ultrafilter on $A\bigsqcup B$ contains exactly one of the complementary summands $A$ and $B$, and restriction gives a unique ultrafilter on that summand. Thereforeso $\mathcal{U}$ preserves finite coproducts.

Let $F:\mathbf{Set}\to \mathbf{Set}$ preserve finite coproducts. For $x\in F(A)$ defineThe decomposition $A=B\bigsqcup (A\setminus B)$ and preservation of coproducts say that $x$ lies in exactly one of the two corresponding images. Thus exactly one of $B$ and its complement belongs to ${\Phi }_A(x)$. Upward closure follows by factoring subset inclusions. If $B,C\in {\Phi }_A(x)$, decompose $A$ into the four disjoint Boolean cells determined by $B$ and $C$. The unique cell containing $x$ must lie inside both sets, so $B\cap C\in {\Phi }_A(x)$. Hence ${\Phi }_A(x)$ is an ultrafilter.

For $f:A\to A^{\prime }$, the decompositionsshow thatThus $\Phi :F\Rightarrow \mathcal{U}$ is natural.

It is the only such natural transformation. For $B\subseteq A$, let ${\chi }_B:A\to \{0,1\}$ be its characteristic function. Sinceand the two ultrafilters on $\{0,1\}$ are the principal ones, naturality with the two singleton inclusions forces any transformation $F\Rightarrow \mathcal{U}$ to send each summand to the corresponding principal ultrafilter. Naturality with ${\chi }_B$ then says that $B$ belongs to the image ultrafilter exactly when $x$ lies in the image of $F(B)\to F(A)$. Hence the transformation must be $\Phi$.

The ultrafilter functor is therefore the terminal finite-coproduct-preserving set endofunctor. The preceding terminal-object argument supplies its unique ultrafilter monad structure.

Let $\mathcal{U}$ be a nonprincipal ultrafilter on an infinite set $I$. It contains no finite set, so a finite $X\subseteq I$ does not belong to $\mathcal{U}$. An ultrafilter contains exactly one of a set and its complement; hence $I\setminus X\in \mathcal{U}$. Equivalently, every nonprincipal ultrafilter contains the cofinite filter.

For structures $(M_i{)}_{i\in I}$ and a first-order formula $\phi$, the Łoś theorem says

Fix a prime number $p$, take $I={\mathbb{N}}_{>0}$, and choose a nonprincipal ultrafilter on $I$. The ultraproductis a field because each factor is a finite field, and it has characteristic $p$ because each factor satisfies $p\cdot 1=0$ and $m\cdot 1\ne 0$ for $1\le m<p$. For every natural number $r$, all sufficiently large factors contain at least $r$ distinct elements. The first-order sentence asserting the existence of $r$ distinct elements therefore holds in $K$. Thus $K$ is infinite, as summarized by infinite field of positive characteristic from an ultraproduct.

The critical point of an elementary embedding $j$ is the least ordinal $\alpha$ for which $j(\alpha )\ne \alpha$.

We first prove by transfinite induction that $j(\alpha )=\alpha$ for every $\alpha <\kappa$. Suppose this is known below $\alpha <\kappa$. If $[f]E[{\mathrm{const}}_{\alpha }]$, thenThe sets $A_{\beta }=\{\xi \in A:f(\xi )=\beta \}$ for $\beta <\alpha$ partition $A$ into fewer than $\kappa$ pieces. A kappa-complete filter that is an ultrafilter must contain one cell $A_{\beta }$: otherwise all their complements would belong to $U$, and their intersection would contradict $A\in U$. Hence $[f]=[{\mathrm{const}}_{\beta }]$. The predecessors of $j(\alpha )$ are consequently exactly the already-fixed ordinals below $\alpha$, so $j(\alpha )=\alpha$.

Now let $\mathrm{id}(\xi )=\xi$. An identical argument shows that the predecessors of $[\mathrm{id}]$ in the ultrapower are exactly $[{\mathrm{const}}_{\beta }]$ for $\beta <\kappa$, soSince $\{\xi <\kappa :\xi <\kappa \}=\kappa \in U$, one has $[\mathrm{id}]E[{\mathrm{const}}_{\kappa }]$. After collapsing, $\kappa \in j(\kappa )$, and therefore $j(\kappa )>\kappa$. Thus $\mathrm{crit}(j)=\kappa$.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of ultrafilters on $\mathbb{N}$, equipped with addition on the Stone-Čech compactification of the natural numbers. This makes $\beta \mathbb{N}$ a compact Hausdorff left-topological semigroup. For completeness, the Ellis–Numakura lemma gives an idempotent in every such semigroup: by the Hausdorff space property and compactness, the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroupis also $K$, and in particular $a+a=a$. Choose the resulting idempotent ultrafilter on the natural numbers $\mathcal{U}$.

One color class $A$ belongs to $\mathcal{U}$. For $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal{U}$ lie in $\mathcal{U}$, and their intersection is $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their closures are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every neighbourhood of $x$ meets every $S_A$; this is the ultrafilter limit of the sequence.

The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal{U}$, because otherwise its complement would belong to $\mathcal{U}$ and the associated $S_A$ would miss $U$. Similarly $I_V\in \mathcal{U}$. Their intersection is empty, contradicting the definition of an ultrafilter.

An ultrafilter is a proper filter maximal under inclusion. Equivalently, for every $A\subseteq \mathbb{N}$, exactly one of $A$ and $\mathbb{N}\setminus A$ belongs to $\mathcal{U}$.

The cofinite subsets of $\mathbb{N}$ form a proper filter. By Zorn lemma, it extends to an ultrafilter $\mathcal{U}$. Since $\mathcal{U}$ contains every cofinite set, it cannot contain a finite set, so it is nonprincipal.

Define the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ to be the set of ultrafilters on $\mathbb{N}$, with basic setsSince $\beta \mathbb{N}\setminus \bar{A}=\overline{\mathbb{N}\setminus A}$, these sets are clopen. If $\mathcal{U}\ne \mathcal{V}$, choose $A\in \mathcal{U}\setminus \mathcal{V}$; then $\bar{A}$ and $\overline{\mathbb{N}\setminus A}$ are disjoint neighbourhoods, proving Hausdorffness. For compactness, a family of basic closed sets with the finite-intersection property corresponds to a family of subsets of $\mathbb{N}$ with the finite-intersection property. Extend that family to an ultrafilter; the resulting point belongs to every closed set. The Alexander subbase theorem now proves compactness.

The Hindman theorem states that every finite colouring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ for which every nonempty finite sum of distinct terms has one colour. Let $\mathcal{U}\in \beta \mathbb{N}$ be an additive idempotent, and choose a colour class $A\in \mathcal{U}$. PutIdempotence gives $A^{\ast }\in \mathcal{U}$, and $A^{\ast }-x\in \mathcal{U}$ whenever $x\in A^{\ast }$. Having selected $x_1,ldots,x_n$ with all finite sums in $A^{\ast }$, chooseThis finite intersection belongs to $\mathcal{U}$ and is nonempty. Induction keeps every finite sum in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Write the ultrafilter as $\mathcal{U}$ to distinguish it from first-order structure domains. On the direct product ${\prod }_{i\in I}{A}_i$, putThe domain of the ultraproduct is the quotient $({\prod }_i{A}_i)/\sim$. Its elements are classes $[\mathbf{a}]$. A constant $c$ is interpreted by $[(c^{{\mathcal{A}}_i}{)}_i]$, and a $k$-ary function symbol byFor a $k$-ary relation symbol,These definitions are independent of representatives: if finitely many input representatives are changed, the coordinates where all old and new inputs agree form a set in the filter. At those coordinates the interpreted outputs and relation truth values agree. Filter closure and upward closure therefore preserve the resulting quotient elements and truth conditions.

A filter on a set $I$ means a proper family $\mathcal{F}\subseteq \mathcal{P}(I)$ with $I\in \mathcal{F}$, $\varnothing \notin \mathcal{F}$, closure under finite intersections, and upward closure under inclusion. An ultrafilter is a maximal proper filter; equivalently, for every $X\subseteq I$, exactly one of $X$ and $I\setminus X$ belongs to it. A principal ultrafilter is $\{X\subseteq I:i\in X\}$ for some $i\in I$. A nonprincipal ultrafilter is one not of that form, equivalently one containing no finite set.

To prove the ultrafilter lemma using Zorn's lemma, order all proper filters containing $\mathcal{F}$ by inclusion. The union of any chain is a proper filter: finitely many members lie together in one filter of the chain, and no chain member contains the empty set. Thus every chain has an upper bound, and Zorn's lemma gives a maximal member $\mathcal{U}$.

For completeness, its decision property follows from maximality. If $X\notin \mathcal{U}$, adjoining $X$ cannot generate a proper filter. Hence some $F\in \mathcal{U}$ has $F\cap X=\varnothing$. Upward closure gives $I\setminus X\in \mathcal{U}$. Both complements cannot be members, since their intersection is empty. Therefore $\boxed{\mathcal{F}\subseteq \mathcal{U}\text{ for an ultrafilter }\mathcal{U}}$. Properness is essential; a family containing the empty set cannot extend to an ultrafilter.

Let $P_1$ and $P_3$ be the sets of primes congruent to one and three modulo four. Both are infinite. For $P_1$, if a finite list contained them all, put $N=(2p_1\cdots p_r{)}^2+1$. Any prime divisor $q$ of this odd number is new and has minus one as a square modulo $q$, hence belongs to $P_1$ by the given criterion. For $P_3$, use $N=4p_1\cdots p_r-1$. It is three modulo four, so at least one of its odd prime divisors is three modulo four; none of the listed primes divides $N$. These are elementary Euclidean proofs of infinitely many primes in each odd residue class modulo four.

On the set of all primes, the cofinite sets together with $P_1$ have the finite intersection property; so do the cofinite sets together with $P_3$. Extend the generated filters to ultrafilters ${\mathcal{U}}_1$ and ${\mathcal{U}}_3$ using the ultrafilter lemma. They are nonprincipal because they contain every cofinite set, including the complement of any singleton. FormEvery sentence of $\Sigma$ has a cofinite truth set, hence belongs to both ultrafilters. The Łoś theorem shows $F_1,F_2\models \Sigma$. In particular they are fields, since each field axiom holds in every factor.

For each fixed positive integer $n$, the sentence $n\cdot 1\ne 0$ holds in ${\mathbb{F}}_p$ unless $p$ divides $n$. Its truth set is cofinite, so both fields satisfy it. Therefore $\boxed{\mathrm{char}{F}_1=\mathrm{char}{F}_2=0}$.

Finally the sentence $\exists x(x^2+1=0)$ holds at exactly the primes in $P_1\cup \{2\}$. This set belongs to ${\mathcal{U}}_1$, whereas its complement contains $P_3$ and belongs to ${\mathcal{U}}_3$. Łoś therefore givesThus minus one is a square in the first characteristic-zero first-order model of $\Sigma$ and is not a square in the second. The construction also shows why eventual truth over all primes need not decide a sentence whose truth splits along two infinite prime subsequences.

A proper filter is an upward-closed family of subsets closed under finite intersections and excluding the empty set. This excludes the improper family of all subsets. An ultrafilter is a maximal proper filter.

For a function $f:X\to Y$, the pushforward of a filter on a set $X$ consists of the subsets $B\subseteq Y$ with $f^{-1}(B)\in \mathcal{F}$. Preimages preserve inclusions and finite intersections, so this is a proper filter. Equivalently it is generated by $f(A)$ for $A\in \mathcal{F}$. Because preimages preserve complements too, the pushforward of an ultrafilter is an ultrafilter.