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The even and odd integers are both absent from the cofinite filter, but their union is present. Thus its filter quantifier need not turn a disjunction into the disjunction of quantified assertions, and failure of a quantified assertion need not imply truth of its quantified negation. An ultrafilter restores both laws by deciding each set against its complement.

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.

For $A\subseteq \mathbb{N}$, the basic set $\hat{A}=\{U:A\in U\}$ in the Stone-Čech compactification of the natural numbers is a clopen set, since its complement is $\widehat{\mathbb{N}\setminus A}$. Distinct ultrafilters are separated by complementary basic sets. A basic cover with no finite subcover would give complements with the finite intersection property, whose generated filter on a set extends to an ultrafilter missing the entire cover. Thus the space is a compact Hausdorff space.

For every infinite set $X$ there is an independent family of sets consisting of $2^{|X|}$ subsets of $X$. One construction uses a set $D$ of pairs $(s,H)$, where $s\subseteq X$ is finite and $H\subseteq \mathcal{P}(s)$. For each $A\subseteq X$, let $I_A=\{(s,H):A\cap s\in H\}$. Given distinct $A_1,\dots ,A_m$, choose a finite $s$ distinguishing all their traces and choose $H$ to contain precisely the traces prescribed to have value one. This realizes every finite Boolean choice. With the axiom of choice, $|D|=|X|$, so a bijection transports this family to $X$. Its independent assignments give $2^{2^{|X|}}$ distinct ultrafilters.

For a proper filter on a set, write ${\forall }_{\mathcal{F}}xp(x)$ when the truth set $\{x:p(x)\}$ belongs to the filter. This quantifier preserves finite conjunctions. For an ultrafilter it also preserves finite disjunctions and obeys classical negation; a general proper filter need not have those latter properties.

An ultrafilter on $P_{\kappa }(\lambda )$ is fine when it contains the set of small subsets containing $\alpha$, for each $\alpha <\lambda$. Fineness makes every coordinate of $\lambda$ represented in the corresponding ultrapower, though no one small subset is required to contain every coordinate.

On an infinite set $D$, take its finite and cofinite subsets with ordinary union, intersection and complement. Singletons are its atoms. Its Stone space consists of the isolated points of $D$ together with one nonisolated point whose ultrafilter contains all cofinite sets. For countably infinite $D$ this is a countable Boolean algebra with countably many atoms, not the uncountable full power set. It describes two-vector definability over a countably infinite field: dependent directions are isolated atoms and the independent-pair type is the additional nonisolated point.

If $A\equiv B$, the elementary diagram of $B$ is finitely satisfiable by interpreting its finitely many named elements in $A$. Choose those finite realizations as coordinates, extend the filter of diagram cones to an ultrafilter, and use the Łoś theorem to embed $B$ elementarily into an ultrapower of $A$. If $A\prec B$, work over the named elements of $A$ and keep their coordinates constant. The resulting embedding extends the diagonal map of $A$. This relative assertion allows an alternating chain of ultrapowers to keep every previously embedded substructure fixed.

For ultrafilters $\mathcal{U}$ on $I$ and $\mathcal{V}$ on $J$, putThis is an ultrafilter on $I\times J$: upward closure and finite intersection follow at both levels, and the ultrafilter dichotomy follows by complementing at both levels. Iterating gives ordered finite products. Projection onto a subsequence of coordinates pushes this product to the product on that subsequence, since omitted coordinates are absent from the tested first-order formula. The order of coordinates matters; these products need not be symmetric.

An additive idempotent ultrafilter satisfies $\mathcal{U}+\mathcal{U}=\mathcal{U}$. On positive integers it is necessarily nonprincipal, since a principal ultrafilter at $n$ adds to itself to give the principal ultrafilter at $2n$. The idempotent-ultrafilter star-set lemma converts this algebraic property into a recursive construction of finite sums. If zero is included, the trivial principal idempotent at zero must be excluded for that increasing-sequence application.

A family of subsets of $X$ is independent when every finite choice of distinct members and their complements has nonempty intersection. Write $A^1=A$ and $A^0=X\setminus A$ in the displayed condition. Every assignment of zero or one to the family then gives a collection with the finite intersection property, which can be extended to an ultrafilter.

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.

Every convergent sequence in the Stone-Čech compactification of the natural numbers is eventually constant. If distinct ultrafilters $U_n$ converged, isolate each from the other sequence terms using convergence and finitely many separating clopen sets. Finite disjointification gives disjoint $B_n\in U_n$. The union of the even-indexed $B_n$ then belongs exactly to the even-indexed ultrafilters, contradicting convergence of membership in a clopen set.

A uniform kappa-complete ultrafilter $U$ on an uncountable cardinal $\kappa$ is normal if every regressive function on a $U$-large set is constant on a $U$-large subset. Equivalently it is closed under diagonal intersections ${\Delta }_{\alpha <\kappa }{X}_{\alpha }=\{\beta <\kappa :(\forall \alpha <\beta )\beta \in X_{\alpha }\}$. Normality supplies homogeneous sets for pair-colourings by combining the chosen large tail at each earlier coordinate.

An ultrafilter $U$ on $P_{\kappa }(\lambda )$ is normal when each function with $f(x)\in x$ on a $U$-large set is constant on a $U$-large subset. This is the small-subset version of normality; together with fineness and kappa-completeness it gives the measure characterization of a supercompact cardinal.

Take formulas whose free variables lie among a fixed $n$-tuple and identify them when their universal equivalence follows from $T$. Conjunction, disjunction and negation induce a Boolean algebra; its order is implication modulo $T$. Complete types correspond to ultrafilters and isolated complete types to atoms. For a complete theory in a countable language, finiteness of these algebras for every finite $n$ characterizes countable categoricity.

Use these standard ultrafilter facts: a space is compact if and only if every ultrafilter on it converges; the image ultrafilter under a map is an ultrafilter; and a convergent ultrafilter contains every neighborhood of its limit.

Let $X={\prod }_{\lambda \in \Lambda }{X}_{\lambda }$ with all factors compact. An empty factor makes $X$ empty and hence compact. Otherwise, given an ultrafilter $\mathcal{U}$ on $X$, push it forward by each product projection to obtain an ultrafilter ${\mathcal{U}}_{\lambda }$ on $X_{\lambda }$. Choose a limit $x_{\lambda }$ of each, and put $x=(x_{\lambda }{)}_{\lambda }$. A basic product neighborhood of $x$ is an intersection of finitely many inverse images of coordinate neighborhoods. Each inverse image belongs to $\mathcal{U}$, so their finite intersection does too. Thus $\mathcal{U}$ converges to $x$, proving compactness of the product. Uniqueness of coordinate limits is not needed, so the factors need not be Hausdorff.

For local compactness, suppose only the factors indexed by a finite set $F$ may be noncompact. At $x\in X$ choose, for each $\lambda \in F$, a compact neighborhood $K_{\lambda }$ of $x_{\lambda }$, with an open neighborhood $U_{\lambda }\subseteq K_{\lambda }$. For other indices put $K_{\lambda }=X_{\lambda }$. The compact product ${\prod }_{\lambda }{K}_{\lambda }$ contains the open product neighborhood restricting just the finitely many coordinates in $F$ to $U_{\lambda }$. Hence every point has a compact neighborhood and the product is locally compact.

The unrestricted assertion is false. Each factor of ${\mathbb{R}}^{\mathbb{N}}$ is locally compact, but its product is not. Any basic neighborhood restricts only finitely many coordinates. Choose a free coordinate $j$; its projection of that neighborhood is all of $\mathbb{R}$. A compact neighborhood containing it would consequently have compact image equal to $\mathbb{R}$, impossible. This shows directly that no point has a compact neighborhood.

A proper filter $\mathcal{F}$ on $X$ is a nonempty family of subsets, closed under finite intersection and passage to supersets, with $\varnothing \notin \mathcal{F}$. An ultrafilter is a maximal proper filter, equivalently one which contains, for each $A\subseteq X$, exactly one of $A$ and $X\setminus A$. The pushforward filter isInverse images preserve intersections and complements. Thus it is a proper filter, and the either-a-set-or-its-complement criterion shows it is an ultrafilter whenever $\mathcal{F}$ is one.

Convergence of a filter $\mathcal{F}\to x$ means that every neighborhood of $x$ belongs to $\mathcal{F}$. A mapping is continuous exactly when $\mathcal{F}\to x$ implies $f_{\ast }\mathcal{F}\to f(x)$ for all proper filters and points. The forward implication follows from inverse images of neighborhoods. For the converse apply the criterion to the neighborhood proper filter at $x$: each inverse image of a neighborhood of $f(x)$ must be a neighborhood of $x$, which is continuity at $x$.

A topological space is compact exactly when every ultrafilter converges to at least one point; equivalently every proper filter has a cluster point. For completeness, in a compact space the closed sets $\overline{A}$, $A\in \mathcal{U}$, have the finite intersection property, hence a common point $x$. Every neighborhood $N$ of $x$ belongs to the ultrafilter. Indeed, if $N$ did not, choose an open $V$ with $x\in V\subseteq N$. Then $V$ would not belong either, so $X\setminus V$ would belong. That set is closed and excludes $x$, contradicting the choice of $x$ in every closure. Conversely, if an open cover has no finite subcover, its closed complements generate a proper filter. An ultrafilter extension cannot converge anywhere, because each point has a covering neighborhood whose complement belongs to that ultrafilter. This proves the compactness criterion using the permitted extension principle.

Now suppose $b\in \overline{f[A]}$. The familyis a filter base of nonempty sets: closure guarantees each intersection is nonempty, and finite intersections correspond to intersections of neighborhoods. Extend its generated proper filter to an ultrafilter $\mathcal{H}$. It contains $A$ and every $f^{-1}[V]$, so $f_{\ast }\mathcal{H}\to b$.

Conversely, if such an ultrafilter exists, each neighborhood $V$ of $b$ has $f^{-1}[V]\in \mathcal{H}$. Its intersection with $A\in \mathcal{H}$ is nonempty. Thus $V$ meets $f[A]$, provingThis part includes the unnumbered proper filter definitions and criteria preceding the printed subparts.

For every ultrafilter $\mathcal{H}$ on $X$, its image is an ultrafilter on the compact space $Y$, so it has a limit. That limit is unique because $Y$ is Hausdorff: distinct limits would have disjoint neighborhoods both belonging to the same proper filter. Therefore $\Lambda$ is well-defined. Since $f[X]$ is dense, every $b\in Y$ lies in $\overline{f[X]}$. Part (a), with $A=X$, supplies an ultrafilter mapping to $b$, proving surjectivity.

Suppose disjoint subsets have disjoint image closures. If $\mathcal{H}\ne \mathcal{K}$, choose $A\in \mathcal{H}$ not in $\mathcal{K}$; the ultrafilter criterion gives $X\setminus A\in \mathcal{K}$. Part (a) givesThe two closures are disjoint by assumption, so the limits differ. Thus $\Lambda$ is injective.

Conversely, suppose $\Lambda$ is injective and disjoint $A,B\subseteq X$ had a common point $b$ in their image closures. Part (a) would supply ultrafilters $\mathcal{H},\mathcal{K}$ containing $A,B$ respectively, both with image limit $b$. They must be distinct, since a proper filter cannot contain the disjoint sets $A$ and $B$ simultaneously. This contradicts injectivity. Hence

Use the Stone-Čech extension property: a discrete topological space $X$ embeds densely into a compact Hausdorff space $\beta X$, and every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Apply it to the characteristic function ${\chi }_A:X\to \{0,1\}$ of an arbitrary subset $A$; discreteness makes it continuous. Its extension ${\tilde{\chi }}_A$ has clopen fibers, andIndeed that fiber is closed and contains $A$, while its openness and density of $X$ show that every point in the fiber is approached by points of $A$.

If $A$ and $B$ are disjoint, the extension of ${\chi }_A$ is one on $\overline{A}$ and zero on $\overline{B}$, so these closures are disjoint. The inclusion $X\to \beta X$ therefore satisfies the criterion from part (b), and its ultrafilter-limit map is both surjective and injective. ThusThis is a bijection of underlying sets; identifying the compatible topology on the ultrafilter topological space would give the familiar stronger homeomorphism statement, but is not needed for the requested conclusion.