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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 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 subsemigroup is a nonempty subset of a semigroup closed under its associative operation. In a compact Hausdorff space carrying a left-topological semigroup, descending chains of nonempty compact subsemigroups have a nonempty compact subsemigroup as their intersection. This supplies the minimal object used in the Ellis–Numakura lemma.