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A subset $A$ of a specified ambient set $X$ is coinfinite if its complement of a set $X\setminus A$ is infinite. In $\mathbb{N}$, the even numbers are both infinite and coinfinite, whereas a cofinite set is not coinfinite.

In the ordinary topology on infinite subsets, the family $D$ of cofinite sets is a countable set, a dense and a meagre set. Countability follows by indexing its members by their finite complements. Each singleton is closed with empty interior, while every basic cylinder with finite initial segment $s$ contains $s\cup \{n:n>\max s\}$. Consequently $D$ is not a nowhere dense set. This example uses the ordinary cylinder topology, rather than the Ramsey cone topology.

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.

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.

LetThis is the dense countable family of cofinite infinite subsets. Every ordinary basic open set $[s]$ contains the cofinite set $s\cup \{n:n>\max s\}$. Thus $D$ is $\tau$-dense: its ordinary closure is the whole space, so it is not $\tau$-nowhere dense.

It is $\ast$-closed. Indeed, for an infinite noncofinite $X$, the Ellentuck topology neighbourhood $[\varnothing ,X]$ contains only noncofinite sets, because every subset of $X$ misses the infinite complement of $X$. This neighbourhood is disjoint from $D$.

Finally every nonempty basic $\ast$-open set $[s,A]$ contains a smaller basic $\ast$-open set disjoint from $D$: enumerate $A$ and let $B$ consist of every second point. Then $A\setminus B$ is infinite, and $[s,B]\subseteq [s,A]$ contains no cofinite set. In particular $D$ has empty $\ast$-interior. As it is already $\ast$-closed, its $\ast$-closure has empty $\ast$-interior, proving that it is $\ast$-nowhere dense.

Take $\tau$ to be the ordinary topology on infinite subsets, whose basic cylinders $[s]$ fix a finite initial segment. LetThis is a dense countable family of cofinite infinite subsets. It is countable because finite complements form a countable family. Each singleton is $\tau$-closed: distinct increasing enumerations differ at some finite coordinate, which separates them by cylinders. No singleton has interior, since every cylinder admits more than one infinite continuation. Thus every singleton is $\tau$-nowhere dense, and $D$ is $\tau$-meagre.

On the other hand, every cylinder $[s]$ contains the cofinite set $s\cup \{n:n>\max s\}$, with $\max \varnothing =0$. Hence $D$ is $\tau$-dense, its closure is all of $X$, and

The paper does not define its symbol $\tau$. If the course instead uses it for the Ramsey cone topology, with basic neighborhoods $[A{]}^{\omega }$ and no finite stem, use $X$ itself as the example. For $D_j=\{Z:\min Z=j\}$, its cone-closure is $\{Z:j\in Z\}$: every cone neighborhood of a set containing $j$ contains an infinite subset of minimum $j$, while a cone based on a set omitting $j$ misses $D_j$. This closure has empty interior, because every infinite reservoir can be thinned to omit $j$. Thus every $D_j$ is cone-nowhere dense, and $X={\bigcup }_{j\ge 1}{D}_j$ is cone-meagre but is not cone-nowhere dense. This supplies the requested example under either convention, without conflating the two topologies.

One form of the Ehrenfeucht-Mostowski theorem says that, for every infinite first-order structure $M$ and every total order $I$, an elementary extension contains distinct elements $(a_i{)}_{i\in I}$ forming an order-indiscernible sequence. After a suitable Skolem expansion, their Skolem hull is a first-order model of $\mathrm{Th}(M)$, and every order automorphism of $I$ extends to a structure automorphism of that hull. If $M$ has a definable infinite linearly ordered subset $P$, the generators may lie in $P$ and their order may agree with that definable order. Uniqueness of the extended structure automorphism is asserted in the chosen Skolem expansion, rather than for every structure automorphism of its reduct.

Here is an ultraproduct proof of the Ehrenfeucht-Mostowski theorem. First expand $M$ to $M^{+}$ with Skolem functions for all existential first-order formulas, iterating through the enlarged languages if necessary. This is done in the ordinary classical metatheory; the choice restriction in question 1 is not a restriction on this question. Fix a nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$.

We can obtain an infinite sequence of distinct elements $b_0,b_1,\dots$ in an elementary extension $M_0$ of $M^{+}$ by one preliminary ultrapower. For each $n\ge 1$ choose a finite list of $n$ distinct elements of $M^{+}$. Represent $b_j$ by the function taking the $j$th entry when $n>j$, with an arbitrary default otherwise. Distinctness holds on a cofinite set, so follows from Łoś theorem. In the definably ordered variant, first name any parameters defining $P$, and choose each finite list increasingly inside $P$ instead; then $M_0\models P(b_j)$ and $b_j<b_k$ for $j<k$. Infinitude of an ordered set supplies every finite increasing chain. This preparatory step uses no Ramsey theorem.

For a finite subset $F=\{i_1<\cdots <i_r\}$ of $I$, define the ultrafilter ${\mathcal{U}}_F$ on ${\mathbb{N}}^F$ by nested membership, in this order:where $(\mathcal{U}n)Q(n)$ means $\{n:Q(n)\}\in \mathcal{U}$. This is the ordered Fubini product of ultrafilters. Closure under intersections and the decision between a set and its complement hold at each nested level, proving it is an ultrafilter. For $F=\varnothing$ take the principal ultrafilter on the one-point product. PutIf $F\subseteq G$, pull a function back along the projection ${\mathbb{N}}^G\to {\mathbb{N}}^F$. A test independent of an omitted coordinate is unchanged by its ultrafilter quantifier, so projection pushes ${\mathcal{U}}_G$ to ${\mathcal{U}}_F$. The induced map $N_F\to N_G$ is therefore an elementary embedding by Łoś theorem. These maps are coherent, giving a directed system indexed by finite subsets of $I$.

Take its directed limit of elementary embeddings $N$. One can verify elementarity directly: representatives of a finite tuple occur at a common stage; functions and atomic relations are interpreted there. In the existential step, a witness in the limit occurs with the parameters at some later common stage, and elementarity pulls the existence statement back. Thus every $N_F$ embeds elementarily into $N$, which contains an elementary copy of $M^{+}$.

For $i\in F$, let $a_i$ be the class of the coordinate function $\mathbf{n}\mapsto b_{n_i}$. Projection coherence makes this independent of $F$. If $i<j$, then for each fixed $n_i$ the set $\{n_j:n_j\ne n_i\}$ is cofinite. The nested test therefore gives $a_i\ne a_j$. In the ordered variant, the same argument with $n_j>n_i$ gives $P(a_i)$ and $a_i<a_j$.

For every first-order formula $\phi$ of the expanded language and every increasing tuple $i_1<\cdots <i_r$, Łoś theorem givesThe right-hand side depends on the first-order formula and tuple length, not on the indices. This proves order indiscernibility in the expanded language. Let $H$ be the Skolem hull of the generators in $N$. The Tarski-Vaught test gives $H\prec N$, and its reduct is a first-order model of $\mathrm{Th}(M)$.

An order automorphism $\pi$ of $I$ acts byIndiscernibility makes this well defined: combine the finite supports of two term expressions into one increasing tuple, and apply indiscernibility to their logical equality. Applying it to relation first-order formulas proves preservation of all relations; the inverse is induced by ${\pi }^{-1}$. Every element of the hull is such a term, so the extension is unique among structure automorphisms preserving the Skolem expansion. MoreoverThis proves the theorem, including its ordered version and structure automorphism conclusion, using ultrapowers and a directed limit throughout.