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For an infinite coinfinite set $A\subseteq \mathbb{N}$, let $E_A=[A{]}^{\omega }$ in the space of infinite subsets of the natural numbers. This cone is a nonempty clopen set in the Ellentuck topology, because it is the basic neighbourhood $[\varnothing ,A]$, and any $X\notin E_A$ has a finite initial segment containing a point outside $A$ and hence a basic neighbourhood disjoint from $E_A$.

The cone is a closed set with empty interior in the ordinary topology on infinite subsets: membership of a point outside $A$ witnesses its complement as an open set, while every ordinary finite-stem neighbourhood can be extended by an arbitrarily large point outside $A$. Thus $E_A$ is an ordinary nowhere dense set, but it is neither ordinary open nor Ellentuck nowhere dense. It provides both distinctions between these two topologies.

A continuous array takes infinite subsets of an infinite $A\subseteq \omega$, in the ordinary topology on infinite subsets, to a discrete preorder $Q$. It is bad if $F(X)\nleqq F(X\setminus \{\min X\})$ for every infinite $X$. A barrier in better-quasi-order theory array induces such a continuous array by evaluating its unique barrier initial segment. The continuous-array and barrier formulations of better-quasi-ordering are equivalent, using refinement of fronts to barriers in the reverse direction.

In the space of infinite subsets of the natural numbers, the families $C_n=\{X:n\notin X\}$ are clopen sets in the ordinary topology on infinite subsets and thus in the finer Ellentuck topology. Their union is $[\mathbb{N}{]}^{\omega }\setminus \{\mathbb{N}\}$. Every basic Ellentuck topology neighborhood of $\mathbb{N}$ contains $\mathbb{N}\setminus \{n\}$ for some sufficiently large $n$, so this union is not a closed set.

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.

For a finite set $s$ and an infinite set $A$ lying above $\max s$, putThese sets form the basic neighborhoods of the Ellentuck topology, also called the star topology. It is finer than the ordinary topology on infinite subsets. Every basic neighborhood is a clopen set in the Ellentuck topology, and a closed set in the ordinary topology.

A finite stem is a finite initial segment $s$ of an infinite subset of $\mathbb{N}$. The associated neighbourhood in the ordinary topology on infinite subsets is $[s]=\{X:s\text{ is an initial segment of }X\}$. For an infinite tail $A$ above $\max s$, the neighbourhood $[s,A]=\{s\cup B:B\in [A{]}^{\omega }\}$ is basic in the Ellentuck topology. Finite stems allow a fusion argument to fix initial choices while successively thinning the remaining infinite tail.

The whole space of infinite subsets of the natural numbers is a meagre set in the Ramsey cone topology. For $D_j=\{X:\min X=j\}$, the closure is $\{X:j\in X\}$, which has empty interior: every infinite ground set can be thinned to omit $j$. Thus all $D_j$ are nowhere dense, while their countable union is the whole space. In contrast, the whole space is not meagre in either the Ellentuck topology or the ordinary topology on infinite subsets.

Every open set $O$ in the ordinary topology on infinite subsets is a Ramsey set of infinite subsets: some infinite $B$ has $[B{]}^{\omega }\subseteq O$ or $[B{]}^{\omega }\cap O=\varnothing$. Use acceptance and rejection of finite stems and fusion to decide every finite stem. If the empty stem is rejected, each rejected stem has only finitely many accepting one-point extensions. Successive avoidance of these finite obstructions gives an infinite set all of whose finite stems are rejected; openness then excludes every infinite subset from $O$. This concerns increasing enumerations of infinite subsets, rather than unrestricted sequences with repetitions.

The cone $[E{]}^{\omega }$ on the even integers is nonempty and Ellentuck topology open, while it is closed and nowhere dense in the ordinary topology on infinite subsets. Choose a finite-symmetric-difference parity colouring on that cone and take one colour class. The class is still ordinarily nowhere dense because its closure lies in the cone, but every stem-free refinement inside the cone meets both colours. Hence it is not completely Ramsey.

Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, with each subset identified with its increasing enumeration. In the homogeneous-cone convention, a family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense when some infinite $M$ satisfiesThe stronger Ramsey set of infinite subsets convention asks for such an infinite $M\subseteq A$ inside every given infinite ground set $A$. We will prove the open-set assertion in this stronger form, and our Ramsey-but-not-completely-Ramsey example will satisfy it as well. Thus the distinction in conventions does not affect any of the conclusions below.

For a non-Ramsey example, declare $X\sim Y$ if their symmetric difference is finite, and choose a representative $R$ from each equivalence class using the axiom of choice. Setfor the representative of the class of $X$. Removing one point reverses this colour, because its symmetric difference with $R$ changes by exactly one point. Consequently every cone $[M{]}^{\omega }$ contains the oppositely coloured sets $M$ and $M\setminus \{\min M\}$. Either colour class of this finite-symmetric-difference parity colouring is not a Ramsey family in the homogeneous-cone sense. No definability is claimed for the representative selection.

The ordinary topology on infinite subsets, denoted $\tau$, has basic cylinders $[s]=\{X:s⊏X\}$, where $s$ is a finite stem of the increasing enumeration. Let $O$ be a $\tau$-open family and fix an arbitrary infinite ground set $A$. We give a complete fusion proof of a homogeneous cone inside $A$.

Use the convention $\max \varnothing =0$. For finite $s$ and an infinite reservoir $B$ above $\max s$, say that $B$ accepts $s$ if $[s,B]\subseteq O$, whereSay that $B$ rejects $s$ if no infinite subset of $B$ accepts $s$. There is always a refinement deciding $s$: choose an accepting subreservoir if one exists, and otherwise retain $B$, which rejects. Both decisions persist under further thinning.

A first fusion produces an infinite $H\subseteq A$ deciding every finite $s\subseteq H$ on its tail $H/s=\{h\in H:h>\max s\}$. First decide the empty stem. Choose the first point, and thin the remaining reservoir successively to decide every subset of the selected prefix. Repeat after choosing each new point. There are finitely many subsets to decide at each stage. The diagonal sequence of chosen points has the required property, since its tail after the largest point of any stem lies in the reservoir where that stem was decided. The empty-stem decision persists too. This is deciding all finite stems by fusion.

If $H$ accepts the empty stem, $[H{]}^{\omega }\subseteq O$ and we are done. Suppose it rejects. For a rejected finite $s\subseteq H$, only finitely many $h\in H/s$ can have $H/h$ accepting $s\cup \{h\}$. Otherwise let $D$ be their infinite set. Every infinite $X\subseteq D$ starts with some such $h$, and its remaining tail lies in $H/h$, so $s\cup X\in O$. Then $D$ accepts $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

A second fusion constructs $K\subseteq H$ whose every finite subset is rejected. If a finite prefix of chosen points has been selected, all its subsets are rejected by induction. Avoid the union of their finitely many accepting successor sets when choosing the next point. The first fusion already decides every successor, so every new subset is rejected. An infinite reservoir remains after each finite exclusion.

If some $X\in [K{]}^{\omega }$ belonged to $O$, openness would give an initial segment $s⊏X$ with $[s]\subseteq O$. Its reservoir $H/s$ would then accept $s$, contradicting the second fusion. Therefore $[K{]}^{\omega }\cap O=\varnothing$. In either case we have a homogeneous cone inside the arbitrary ground set $A$. Every $\tau$-open family is a Ramsey set of infinite subsets, under both conventions.

A completely Ramsey set requires more: for every finite $s$ and infinite reservoir $A$ above $\max s$, there is an infinite $B\subseteq A$ such thatThe finite stem must be kept fixed. These $[s,A]$ are the basic neighbourhoods of the Ellentuck topology, also called the star topology.

Let $E$ be the even positive integers, and construct the finite-symmetric-difference parity colouring $\epsilon$ on $[E{]}^{\omega }$. DefineFor any infinite ground set $A$, the set $A\setminus \{1\}$ is still infinite and its whole cone misses $\mathcal{R}$. Thus $\mathcal{R}$ is a Ramsey set of infinite subsets even in the stronger convention. But inside $[\{1\},E]$, every infinite refinement $B\subseteq E$ gives the two sets $\{1\}\cup B$ and $\{1\}\cup (B\setminus \{\min B\})$ of opposite colours. Hence no such refinement is homogeneous. The family $\mathcal{R}$ is Ramsey but not completely Ramsey. The stem-supported Ramsey family need not be completely Ramsey construction explains this example: forgetting the stem $\{1\}$ is precisely what loses the obstruction.

Finally take $D=[E{]}^{\omega }$. It is nonempty and star-open, since it equals $[\varnothing ,E]$. It is $\tau$-closed: if an infinite set contains an odd integer, an initial segment witnessing that integer has a cylinder disjoint from $D$. Its $\tau$-interior is empty: after any finite stem of even integers, append an arbitrarily large odd integer and then infinitely many further points. ThusThis is the cone on an infinite coinfinite ground set. Choose its subset $Y=\{X\in [E{]}^{\omega }:\epsilon (X)=0\}$. Since ${\overline{Y}}^{\tau }\subseteq D$, $Y$ is also $\tau$-nowhere dense. Nevertheless every $[\varnothing ,B]$ with infinite $B\subseteq E$ contains both $B$ and $B\setminus \{\min B\}$, of opposite colours. Hence $Y$ is not completely Ramsey. We have proved explicitly that ordinarily nowhere-dense sets need not be completely Ramsey; the topology in the nowhere-dense hypothesis cannot be silently changed to the star topology.

Identify the space of infinite subsets of the natural numbers with their increasing enumerations, with the ordinary topology on infinite subsets inherited from the discrete product ${\omega }^{\omega }$. This is the sequence convention in the Open Ramsey theorem. For a general infinite ground set, first choose a countably infinite subset and use its enumeration. The theorem cannot mean unrestricted sequences with repetitions: the clopen partition according as the first two entries agree or differ has both colours over every infinite ground set.

For a finite finite stem $s$ and an infinite tail $B$ above $\max s$, writewhere the enumeration of $s$ precedes the tail. Let $O$ be open. The tail $B$ accepts $s$ if $[s,B]\subseteq O$; it rejects $s$ if no infinite subtail accepts it. Both properties persist on infinite subtails, and any stem can be decided by thinning: choose an accepting subtail if one exists; otherwise the current tail already rejects it.

Use deciding all finite stems by fusion. At stage $n$, choose a new element $b_n$ from the current tail, then thin the remaining tail to decide each of the finitely many subsets of $\{b_0,\dots ,b_n\}$. First decide the empty stem as well. The final selected set $B$ decides every finite stem from $B$, because all subsequent selections lie in the tail that decided that stem. If $B$ accepts the empty stem, $[B{]}^{\omega }\subseteq O$ and we are done.

Otherwise the empty stem is rejected. If a stem $s\subset B$ is rejected, only finitely many $a\in B$ above $\max s$ can have the tail $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed $D$, then every $X\in [D{]}^{\omega }$ starts with one of them and has its remaining tail in $B/a$. It follows that $[s,D]\subseteq O$, contradicting rejection of $s$. All other one-point extensions are rejected, since the first fusion has already decided them.

Build an infinite $C\subseteq B$ recursively. Maintain that every subset of the finite selected prefix is rejected. To select its next element, avoid the finite union of the finitely many accepting-extension obstructions for these rejected stems. Such an element exists, and all the new stems remain rejected. Hence every finite stem from $C$ is rejected. If some $X\in [C{]}^{\omega }$ were in $O$, openness would supply a finite initial segment $s$ of $X$ whose entire ordinary cylinder is contained in $O$. The tail $C/\max s$ would accept $s$, a contradiction. ThusIn particular every open payoff has an infinite homogeneous set. A finite partition into open colour classes can be handled by successive application of this two-colour assertion.

For an arbitrary partition use finite-symmetric-difference parity colouring. On the infinite subsets of the ground set, declare $X\sim Y$ when $X△Y$ is finite, and choose a representative $R$ from each equivalence class. Colour $X$ byFor any infinite $H$, removing one member gives an infinite $H^{\prime }$ in the same class and reverses the parity. Thus $H$ and $H^{\prime }$ have opposite colours, so neither colour contains all infinite subsets of any infinite ground set. This two-colouring has no infinite homogeneous set. Representative selection uses choice; the construction is deliberately unrestricted by topological regularity.

Write $[M{]}^{\omega }$ for the set of all infinite subsets of $M$. A family $Y\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense if there is an infinite $M$ such that either $[M{]}^{\omega }\subseteq Y$ or $[M{]}^{\omega }\cap Y=\varnothing$. Thus membership in $Y$ is constant on an infinite homogeneous cone. We will prove the stronger version for open families: such a cone exists inside every prescribed infinite ground set, as in the definition of a Ramsey set of infinite subsets.

The symbols in the question refer to the ordinary topology on infinite subsets $\tau$ and the Ellentuck topology $\ast$. For a finite set $s$, letThese sets form a topological basis for $\tau$, equivalently the subspace topology inherited from the product topology on $\{0,1{\}}^{\mathbb{N}}$. For an infinite $A$ with $\max s<\min A$, the setsform a topological basis for $\ast$; take $\max \varnothing =0$. In particular $[s]=[s,\{n:n>\max s\}]$, so $\ast$ is finer than $\tau$.

For a non-Ramsey example use the finite-symmetric-difference parity colouring. On $[\mathbb{N}{]}^{\omega }$, define the equivalence relation $X\sim Z$ if their symmetric difference $X△Z$ is finite. By the axiom of choice, select a representative $R$ in every equivalence class. Give $X$ the colourwhere $R_X$ is its class representative. Deleting one point stays in the same class and toggles the parity. Thus for every infinite $M$, the sets $M$ and $M\setminus \{\min M\}$ are both in $[M{]}^{\omega }$ but have opposite colours. Either colour class, regarded as a family $Y$, is therefore not Ramsey.

Now let $U$ be $\tau$-open. We prove that for every infinite $A_0$ there is an infinite $C\subseteq A_0$ with $[C{]}^{\omega }\subseteq U$ or $[C{]}^{\omega }\cap U=\varnothing$. The proof uses acceptance and rejection of finite stems, which we define explicitly. For a finite stem $s$ and an infinite tail $A$ above $\max s$, say that $A$ accepts $s$ if $[s,A]\subseteq U$, and rejects $s$ if no infinite subset of $A$ accepts $s$. Both properties persist on taking infinite subsets. Every tail has an infinite subset that decides $s$: take an accepting subset if one exists, and otherwise it already rejects $s$.

First thin $A_0$ to decide the empty stem. Then construct $b_1<b_2<\cdots$ and nested infinite tails. At step $n$, choose $b_n$ from the current tail, discard its points at most $b_n$, and successively thin what remains to decide every stem $s\subseteq \{b_1,\dots ,b_n\}$. There are only finitely many such stems, so an infinite tail remains. Let $B=\{b_1,b_2,\dots \}$ and write $B/s=\{b\in B:b>\max s\}$. For every finite $s\subseteq B$, the tail $B/s$ decides $s$, since it lies in the tail obtained at the stage when the largest point of $s$ was selected. The empty stem is also decided. This is deciding all finite stems by fusion.

If $B$ accepts the empty stem, then $[B{]}^{\omega }\subseteq U$, as required. Otherwise $B$ rejects it. We need the finitely many accepting extensions of a rejected stem observation. Suppose a tail $B/s$ rejects $s$. Only finitely many $a\in B/s$ can have $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed a set $D$, every member of $[s,D]$ would have some first new point $a$ and all later points in $B/a$, hence would lie in $U$. This would make $D$ an accepting subset of $B/s$, contradicting rejection.

Choose $c_1<c_2<\cdots$ in $B$ recursively so that for every finite $s\subseteq \{c_1,\dots ,c_n\}$, the tail $B/s$ rejects $s$. The empty stem satisfies this initially. At the next step, each of the finitely many previous stems has only finitely many accepting one-point extensions by the observation. Choose $c_{n+1}$ larger than all previous points and outside all those finite exceptional sets. The first fusion ensured that the tail for every new stem decides it; since it does not accept it, it rejects it. This maintains the recursion.

Put $C=\{c_1,c_2,\dots \}$. If some $X\in [C{]}^{\omega }$ belonged to $U$, ordinary openness would give a finite initial segment $s$ of $X$ with $[s]\subseteq U$. In particular $[s,C/s]\subseteq U$. But $C/s$ is an infinite subset of $B/s$, which rejects $s$, a contradiction. Hence $[C{]}^{\omega }\cap U=\varnothing$. We have proved

For $n\ge 1$, let $C_n=\{Z\in X:n\notin Z\}$. The presence or absence of $n$ is determined by a sufficiently long finite initial segment, so $C_n$ and its complement are open sets in the ordinary topology on infinite subsets. The Ellentuck topology is finer, hence each $C_n$ is star-clopen and in particular star-closed. Their union isEvery basic star-neighborhood $[s,A]$ of $\mathbb{N}$ has $s=\{1,\dots ,k\}$ for some $k\ge 0$, and its reservoir $A$ contains every integer greater than $k$. Choosing $n>k$ gives $\mathbb{N}\setminus \{n\}\in [s,A]\cap C_n$. Thus $\mathbb{N}$ is in the star-closure of the union, but not in the union itself. Therefore

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.

Write $\Omega =[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, identifying a set with its increasing enumeration. If $s$ is finite and $A$ is infinite with every element of $A$ above $\max s$, setFor the empty stem there is no lower-bound restriction. These sets form the basis of the Ellentuck topology, also called the star topology. The stem $s$ is fixed, while the infinite tail may be thinned.

A completely Ramsey set $E\subseteq \Omega$ is one for which every $[s,A]$ admits an infinite $B\subseteq A$ with $[s,B]\subseteq E$ or $[s,B]\cap E=\varnothing$. This keeps the same stem. A Ramsey set of infinite subsets requires that every infinite $A$ admit an infinite $B\subseteq A$ with a homogeneous empty-stem cone $[B{]}^{\omega }$. It does not require preserving a nonempty finite stem.

A set is a star-Baire set if it differs from a star-open set by a star-meagre set, equivalently $E△U$ is star-meagre for some star-open $U$. A nowhere dense set has closure with empty interior, and a meagre set is a countable union of such sets, with both notions interpreted in the specified topology.

For a non-Ramsey example, put $M=\{2,3,\dots \}$ and well-order all infinite subsets of $M$ as $(A_{\alpha }{)}_{\alpha <\mathfrak{c}}$, where $\mathfrak{c}=2^{{\aleph }_0}$. Recursively choose two fresh points $X_{\alpha },Y_{\alpha }\in [A_{\alpha }{]}^{\omega }$, not used at earlier stages. This is possible because each cone has cardinality $\mathfrak{c}$ and fewer than $\mathfrak{c}$ points have been used before any stage. Letkeeping all the $Y_{\alpha }$ outside $C$. Every infinite-subset cone on $M$ meets both $C$ and its complement. Thus $C$ is not Ramsey, even when regarded as a subset of $\Omega$: any infinite set can first be thinned to avoid $1$. This uses the axiom of choice, rather than claiming a Borel counterexample.

Now setEvery infinite $A$ has an infinite subset $B$ avoiding $1$, and then $[B{]}^{\omega }\cap E=\varnothing$. Hence $E$ is Ramsey. But no stem-preserving thinning of $[\{1\},M]$ is homogeneous for $E$, because $C$ splits every tail cone. Hence $E$ is not completely Ramsey.

To prove the topological equivalence, we first establish the Ellentuck meagre-set fusion lemma: every star-meagre set can be avoided in a refinement $[s,B]$ with the original stem. Call a set completely Ramsey-null if this avoidance holds in every $[s,A]$.

If $N$ is star-nowhere dense, then $U=\Omega \setminus \overline{N}$ is star-open and dense. By the granted complete-Ramsey property of star-open sets, every $[s,A]$ has a refinement $[s,B]$ either contained in $U$ or disjoint from $U$. The latter alternative would put a nonempty open set inside $\overline{N}$, contradicting density of $U$. Thus $N$ is completely Ramsey-null.

For a countable union ${\bigcup }_j{N}_j$ of completely Ramsey-null sets, use fusion, taking care of every possible finite stem. Starting with $A_0=A$, at stage $j$ choose $b_j\in A_{j-1}$ above all previous choices. From the part of $A_{j-1}$ above $b_j$, successively thin an infinite tail for each of the finitely many subsets $t\subseteq \{b_1,\dots ,b_j\}$ so thatEach thinning keeps the stem $s\cup t$ fixed; subsequent thinnings preserve earlier avoidance. Let $B=\{b_1,b_2,\dots \}$. For any $X\in [s,B]$ and fixed $j$, put $t=(X\setminus s)\cap \{b_1,\dots ,b_j\}$. The rest of $X$ lies in $A_j$, because all later selected points do. Therefore $X\in [s\cup t,A_j]$ and $X\notin N_j$. Since $j$ was arbitrary,This proves the fusion lemma. Considering all subsets $t$, not merely the single selected prefix, is essential: elements of $[s,B]$ need not use every $b_i$.

Suppose first that $E$ is star-Baire, with $E△U$ star-meagre and $U$ star-open. Thin $[s,A]$ to $[s,B]$ avoiding the meagre difference by the lemma. Then apply the granted complete-Ramsey property of $U$ to find $[s,D]$ homogeneous for $U$, with $D\subseteq B$. On this refinement $E$ and $U$ agree, so it is homogeneous for $E$. Thus $E$ is completely Ramsey.

Conversely, let $E$ be completely Ramsey and let $U$ be the union of all basic neighborhoods wholly contained in $E$, namely its star-interior. Every $[s,A]$ has a homogeneous refinement. If that refinement lies in $E$, it lies in $U$; if it avoids $E$, it also avoids $E\setminus U$. Consequently every basic neighborhood contains a nonempty open refinement disjoint from $E\setminus U$. Since such a refinement is also disjoint from the closure of $E\setminus U$, this difference is star-nowhere dense. Thus $E△U=E\setminus U$ is star-meagre, and $E$ is star-Baire. We have provedFinally, $\Omega$ is not star-meagre: otherwise apply the fusion lemma to its purported meagre cover inside any nonempty basic neighborhood, obtaining a nonempty $[s,B]$ disjoint from $\Omega$, a contradiction.

The printed paper does not define $\tau$. In the coarser Ramsey cone topology, whose basic open sets are $[A{]}^{\omega }$ without finite stems, $\Omega$ is $\tau$-meagre. To prove this, let $D_j=\{X:\min X=j\}$. Its cone-topology closure is $H_j=\{X:j\in X\}$: every cone neighborhood of a point containing $j$ has a subset with least element $j$, whereas if $j\notin X$ the neighborhood $[X{]}^{\omega }$ avoids $D_j$. No nonempty cone lies in $H_j$, since its infinite ground set can be thinned to remove $j$. Therefore each $D_j$ is nowhere dense, but $\Omega ={\bigcup }_{j\ge 1}{D}_j$. This proves meagreness of the Ramsey cone topology and exhibits the contrast with the star topology.

If $\tau$ instead denotes the ordinary topology on infinite subsets, with basic cylinders $[s]=\{X:s\text{ is an initial segment of }X\}$, the answer is not $\tau$-meagre. Indeed, given countably many nowhere dense sets, extend a finite increasing prefix successively so that its cylinder avoids the closure of the next set. Make the prefix longer at each step. Their union is an infinite increasing enumeration belonging to every chosen cylinder and avoiding the whole proposed cover. The same construction starts inside any nonempty cylinder, so this alternative product topology is also a Baire space. The two conventions have different answers, so the distinction must be made explicitly rather than inferred from the symbol alone.

Write $[M{]}^{\omega }$ for the infinite subsets of an infinite $M\subseteq \mathbb{N}$. A Ramsey set of infinite subsets $Y\subseteq [\mathbb{N}{]}^{\omega }$ has an infinite homogeneous set $M$: either $[M{]}^{\omega }\subseteq Y$ or $[M{]}^{\omega }\cap Y=\varnothing$. The proof below actually supplies homogeneous refinements within every infinite reservoir, and with any fixed initial finite stem; this stronger property is that of a completely Ramsey set.

For a non-Ramsey example, identify two infinite sets when their symmetric difference is finite, and choose one representative $R$ for each equivalence class. Define a two-colouring bywhere $R$ is the representative of $A$'s class. Removing one point changes this parity. Thus, for every infinite $M$, the two subsets $M$ and $M\setminus \{\min M\}$ have opposite colours. The zero-colour family meets every $[M{]}^{\omega }$ and so does its complement. Hence the zero-colour family is not Ramsey. This finite-symmetric-difference parity colouring uses a choice of representatives; no definability or regularity is being claimed for that first example.

To prove the open-set assertion, for a finite increasing set $s$ and an infinite tail $A$ above $\max s$ defineThe star topology, or Ellentuck topology, has these sets as basic open sets. For an empty stem there is no lower-bound restriction. More generally $[s,A]$ always uses only the points of $A$ above $\max s$. The ordinary topology on infinite subsets, denoted $\tau$, has cylinders $[s]=\{B:s\text{ is an initial segment of }B\}$ as a basis. The two topologies must not be confused.

Fix any family $Y$, initially without any openness assumption. An infinite $A$ accepts a stem $s$ if $[s,A]\subseteq Y$, and rejects it if no infinite subset of its tail accepts $s$. Every infinite reservoir has an infinite refinement deciding $s$: take an accepting refinement if one exists, and otherwise the reservoir itself rejects. Acceptance and rejection are both inherited by infinite refinements. These facts follow from the definitions, giving acceptance and rejection of finite stems.

First refine the initial reservoir to decide the empty stem. If it accepts, its infinite subsets already lie in $Y$. Otherwise start from an infinite reservoir rejecting the empty stem. Choose an increasing point $a_1$, then thin its remaining tail finitely many times to decide all subsets of $\{a_1\}$. Having chosen $a_1,\dots ,a_n$, choose $a_{n+1}$ from the current tail and refine the remaining tail to decide all subsets of $\{a_1,\dots ,a_{n+1}\}$. Each stage has only finitely many stems to handle. Let $L=\{a_1,a_2,\dots \}$.

For every finite $s\subset L$, at the stage of its largest point the reservoir was made to decide $s$, and all later chosen points remain inside that reservoir. Heredity therefore makes the tail of $L$ decide $s$. Moreover $L$ still rejects the empty stem. This proves deciding all finite stems by fusion without assuming a separate fusion theorem.

Suppose $L$ rejects a finite $s\subset L$. There are only finitely many $a\in L$ above $\max s$ for which $L$'s tail above $a$ accepts $s\cup \{a\}$. For if there were infinitely many, collect them into $C$. Every infinite $B\subset C$ has least point $a$ of this kind, and its remaining tail lies in the accepting tail of $L$. Hence $s\cup B\in Y$ for every such $B$, so $C$ would accept $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

Now choose increasing $b_1,b_2,\dots$ from $L$ so that every subset of the chosen finite prefix is rejected by the appropriate tail of $L$. The empty prefix is rejected. At the next step, for each of the finitely many already chosen subsets $s$, avoid the finitely many accepting successors just identified. All other successors are rejected, since the previous fusion made every finite stem in $L$ decided. Thus one can choose the next point beyond the finite forbidden union. With $M=\{b_1,b_2,\dots \}$, every finite $s\subset M$ is rejected.

Now assume $Y$ is star-open. If some infinite $B\subset M$ lay in $Y$, there would be a basic neighbourhood $[s,A]\subseteq Y$ containing $B$. Here $s$ is a finite initial segment of $B$, and its infinite tail $C=B\setminus s$ lies in $A$ and in the relevant tail of $L$. Then $[s,C]\subseteq [s,A]\subseteq Y$, so $C$ accepts $s$, contradicting rejection. Therefore $[M{]}^{\omega }\cap Y=\varnothing$. In the earlier acceptance case, $[M{]}^{\omega }\subseteq Y$. We have provedThe construction works with any infinite starting reservoir. Keeping an initial finite stem fixed and applying the same acceptance/rejection argument to its tails proves the completely Ramsey set conclusion as well. This supplies the full fusion proof for open Ellentuck sets; no unproved course combinatorial lemma has been used.

For the final request, a family has the Baire property in the ordinary infinite-subset topology if it differs from a $\tau$-open family by a $\tau$-meagre set. We construct a non-Ramsey family which is already $\tau$-nowhere dense. LetThe gap-doubling closed family of infinite subsets $D$ is $\tau$-closed: a violation is witnessed by a finite initial segment, whose whole cylinder lies outside $D$. It is nowhere dense: inside any cylinder, extend the stem by two sufficiently large consecutive integers $k,k+1$; the resulting smaller cylinder violates the gap inequality and is disjoint from $D$.

For every infinite $M$, the set $D\cap [M{]}^{\omega }$ has cardinality $\mathfrak{c}=2^{{\aleph }_0}$. To prove the lower bound, build a binary tree of finite gap-doubling sequences using points of $M$, choosing two distinct next points above twice the previous point at every node. Distinct infinite binary branches give distinct increasing sets. The upper bound follows because all these sets are subsets of the countable set $\mathbb{N}$.

Well-order all infinite subsets as $(M_{\alpha }{)}_{\alpha <\mathfrak{c}}$, using the initial ordinal of that cardinality. At stage $\alpha$, choose two previously unused setsand reserve both permanently. There are fewer than $\mathfrak{c}$ previously reserved sets but $\mathfrak{c}$ candidates, so this recursion always continues. This argument does not assume that the continuum is a regular cardinal. Set $X=\{R_{\alpha }:\alpha <\mathfrak{c}\}$. Every cone $[M_{\alpha }{]}^{\omega }$ contains its red choice in $X$ and its blue choice outside $X$; the blue choices can never become later red choices. Therefore $X$ is not Ramsey.

But $X\subseteq D$, and $D$ is closed nowhere dense, so the closure of $X$ also has empty interior. Thus $X$ is nowhere dense, hence meagre, and $X△\varnothing =X$ proves its ordinary Baire property. We have obtained a meagre set meeting every Ramsey cone in both colours:This explains why ordinary Baire regularity cannot replace the star-topology regularity in the open-set theorem. A family supported only on the infinite subsets of a fixed sparse set would not suffice for this counterexample: another infinite set could give an entirely disjoint cone. The gap-doubling family used here instead meets every cone in continuum many candidates.

Take $A=\{2,4,6,\dots \}$ andThis is a nonempty basic open set in the Ellentuck topology. It is a closed set in the ordinary topology on infinite subsets: an infinite set outside $O$ contains an odd integer, and its finite initial segment through that integer gives an ordinary cylinder disjoint from $O$.

Its ordinary interior is empty. In any nonempty ordinary cylinder $[s]$, choose an odd integer $b>\max s$ and continue $s\cup \{b\}$ by any infinite increasing tail. The resulting point belongs to $[s]$ and lies outside $O$. Indeed the smaller cylinder $[s\cup \{b\}]$ is disjoint from $O$. Since $O$ is ordinary-closed with empty ordinary interior, it is a nowhere dense set for $\tau$. Hence

The two Ramsey properties and the topologies. Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, and $[A{]}^{\omega }$ for the infinite subsets of $A$. The positive integers are used throughout. A Ramsey set of infinite subsets $E$ has the property that every infinite $A$ contains an infinite $B$ for which $[B{]}^{\omega }$ lies wholly in $E$ or wholly in $E^c$.

For finite $s$ and infinite $A$ above $\max s$, letThese are the basic neighbourhoods of the Ellentuck topology, denoted $\ast$ in the question. They form a basis: if $X$ belongs to two such sets, their stems are comparable initial segments of $X$; take the longer stem and the intersection of the two tails beyond that stem. This intersection contains the infinite remaining tail of $X$ and yields a basic set contained in both.

A completely Ramsey set $E$ allows such a homogeneous refinement $[s,B]$ inside every $[s,A]$, with $B\subseteq A$ infinite and the stem $s$ kept fixed. Taking $s=\varnothing$ shows that complete Ramseyness implies Ramseyness. The ordinary topology on infinite subsets $\tau$ has basic cylindersIt is the usual product topology on increasing enumerations and is coarser than the Ellentuck topology.

Examples distinguishing the properties. A non-Ramsey example uses the axiom of choice. Let $\kappa =2^{{\aleph }_0}$ and enumerate all infinite sets as $(A_{\alpha }{)}_{\alpha <\kappa }$. Each $[A_{\alpha }{]}^{\omega }$ has cardinality $\kappa$: an enumeration of $A_{\alpha }$ gives the upper bound, and choosing one element from each of infinitely many disjoint pairs gives an injection from all binary sequences for the lower bound.

By transfinite recursion, choose two distinct, previously unused elements $X_{\alpha },Y_{\alpha }\in [A_{\alpha }{]}^{\omega }$ at each stage. Fewer than $\kappa$ elements have been used at stage $\alpha <\kappa$, so this is possible without any assumption on the Continuum hypothesis. Put $R=\{X_{\alpha }:\alpha <\kappa \}$. All the $Y_{\alpha }$ remain outside $R$, since subsequent choices also avoid previously chosen elements. Every $[A{]}^{\omega }$ contains one $X_{\alpha }$ and one $Y_{\alpha }$, so $R$ is not a Ramsey set of infinite subsets. This is the non-Ramsey set from transfinite selection construction.

Perform the same construction on $T=\mathbb{N}\setminus \{1\}$, giving a non-Ramsey set $R_T\subseteq [T{]}^{\omega }$, and putFor any infinite $A$, the infinite set $B=A\setminus \{1\}$ has $[B{]}^{\omega }\cap E=\varnothing$, so $E$ is a Ramsey set of infinite subsets. But no $[\{1\},B]$ with infinite $B\subseteq T$ is homogeneous, because $[B{]}^{\omega }$ meets both $R_T$ and its complement. Therefore $E$ is not a completely Ramsey set.

Every star-open set is completely Ramsey. We give the full fusion proof for open Ellentuck sets, without quoting an infinite Ramsey theorem. Fix a star-open set $O$ and a basic set $[s,A]$. For finite $a\subseteq A$ and an infinite tail $D\subseteq A$ above $\max a$, say $D$ accepts $a$ if $[s\cup a,D]\subseteq O$; it rejects $a$ if no infinite subset of $D$ accepts $a$. Both properties persist under passing to infinite subsets. Every infinite tail can be thinned to decide $a$: take an accepting subset if one exists, and otherwise the original tail rejects it.

First thin $A$ to decide $a=\varnothing$. If it accepts, the required homogeneous neighborhood has already been found. Otherwise keep a rejecting tail. We now fuse to an infinite set $B$ deciding every finite $a\subseteq B$, where the decision for $a$ is made by $B/a=\{b\in B:b>\max a\}$.

To do this, choose $b_1<b_2<\cdots$ successively from an infinite current tail. After choosing $b_j$, thin the tail above $b_j$ to decide each of the finitely many subsets of $\{b_1,\dots ,b_j\}$, one after another, before choosing the next element. For any finite $a\subseteq B$, at the stage when its largest element is selected, the current tail decides $a$ and contains every later selected element. Heredity therefore gives a decision on $B/a$. For $a=\varnothing$, rejection persists from the initial tail.

If $B/a$ rejects $a$, only finitely many $n\in B/a$ can have $B/n$ accepting $a\cup \{n\}$. Otherwise let $D$ be the infinite set of those $n$. Every member of $[s\cup a,D]$ has a first new element $n$, and its remaining tail lies in $B/n$. It therefore belongs to $O$ by acceptance of $a\cup \{n\}$. This would make $D$ accept $a$, contradicting rejection.

Thin $B$ a second time to $C=\{c_1<c_2<\cdots \}$ so that every finite $a\subseteq C$ is rejected. Start with the rejected empty set. At a finite stage, all subsets of the selected elements are rejected. For each of these finitely many subsets, avoid its finitely many next elements that would give an accepted extension, and choose the next $c_j$ beyond all excluded elements. Since $B$ decides every finite extension, each new extension is rejected. Induction gives the asserted property of $C$.

We claim $[s,C]\cap O=\varnothing$. If $X\in [s,C]\cap O$, openness provides a basic set $[t,D]\subseteq O$ containing $X$. Extend $t$, if necessary, to an initial segment $u$ of $X$ containing $s$. Write $u=s\cup a$ with finite $a\subseteq C$. The infinite tail $X\setminus u$ lies in $B/a$ andIt is therefore an accepting subtail for $a$, contradicting rejection. We have found either an accepting neighborhood or a neighborhood disjoint from $O$. Thus

Basic neighborhoods are clopen. Consider $[s,A]$ and a point $X$ outside it. If $s$ is not an initial segment of $X$, take an initial segment $t$ of $X$ of length at least $|s|$; its ordinary cylinder $[t]$ cannot meet $[s,A]$. Otherwise some element of $X\setminus s$ is outside $A$; take $t$ through that element, and again $[t]$ is disjoint from $[s,A]$.

Thus the complement of $[s,A]$ is already open in the ordinary topology on infinite subsets, and hence in the finer Ellentuck topology. A basic neighbourhood is open by definition, so

This topology has basic open sets $[A{]}^{\omega }$ for infinite ground sets $A$, without fixed finite stems. It is coarser than the Ellentuck topology and differs from the ordinary topology on infinite subsets, whose basic cylinders fix an initial segment. The symbol $\tau$ is sometimes used for either coarser convention, so it should be accompanied by a definition.

A family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is Ramsey in the homogeneous-cone sense if there is an infinite $M\subseteq \mathbb{N}$ such that either $[M{]}^{\omega }\subseteq \mathcal{A}$ or $[M{]}^{\omega }\cap \mathcal{A}=\varnothing$. This is the existence of one infinite homogeneous cone. Requiring such a refinement inside every infinite initial ground set gives the stronger definition used for a Ramsey set of infinite subsets; requiring a homogeneous refinement in every finite-stem neighbourhood gives a completely Ramsey set.

These quantifiers should be distinguished. For example, the existence of a homogeneous cone on the even numbers does not constrain a family's restriction to the odd numbers. If $\mathcal{E}$ is the finite-symmetric-difference parity colouring class on the odd numbers, the family consisting of $\mathcal{E}$ alone has a homogeneous cone disjoint from it on the even numbers, but no homogeneous cone on the odd numbers. Every open set in the ordinary topology on infinite subsets has homogeneous cones inside all infinite ground sets, as follows by deciding finite stems through fusion.