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A subset of the space of infinite subsets of the natural numbers has this property when it differs from an Ellentuck topology open set by a meagre set in the same topology. By the complete-Ramsey characterization, these are precisely the completely Ramsey sets. Using meagreness from a different topology changes the definition.

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.

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.

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.

Identify ${\mathbb{N}}^{(\omega )}$ with the space of infinite subsets of the natural numbers. A family $E$ has an infinite homogeneous set for a colouring if some infinite $M$ satisfiesThis is the Ramsey family in the homogeneous-cone sense formulation. The relative Ramsey set of infinite subsets formulation asks for such an $M$ inside every prescribed infinite ground set. We prove the stronger completely Ramsey assertion, keeping an arbitrary finite stem fixed.

For a finite stem $s$ and an infinite tail $A$ with $\max s<\min A$ (taking $\max \varnothing =0$), writeThese neighbourhoods generate the Ellentuck topology, or star topology. A star-Borel set belongs to the sigma-algebra generated by the open sets of this topology. It is completely Ramsey if every $[s,A]$ has an infinite $B\subseteq A$ with $[s,B]\subseteq E$ or $[s,B]\cap E=\varnothing$. Taking $s=\varnothing$ gives both preceding Ramsey formulations.

We first prove the fusion proof for open Ellentuck sets. Fix a star-open set $U$ and a starting neighbourhood $[s,A]$. A reservoir $C$ above a finite extension $s\cup t$ accepts $t$ when $[s\cup t,C]\subseteq U$. It rejects $t$ when no infinite subreservoir accepts it. Every infinite reservoir has a refinement deciding $t$: if an accepting subreservoir exists choose one, and otherwise the reservoir already rejects. Both acceptance and rejection persist under further infinite thinning. These are acceptance and rejection of finite stems.

First refine $A$ to decide the empty extension. Choose $d_0$ from the refined reservoir and thin the remaining tail to decide every subset of $\{d_0\}$. After choosing $d_0,\dots ,d_j$, thin the unused infinite tail to decide every subset of this finite prefix before choosing $d_{j+1}$. Only finitely many decisions occur at each stage. Let $D=\{d_0<d_1<\cdots \}$. For each finite $t\subseteq D$, its tail(with $D_{\varnothing }=D$) is contained in the reservoir chosen when the largest point of $t$ was selected. Thus $D_t$ decides $t$. This is deciding all finite stems by fusion.

If $D$ accepts the empty extension, $[s,D]\subseteq U$ and we are done. Otherwise it rejects it. Call a finite $t\subseteq D$ accepted or rejected according to the decision of $D_t$. For each rejected $t$, the setis finite. Indeed, if $G_t$ were infinite, every $X\in [G_t{]}^{\omega }$ would have a least element $a$ with $t\cup \{a\}$ accepted, and its remaining tail lies in $D_{t\cup \{a\}}$. Hence $s\cup t\cup X\in U$ for every such $X$, making $G_t$ an accepting subreservoir for $t$, a contradiction. This proves the finitely many accepting extensions of a rejected stem fact.

Build $B=\{b_0<b_1<\cdots \}\subseteq D$ by choosing each new point outside all the finitely many sets $G_t$ for subsets $t$ of the already selected prefix. Begin with the rejected empty extension. Inductively every finite subset of the new prefix is rejected: the old subsets stay rejected, and each new subset is a rejected $t$ extended by a point outside $G_t$, whose decision must therefore be rejection. Consequently every finite $t\subseteq B$ is rejected.

If a point $X\in [s,B]$ belonged to $U$, star-openness would give a basic neighbourhood $[u,C]\subseteq U$ containing it. Extend $u$ along the increasing enumeration of $X$, if necessary, until its stem is $s\cup t$ for a finite $t\subseteq B$. The infinite tail of $X$ above that stem then provides an accepting subreservoir for $t$, contradicting rejection. Therefore $[s,B]\cap U=\varnothing$. We have proved that every star-open set is completely Ramsey.

A family $N$ is completely Ramsey-null when every $[s,A]$ has a stem-preserving refinement disjoint from $N$. Every star-nowhere dense set has this property. Apply the open-set result to the dense open complement of its star-closure. The disjoint alternative would put an entire basic neighbourhood inside a closure with empty interior, which is impossible. The other alternative gives the required avoidance.

We also need the full Ellentuck meagre-set fusion lemma, not a diagonal argument checking only one prefix. Let $N_0,N_1,\dots$ be completely Ramsey-null, and start inside $[s,A]$. At stage $j$, suppose $b_0,\dots ,b_{j-1}$ have been chosen. Refine the remaining infinite tail successively so thatThe null property permits each refinement with its stem unchanged; previous avoidance survives thinning. There are finitely many $t$ at this stage. Choose $b_j\in C_j$ and continue above it. Let $B=\{b_j:j\ge 0\}$.

For any infinite $Y\subseteq B$ and any fixed $j$, put $t=Y\cap \{b_0,\dots ,b_{j-1}\}$. Its remaining points form an infinite subset of $C_j$, so $s\cup Y\in [s\cup t,C_j]$ and hence $s\cup Y\notin N_j$. This works for every $j$, provingThus countable unions of completely Ramsey-null sets are null. Subsets of null sets are null as well. In particular every star-meagre set, being a countable union of star-nowhere-dense sets, is completely Ramsey-null.

Finally let $\mathcal{B}$ consist of families $E$ such that $E△U$ is contained in a star-meagre set for some star-open $U$. This is the Baire property in the Ellentuck topology. We verify that it forms a sigma-algebra. Open sets belong with empty error. For a countable union, use $U={\bigcup }_j{U}_j$; its error is contained in the countable union of the meagre errors. For complements, replace $U^c$ by its interior. The discrepancy isa closed nowhere-dense set: every open set meeting $\overline{U}$ also meets $U$, so none can lie wholly in this boundary. Adding it to the original meagre error handles complementation. Consequently every star-Borel set belongs to $\mathcal{B}$.

For such an $E$, start with any $[s,A]$. First apply null avoidance to its meagre error to find $[s,B]$ on which $E$ agrees with $U$. Then apply open-set homogeneity to refine it to $[s,C]$ wholly inside or outside $U$, and therefore wholly inside or outside $E$. This is the Baire-property reduction for completely Ramsey families. We concludeAll closures, interiors, openness and meagreness in this argument refer to the star topology; replacing it midway by the ordinary finite-prefix topology would not justify the null-avoidance step.

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 $X=[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, and $[M{]}^{\omega }$ for all infinite subsets of an infinite $M$. A family $Y\subseteq X$ is a Ramsey set of infinite subsets if for every infinite $M$ there is an infinite $L\subseteq M$ with either $[L{]}^{\omega }\subseteq Y$ or $[L{]}^{\omega }\cap Y=\varnothing$.

A non-Ramsey set of infinite subsets is obtained by finite-symmetric-difference parity colouring. Declare $A\sim B$ when their symmetric difference is finite, and use the axiom of choice to select a representative $R$ for each equivalence class. PutFor every infinite $L$, both $L$ and $L\setminus \{\min L\}$ belong to $[L{]}^{\omega }$ and have the same representative. Toggling one element changes the parity of the finite symmetric difference, so these two sets have opposite colors. Thus no $[L{]}^{\omega }$ is homogeneous, and $Y$ is not a Ramsey set of infinite subsets. Only finite differences are counted; no parity is being assigned to an infinite cardinal.

We use the Ellentuck topology for the star notation. Its basic neighborhoods arewhere $s$ is a finite stem (the initial segment of each member of the neighborhood) and every element of the infinite reservoir $A$ exceeds $\max s$; the empty stem is allowed. A star-nowhere dense set $N$ has star-closure with empty star-interior; a star-meagre set is a countable union of star-nowhere dense sets.

We need the following stem-preserving open-set conclusion: a star-open set $G$ has, in every $[s,A]$, an infinite $B\subseteq A$ such that $[s,B]\subseteq G$ or $[s,B]\cap G=\varnothing$. Here is the fusion proof for open Ellentuck sets, so that the precise form being used is explicit. A reservoir $T$ accepts a finite extension $t$ of $s$ when $[t,T]\subseteq G$, and rejects it when no infinite subreservoir accepts it. Every reservoir can be thinned to decide a prescribed stem: take an accepting subreservoir if one exists, and otherwise it already rejects. Acceptance and rejection are hereditary under further infinite thinning.

First thin $A$ to decide $s$. If it accepts, we have finished. Otherwise, choose successive points $b_0<b_1<\cdots$, thinning the unused tail after each selection to decide every stem $s\cup t$ with $t$ a subset of the finitely many points already selected. Each stage requires only finitely many decisions. The final set $B$ therefore decides every such finite extension, using the tail above its largest point. It rejects $s$.

For any rejected stem $s\cup t$, only finitely many possible next points $b$ of $B$ have accepting tails for $s\cup t\cup \{b\}$. Otherwise collect infinitely many such points into $C$. Every member of $[s\cup t,C]$ starts with some such $b$, and its remaining points lie in the accepting tail above $b$. Hence $[s\cup t,C]\subseteq G$, contradicting rejection. A second selection now gives $C\subseteq B$ all of whose finite extensions of $s$ are rejected: at each stage avoid the finitely many forbidden next points for each subset of the chosen prefix. Previous rejections persist on thinning. If $Z\in [s,C]\cap G$, star-openness gives an initial stem $s\cup t$ of $Z$ and an infinite tail of $Z$ whose entire neighborhood lies in $G$. This is an accepting subreservoir for a stem we made rejected, a contradiction. Thus $[s,C]\cap G=\varnothing$, proving the needed open-set conclusion.

Apply this conclusion to $G=X\setminus {\overline{N}}^{\ast }$ when $N$ is star-nowhere dense. This $G$ is star-open and star-dense. The alternative $[s,B]\cap G=\varnothing$ would give a nonempty star-open set inside ${\overline{N}}^{\ast }$, contradicting empty interior. Consequently every star-nowhere dense set can be avoided without extending the finite stem.

Now let $N={\bigcup }_{j\ge 0}{N}_j$, each $N_j$ star-nowhere dense, and fix $[s,A]$. Construct $b_0<b_1<\cdots$ with nested infinite unused reservoirs. At stage $j$, let $P_j=\{b_0,\dots ,b_{j-1}\}$. For every $t\subseteq P_j$, in turn, thin the current reservoir using the preceding avoidance property untilThere are only finitely many such $t$, and subsequent thinning preserves all previous avoidances. Choose $b_j=\min C_j$ and retain the part of $C_j$ above $b_j$ for the next stage. Put $B=\{b_j:j\ge 0\}$. For any $Z\in [s,B]$ and any $j$, the finite part $t=(Z\setminus s)\cap P_j$ is one of the stems tested at stage $j$, and the rest of $Z\setminus (s\cup t)$ lies in $C_j$. Hence $Z\in [s\cup t,C_j]$ and $Z\notin N_j$. This proves $[s,B]\cap N=\varnothing$.

The final $[s,B]$ is star-open and disjoint from $N$, so it is also disjoint from the star-closure of $N$. Every basic neighborhood therefore has a nonempty open refinement missing that closure. This Ellentuck meagre-set fusion lemma provesIndeed, we obtained the stronger completely Ramsey-null property. Testing all subsets of the selected prefix was essential; testing just the entire prefix would not cover every infinite subset of $B$.

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.

The answer is no. Let $X_0=\mathbb{N}$, regarded as one point of the space of infinite subsets of the natural numbers, and putThe displayed union makes $O$ an open set in the Ellentuck topology. On the other hand, every basic neighbourhood $[s,A]$ of $X_0$ contains a different infinite set: keep $s$ and remove one element from its infinite remaining tail. This other point belongs to $O$. Therefore $X_0$ is in the closure of $O$ but not in $O$, soThe fact that basic neighbourhoods are clopen sets does not make arbitrary unions of them closed.

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