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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

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 $[M{]}^{\omega }$ 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.

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.

Let E be the even positive integers, and take a finite-symmetric-difference parity colouring $\epsilon$ of $[E{]}^{\omega }$. The family $\{\{1\}\cup X:X\in [E{]}^{\omega },\epsilon (X)=0\}$ is Ramsey even under the every-ground-set convention, since every infinite ground set can be thinned to omit 1. It is not completely Ramsey: inside $[\{1\},E]$, every infinite tail has two subfamilies of opposite parity. Its supporting neighbourhood is closed and nowhere dense in the ordinary topology on infinite subsets, but open in the Ellentuck topology.