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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 $[A{]}^{\omega }$ for the infinite subsets of an infinite $A\subseteq \mathbb{N}$. A Ramsey set of infinite subsets is a family $E\subseteq [\mathbb{N}{]}^{\omega }$ such that each infinite ground set $A$ has an infinite subset $B$ withThe convention asking only for one such homogeneous cone is weaker; the proof below establishes the displayed property and, in fact, the stronger completely Ramsey set property.

For a finite stem $s$ and an infinite reservoir $A$ lying above every element of $s$, putThese sets form the basic open neighbourhoods of the Ellentuck topology, or star topology. Here $s$ is an initial segment of every member of $[s,A]$. A star-Borel set means a member of the sigma-algebra generated by these open sets. A family is completely Ramsey if every $[s,A]$ has a homogeneous refinement $[s,B]$ with $B\subseteq A$ infinite and the same stem $s$. We prove this for every star-Borel set, beginning with open sets.

Fix a star-open family $O$ and a basic neighbourhood $[s,A]$. An infinite reservoir $T$ above a finite stem $t$ accepts $t$ if $[t,T]\subseteq O$. It rejects $t$ if no infinite subset of $T$ accepts $t$. A reservoir decides $t$ if it accepts or rejects it. Any reservoir has a refinement deciding a prescribed stem: take an accepting refinement if there is one, and otherwise the current reservoir rejects. Both kinds of decision persist under infinite thinning.

First refine $A$ to decide $s$. If the decision is acceptance, the required homogeneous neighbourhood has already been found. Suppose it is rejection. We use deciding all finite stems by fusion to selectfrom nested infinite reservoirs. After selecting $b_n$, successively thin the remaining reservoir to decide every stem $s\cup a$ with $a\subseteq \{b_1,\dots ,b_n\}$. There are only finitely many such $a$, so every stage can be completed. Choose $b_{n+1}$ from that thinned reservoir and repeat. Let $B=\{b_1,b_2,\dots \}$. For every finite $a\subseteq B$, the taildecides $s\cup a$; for $a=\varnothing$, use all of $B$. Indeed, the stem was decided at the stage when its largest point was selected, and its final tail is a subset of the reservoir retained at that stage. The empty stem remains rejected by the initial decision.

The crucial finite-obstruction fact is that a rejected stem has only finitely many accepting one-point extensions. More precisely, suppose $B/a$ rejects $s\cup a$. If infinitely many $b\in B/a$ had $B/b$ accepting $s\cup a\cup \{b\}$, collect those $b$ into an infinite $C$. Any $X\in [C{]}^{\omega }$ has a least point $b$, and its remaining tail lies in $B/b$. Acceptance at $b$ would then imply $s\cup a\cup X\in O$. Thus $C$ would accept $s\cup a$, contradicting rejection. This proves the finitely many accepting extensions of a rejected stem assertion with the reservoirs specified.

Thin $B$ a second time. Having selected a finite prefix of $C$, maintain that every stem $s\cup a$ with $a$ a subset of that prefix is rejected relative to its tail in $B$. Each such stem has finitely many forbidden next points leading to acceptance. Choose the next point of $C$ above the prefix and outside the union of these finitely many finite forbidden sets. The deciding property of $B$ then makes every newly created stem rejected. Starting with the rejected empty extension gives an infinite $C\subseteq B$ all of whose finite extensions of $s$ are rejected.

We claim $[s,C]\cap O=\varnothing$. Otherwise take $Y\in [s,C]\cap O$. Star-openness supplies a basic neighbourhood inside $O$ containing $Y$. Extend its finite stem along $Y$ until it contains $s$; the new stem has the form $s\cup a$ for a finite initial segment $a$ of $Y\setminus s$. The remaining infinite tail of $Y$ is a subset of $B/a$ and accepts that stem, contradicting rejection. This proves the fusion proof for open Ellentuck sets: every star-open set is completely Ramsey.

Next call $N$ completely Ramsey-null if every $[s,A]$ has a same-stem refinement disjoint from $N$. Every star-nowhere dense set is completely Ramsey-null. Indeed, the complement $D$ of its star-closure is star-open and dense. Apply the just-proved open-set result inside $[s,A]$. The homogeneous alternative disjoint from $D$ would be a nonempty open set contained in the closure of a nowhere-dense set, which is impossible. Hence some $[s,B]$ lies entirely in $D$ and avoids $N$.

Countable unions of completely Ramsey-null sets are again completely Ramsey-null. Here is the full Ellentuck meagre-set fusion lemma. Given $N_1,N_2,\dots$ and $[s,A]$, construct selected prefixes $P_j=\{b_1,\dots ,b_j\}$ and nested unused infinite reservoirs. At stage $j$, refine the reservoir, still lying above $\max P_j$, to arrangeEach constraint can be imposed by complete Ramsey-nullness, with all earlier constraints preserved by thinning; there are finitely many constraints. Choose $b_{j+1}\in T_j$ and retain the tail above it for the next stage. Start with $P_0=\varnothing$ and the reservoir $A$.

Let $B=\{b_1,b_2,\dots \}$. If $Y\in [s,B]$ and $i$ is fixed, take any $j\ge i-1$ and put $a=(Y\setminus s)\cap P_j$. Every remaining point of $Y\setminus s$ lies in $T_j$, so $Y\in [s\cup a,T_j]$ and $Y\notin N_i$. ThereforeThis construction handles all subsets of a selected prefix, not merely the full prefix: members of $[s,B]$ may skip arbitrarily many selected points. Since a meagre set is a countable union of nowhere-dense sets, every star-meagre set is completely Ramsey-null.

It remains to pass from star-open sets to star-Borel sets. Let $\mathcal{C}$ consist of the families $E$ expressible aswhere $O$ is star-open and $M$ is star-meagre. We verify directly that $\mathcal{C}$ is a sigma-algebra. It contains every open set. Countable unions remain in $\mathcal{C}$, becauseand a countable union of meagre sets is meagre. For complements, replace $O^c$ by the open set $\mathrm{int}(O^c)$. The omitted part is $\overline{O}\setminus O$, the boundary of the open set $O$, which is closed and nowhere dense: a nonempty open subset of $\overline{O}$ must meet $O$, so it cannot lie in this boundary. Thusa meagre set. Subsets of meagre sets are meagre as well, since taking a subset preserves nowhere denseness. These observations prove the claimed sigma-algebra property. Hence every star-Borel set belongs to $\mathcal{C}$.

Finally, for $E△O\subseteq M$ and any $[s,A]$, first use complete Ramsey-nullness to obtain $[s,B]$ avoiding $M$. On that neighbourhood $E$ and $O$ coincide. Apply the open-set result to $O$ within $[s,B]$ to get a homogeneous $[s,C]$. ThereforeTaking $s=\varnothing$ gives the requested homogeneous infinite-subset cone inside any given infinite ground set. Every closure, interior, open set and meagreness used in this proof refers to the Ellentuck topology; substituting the coarser ordinary topology at an intermediate step would change the assertion.

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.

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