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A family $Y\subseteq [\mathbb{N}{]}^{\omega }$ is $\tau$-Baire when $Y△O$ is meagre for some ordinarily open family $O$. This is weaker than the corresponding regularity in the star topology. A meagre set meeting every Ramsey cone in both colours has the ordinary Baire property while failing to be a Ramsey set of infinite subsets.

A set $E\subseteq [\mathbb{N}{]}^{\omega }$ is completely Ramsey if every basic Ellentuck topology neighborhood $[s,A]$ admits an infinite $B\subseteq A$ with $[s,B]\subseteq E$ or $[s,B]\cap E=\varnothing$. Keeping the finite stem $s$ makes this stronger than being a Ramsey set of infinite subsets.

Choose one representative $R$ for each equivalence class of infinite sets under finite symmetric difference. The colour $|A△R|\mathrm{mod}2$ changes on removing one point. Every cone $[M{]}^{\omega }$ therefore contains opposite-coloured sets $M$ and $M\setminus \{\min M\}$. Either colour family is a non-Ramsey set of infinite subsets. Choice of representatives is essential to this construction, which does not claim a definable regular family.

Using the axiom of choice, enumerate all infinite subsets $A_{\alpha }$ of the positive integers in order type $2^{{\aleph }_0}$. By transfinite recursion, choose two previously unused infinite subsets $X_{\alpha },Y_{\alpha }$ of each $A_{\alpha }$. This is possible because each $[A_{\alpha }{]}^{\omega }$ has cardinality $2^{{\aleph }_0}$ and fewer choices have been made at stage $\alpha$. The set $\{X_{\alpha }\}$ is not a Ramsey set of infinite subsets, since every $[A_{\alpha }{]}^{\omega }$ meets both it and its complement.

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

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 $[\mathbb{N}{]}^{\omega }$ 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.

The Rosenthal l1 theorem states that every bounded sequence in a real or complex Banach space has either a weakly Cauchy subsequence or a subsequence equivalent to the unit vector basis of ${\ell }^1$. Equivalence means that there are $0<c\le C<\infty$ such that $c\sum |a_i|\le \Vert \sum a_i{x}_{n_i}\Vert \le C\sum |a_i|$ for every finite scalar combination.

We use the infinite Ramsey result that every closed subset of $[\mathbb{N}{]}^{\omega }$ is a Ramsey set of infinite subsets. It follows by taking complements in the open-set theorem proved in Question 6(a), and is also among the infinite Ramsey results permitted here.

Let $E$ be the closed span of the sequence and $K=B_{E^{\ast }}$ with its weak-star topology. It is compact by the Banach-Alaoglu theorem. The functions $f_n(\varphi )=\varphi (x_n)$ are continuous and uniformly bounded. Pointwise convergence of a subsequence on $K$ is precisely the weakly Cauchy conclusion.

Choose a countable collection of pairs $(Q_0,Q_1)$ of disjoint closed rational boxes in the scalar plane, or intervals for real scalars. Write $u,v$ for their centres, $d=|u-v|>0$, and require their diameters to be at most some $\eta <d/(4\pi )$. Include arbitrarily small such boxes, so any two distinct scalar cluster points lie in the interiors of one such pair. For a fixed pair, say an infinite set $M$ oscillates if some $\varphi \in K$ has $f_n(\varphi )\in Q_0$ and $f_n(\varphi )\in Q_1$ each for infinitely many $n\in M$.

There are two possibilities. If at some pair and some infinite $M$ every infinite subset oscillates, retain them. Otherwise enumerate the box pairs and successively thin so that no point oscillates for the current pair. A diagonal subsequence is eventually in each refined set, so has no point oscillating for any pair. A bounded scalar sequence with two cluster points would oscillate for a pair of boxes enclosing those points. Hence every scalar evaluation on that subsequence converges, giving the weakly Cauchy alternative.

In the remaining case put $A_n=f_n^{-1}(Q_0)$ and $B_n=f_n^{-1}(Q_1)$, closed subsets of $K$. ConsiderThis is closed. If the infinite intersection is empty, compactness gives an empty finite intersection, and every infinite sequence with the same sufficiently long initial segment also has an empty intersection. The complement is therefore open. Apply closed-set Ramsey to obtain an infinite $L\subseteq M$ homogeneous for $\mathcal{C}$.

It cannot be homogeneous outside $\mathcal{C}$: the retained oscillation property provides a point visiting both level sets infinitely often on $L$, from which we can select an alternating sequence belonging to $\mathcal{C}$. Thus every infinite subset of $L$ is in $\mathcal{C}$.

Pass to the even-indexed members $L^{\prime }=(l_2,l_4,\dots )$. The pairs $(A_n,B_n)$ for $n\in L^{\prime }$ are independent: every assignment of either $A$ or $B$ to finitely many chosen indices has a common point. Indeed insert zero or one unused members of $L$ before and between the specified indices to arrange their required odd or even positions, then complete to an infinite sequence. The gaps between even-indexed members supply these spare indices. Membership of that infinite sequence in $\mathcal{C}$ supplies the point realizing the finite assignment.

Take any finite complex coefficients $a_i$. Some unit scalar $\zeta$ satisfiesbecause the average over the unit circle is exactly the right side. Put ${\epsilon }_i=\mathrm{sign}\mathrm{Re}(\zeta a_i)$, choosing either sign at zero. Independence supplies functionals $\varphi ,\psi \in K$ such that $f_{n_i}(\varphi )$ lies near $u$ and $f_{n_i}(\psi )$ near $v$ when ${\epsilon }_i=1$, with the assignments reversed when ${\epsilon }_i=-1$. ThusAlso $|\sum {\epsilon }_i{a}_i|\ge \sum |\mathrm{Re}(\zeta a_i)|$. Since $\Vert \varphi -\psi \Vert \le 2$, we concludeThe coefficient is positive by the box choice, and boundedness of the original sequence gives the upper estimate. This proves the ${\ell }^1$ alternative over complex scalars. For real scalars simply partition by the signs of the coefficients; the same calculation works and gives the stronger lower constant $d/2-\eta$. This completes the proof.

Identify $[\mathbb{N}{]}^{\omega }$ with increasing infinite sequences, with the product topology. A Ramsey set of infinite subsets $\mathcal{A}$ is one such that for every infinite $M$, some infinite $L\subseteq M$ satisfies $[L{]}^{\omega }\subseteq \mathcal{A}$ or $[L{]}^{\omega }\cap \mathcal{A}=\varnothing$. We prove this for open $\mathcal{A}$.

For a finite increasing stem $s$ and an infinite tail $B$ above it, say that $B$ accepts $s$ if every infinite $N\subseteq B$ has $s^{⌢}N\in \mathcal{A}$. Say that $B$ rejects $s$ if no infinite subset of $B$ accepts it. Either a tail has an accepting refinement, or it rejects. Both acceptance and rejection persist under passage to infinite subsets.

Starting with any infinite ground set $M$, use fusion to obtain an infinite $B$ deciding every finite subset of $B$: after selecting its first $n$ elements, refine the remaining tail successively to decide each of the finitely many subsets of that prefix. Then select the next element from the refined tail. The final tail beyond a decided stem is contained in the tail which decided it, so the decision survives. In particular $B$ decides the empty stem.

If $B$ accepts the empty stem, it is homogeneous inside $\mathcal{A}$. Otherwise it rejects. We thin it further so that every finite subset is rejected. The key observation is that if $B/s$ rejects $s$, there can be only finitely many $n\in B$ above $s$ for which $B/n$ accepts $s^{⌢}n$. If there were infinitely many, call their set $C$. For any infinite $N\subseteq C$, its first element $n$ and the accepting tail $B/n$ would imply $s^{⌢}N\in \mathcal{A}$. Thus $C$ would accept $s$, contradicting rejection.

Build $L$ inductively. Given a finite prefix whose subsets are all rejected, exclude the finitely many bad next elements for each of those subsets, and choose the next element outside their union. All new stems are decided by the first fusion and are not accepted, hence are rejected. This preserves the induction.

If an infinite $N\subseteq L$ belonged to the open set $\mathcal{A}$, some finite initial segment $s$ of $N$ would have every infinite extension above $\max s$ in $\mathcal{A}$. Then $B/s$ would accept $s$, contradicting the rejection just arranged. Therefore $[L{]}^{\omega }$ misses $\mathcal{A}$. We have provedTaking complements gives the same result for closed subsets. The proof also works with any fixed finite stem, yielding the stronger finite-stem version when desired.

LetThis scalar function is continuous. The first element is locally constant; deleting it is continuous on increasing sequences; and evaluation on a fixed $x_m$ is continuous in the weak-star topology. Thus $\mathcal{U}=\{M:|h(M)|<\delta \}$ is open.

By part (a), choose an infinite $L_0$ homogeneous for $\mathcal{U}$. If all its infinite subsets are in $\mathcal{U}$, the required conclusion follows. Suppose instead every one has $|h(M)|\ge \delta$. Choose finitely many unit scalars ${\zeta }_1,\dots ,{\zeta }_q$ such that every scalar $z$ with $|z|\ge \delta$ has $\mathrm{Re}({\zeta }_{j}z)>\delta /2$ for some $j$; for real scalars the two signs suffice. The open setscover $[L_0{]}^{\omega }$. Apply the open Ramsey set of infinite subsets result successively to them. After finitely many refinements, at least one, say ${\mathcal{U}}_j$, contains every infinite subset of a remaining infinite $L$; otherwise their cover would be violated.

For any finite $F\subseteq L$, choose an infinite $N\subseteq L$ after $\max F$. For each $m\in F$, apply this homogeneity to $M=\{m\}\cup N$. Its deleted tail is always the same $N$, so the same functional satisfiesBut $(x_m{)}_{m\in L}$ is a weakly null sequence. The Mazur lemma gives a finite convex combination $z={\sum }_{m\in F}{\lambda }_m{x}_m$ with $\Vert z\Vert <\delta /2$. The displayed inequalities give $\mathrm{Re}({\zeta }_j{x}_N^{\ast }(z))>\delta /2$, contradicting $\Vert x_N^{\ast }\Vert \le 1$. The bad homogeneous case is impossible. ThereforeContinuity is used to make the scalar tests open; weak nullity is used only for the finite convex-combination contradiction. The argument works over both real and complex scalars.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.