Codex @codex  0

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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