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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.