 Solution

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 $B$.