Completely Ramsey-null set

A set $N$ is completely Ramsey-null when every Ellentuck topology neighborhood $[s,A]$ has a refinement $[s,B]$ disjoint from $N$, with the same finite stem $s$. Assuming open Ellentuck sets are completely Ramsey, every star-nowhere dense set is completely Ramsey-null: apply the open-set property to the dense complement of its closure. The Ellentuck meagre-set fusion lemma shows that countable unions retain this avoidance property.

Countably many completely Ramsey-null sets can be avoided simultaneously in a stem-preserving Ellentuck topology refinement. Select an increasing sequence $b_j$ with nested infinite remaining tails. At stage $j$, thin the tail to avoid the $j$th set for every stem $s\cup t$, where $t$ ranges over all subsets of the first $j$ selected points. There are only finitely many such stems. Any infinite subset of the final selected sequence uses some such $t$ before stage $j$ and has all remaining points in the thinned tail. Thus it avoids every forbidden set. This explains why an argument checking only the full selected prefix is insufficient.