Completely Ramsey set

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.

Every open set in the Ellentuck topology is a completely Ramsey set. Fix a finite stem $s$. An infinite tail accepts a finite extension $a$ if its entire basic neighborhood $[s\cup a,A]$ lies in the open set; it rejects $a$ if no infinite subtail accepts it. Decisions persist under thinning. A fusion chooses an infinite set deciding every finite extension. For any rejected extension, only finitely many possible next elements can yield accepted extensions, since infinitely many would themselves provide an accepting tail. A second thinning therefore makes all finite extensions rejected. Openness then rules out any point of the set in the resulting neighborhood.