Completely Ramsey set
= 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>.