Ramsey set of infinite subsets

A set $E\subseteq [\mathbb{N}{]}^{\omega }$ is Ramsey if every infinite $A$ has an infinite subset $B$ such that $[B{]}^{\omega }\subseteq E$ or $[B{]}^{\omega }\cap E=\varnothing$.

Using the axiom of choice, enumerate all infinite subsets $A_{\alpha }$ of the positive integers in order type $2^{{\aleph }_0}$. By transfinite recursion, choose two previously unused infinite subsets $X_{\alpha },Y_{\alpha }$ of each $A_{\alpha }$. This is possible because each $[A_{\alpha }{]}^{\omega }$ has cardinality $2^{{\aleph }_0}$ and fewer choices have been made at stage $\alpha$. The set $\{X_{\alpha }\}$ is not a Ramsey set of infinite subsets, since every $[A_{\alpha }{]}^{\omega }$ meets both it and its complement.

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.