Fusion proof for open Ellentuck sets

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.

Relative to a family $Y$, an infinite reservoir accepts a finite stem $s$ when its entire Ellentuck topology neighbourhood $[s,A]$ lies in $Y$. It rejects $s$ when no infinite refinement accepts it. Every reservoir has a refinement deciding a prescribed stem, and both outcomes are hereditary under further infinite thinning. These definitions turn open-set homogeneity into a fusion construction.

Suppose a reservoir rejects $s$ and decides every extension $s\cup \{a\}$. Only finitely many of those one-point extensions can have accepting tails. If infinitely many did, collect their new points into $C$; every infinite subset of $C$ begins with an accepting successor, so $[s,C]$ would lie in the family. That would make $C$ accept $s$, contradicting rejection. This finite-obstruction fact lets a second fusion preserve rejection of every finite stem.

Choose successive points and, after each choice, thin the unused reservoir to decide all subsets of the finite chosen prefix. There are only finitely many stems to handle at each stage. The final diagonal infinite set decides every one of its finite stems, because its relevant tail lies in the reservoir chosen when the stem's largest point was selected. The argument uses the hereditary decisions from acceptance and rejection of finite stems.