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 $[B{]}^{\omega }\cap E=\varnothing$.

Every open set $O$ in the ordinary topology on infinite subsets is a Ramsey set of infinite subsets: some infinite $B$ has $[B{]}^{\omega }\subseteq O$ or $[B{]}^{\omega }\cap O=\varnothing$. Use acceptance and rejection of finite stems and fusion to decide every finite stem. If the empty stem is rejected, each rejected stem has only finitely many accepting one-point extensions. Successive avoidance of these finite obstructions gives an infinite set all of whose finite stems are rejected; openness then excludes every infinite subset from $O$. This concerns increasing enumerations of infinite subsets, rather than unrestricted sequences with repetitions.

Choose one representative $R$ for each equivalence class of infinite sets under finite symmetric difference. The colour $|A△R|\mathrm{mod}2$ changes on removing one point. Every cone $[M{]}^{\omega }$ therefore contains opposite-coloured sets $M$ and $M\setminus \{\min M\}$. Either colour family is a non-Ramsey set of infinite subsets. Choice of representatives is essential to this construction, which does not claim a definable regular family.

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.

Let E be the even positive integers, and take a finite-symmetric-difference parity colouring $\epsilon$ of $[E{]}^{\omega }$. The family $\{\{1\}\cup X:X\in [E{]}^{\omega },\epsilon (X)=0\}$ is Ramsey even under the every-ground-set convention, since every infinite ground set can be thinned to omit 1. It is not completely Ramsey: inside $[\{1\},E]$, every infinite tail has two subfamilies of opposite parity. Its supporting neighbourhood is closed and nowhere dense in the ordinary topology on infinite subsets, but open in the Ellentuck topology.

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.

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.