Space of infinite subsets of the natural numbers ($[\mathbb{N}{]}^{\omega }$)

The space $[\mathbb{N}{]}^{\omega }$, also written ${\mathbb{N}}^{(\omega )}$, consists of all infinite subsets of the positive integers. An element is identified with its strictly increasing enumeration. If $s$ is finite, write $s⊏X$ when $s$ is the finite initial segment of this enumeration.

A finite stem is a finite initial segment $s$ of an infinite subset of $\mathbb{N}$. The associated neighbourhood in the ordinary topology on infinite subsets is $[s]=\{X:s\text{ is an initial segment of }X\}$. For an infinite tail $A$ above $\max s$, the neighbourhood $[s,A]=\{s\cup B:B\in [A{]}^{\omega }\}$ is basic in the Ellentuck topology. Finite stems allow a fusion argument to fix initial choices while successively thinning the remaining infinite tail.

A family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is Ramsey in the homogeneous-cone sense if there is an infinite $M\subseteq \mathbb{N}$ such that either $[M{]}^{\omega }\subseteq \mathcal{A}$ or $[M{]}^{\omega }\cap \mathcal{A}=\varnothing$. This is the existence of one infinite homogeneous cone. Requiring such a refinement inside every infinite initial ground set gives the stronger definition used for a Ramsey set of infinite subsets; requiring a homogeneous refinement in every finite-stem neighbourhood gives a completely Ramsey set.

These quantifiers should be distinguished. For example, the existence of a homogeneous cone on the even numbers does not constrain a family's restriction to the odd numbers. If $\mathcal{E}$ is the finite-symmetric-difference parity colouring class on the odd numbers, the family consisting of $\mathcal{E}$ alone has a homogeneous cone disjoint from it on the even numbers, but no homogeneous cone on the odd numbers. Every open set in the ordinary topology on infinite subsets has homogeneous cones inside all infinite ground sets, as follows by deciding finite stems through fusion.

This topology has basic open sets $[A{]}^{\omega }$ for infinite ground sets $A$, without fixed finite stems. It is coarser than the Ellentuck topology and differs from the ordinary topology on infinite subsets, whose basic cylinders fix an initial segment. The symbol $\tau$ is sometimes used for either coarser convention, so it should be accompanied by a definition.

The whole space of infinite subsets of the natural numbers is a meagre set in the Ramsey cone topology. For $D_j=\{X:\min X=j\}$, the closure is $\{X:j\in X\}$, which has empty interior: every infinite ground set can be thinned to omit $j$. Thus all $D_j$ are nowhere dense, while their countable union is the whole space. In contrast, the whole space is not meagre in either the Ellentuck topology or the ordinary topology on 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.

For a finite set $s$ and an infinite set $A$ lying above $\max s$, putThese sets form the basic neighborhoods of the Ellentuck topology, also called the star topology. It is finer than the ordinary topology on infinite subsets. Every basic neighborhood is a clopen set in the Ellentuck topology, and a closed set in the ordinary topology.

Every Borel set for the Ellentuck topology is completely Ramsey. First, an acceptance-and-rejection fusion makes every open set homogeneous after thinning the infinite reservoir while keeping its finite stem. Every nowhere-dense set is consequently completely Ramsey-null, and a fusion handling all subsets of each finite selected prefix shows that countable unions of such null sets remain null. Finally, sets differing from an open set by a meagre set form a sigma-algebra containing the open sets. Thus every Borel set has the required stem-preserving homogeneous refinement.

In the Ellentuck topology, suppose every open set is completely Ramsey and every meagre set is completely Ramsey-null. If $E△U$ is contained in a meagre set for some open $U$, first refine a basic neighbourhood without changing its stem to avoid that exceptional set, and then refine again to be homogeneous for $U$. The resulting neighbourhood is homogeneous for $E$. Sets with such an open representative form a sigma-algebra: countable unions collect the meagre errors, and complements replace $U^c$ by its interior, differing only on the nowhere-dense boundary of $U$. This proves that all star-Borel sets are completely Ramsey.

For an infinite coinfinite set $A\subseteq \mathbb{N}$, let $E_A=[A{]}^{\omega }$ in the space of infinite subsets of the natural numbers. This cone is a nonempty clopen set in the Ellentuck topology, because it is the basic neighbourhood $[\varnothing ,A]$, and any $X\notin E_A$ has a finite initial segment containing a point outside $A$ and hence a basic neighbourhood disjoint from $E_A$.

The cone is a closed set with empty interior in the ordinary topology on infinite subsets: membership of a point outside $A$ witnesses its complement as an open set, while every ordinary finite-stem neighbourhood can be extended by an arbitrarily large point outside $A$. Thus $E_A$ is an ordinary nowhere dense set, but it is neither ordinary open nor Ellentuck nowhere dense. It provides both distinctions between these two topologies.

The cone $[E{]}^{\omega }$ on the even integers is nonempty and Ellentuck topology open, while it is closed and nowhere dense in the ordinary topology on infinite subsets. Choose a finite-symmetric-difference parity colouring on that cone and take one colour class. The class is still ordinarily nowhere dense because its closure lies in the cone, but every stem-free refinement inside the cone meets both colours. Hence it is not completely Ramsey.

In the space of infinite subsets of the natural numbers, the families $C_n=\{X:n\notin X\}$ are clopen sets in the ordinary topology on infinite subsets and thus in the finer Ellentuck topology. Their union is $[\mathbb{N}{]}^{\omega }\setminus \{\mathbb{N}\}$. Every basic Ellentuck topology neighborhood of $\mathbb{N}$ contains $\mathbb{N}\setminus \{n\}$ for some sufficiently large $n$, so this union is not a closed set.

A subset of the space of infinite subsets of the natural numbers has this property when it differs from an Ellentuck topology open set by a meagre set in the same topology. By the complete-Ramsey characterization, these are precisely the completely Ramsey sets. Using meagreness from a different topology changes the definition.

The ordinary topology on $[\mathbb{N}{]}^{\omega }$ has basic sets $[s]=\{X:s⊏X\}$ for finite $s$. It is the product topology on increasing enumerations, and also the subspace topology on infinite-subset indicator functions in $\{0,1{\}}^{\mathbb{N}}$.

In the ordinary topology on infinite subsets, the family $D$ of cofinite sets is a countable set, a dense and a meagre set. Countability follows by indexing its members by their finite complements. Each singleton is closed with empty interior, while every basic cylinder with finite initial segment $s$ contains $s\cup \{n:n>\max s\}$. Consequently $D$ is not a nowhere dense set. This example uses the ordinary cylinder topology, rather than the Ramsey cone topology.

The family $D=\{\{a_1<a_2<\cdots \}:a_{n+1}>2a_n\text{ for all }n\}$ is ordinarily closed and nowhere dense. A failed gap is witnessed by a finite prefix, and any cylinder can be refined by adjoining consecutive large integers that fail the gap. Nevertheless $D$ intersects every infinite-subset cone in continuum many sets: a binary tree chooses two sufficiently large next points inside its reservoir at each step.

A closed nowhere dense family can meet every cone $[M{]}^{\omega }$ in continuum many sets, as the gap-doubling closed family of infinite subsets does. Well-order the cones and choose two fresh candidates in each, permanently reserving one of each colour. The red family meets every cone and its complement contains every reserved blue candidate, so it is not Ramsey. As a subset of the closed nowhere dense family, it is still nowhere dense and has the Baire property in the ordinary infinite-subset topology. This recursion requires no regularity assumption on the continuum cardinal.

A family $Y\subseteq [\mathbb{N}{]}^{\omega }$ is $\tau$-Baire when $Y△O$ is meagre for some ordinarily open family $O$. This is weaker than the corresponding regularity in the star topology. A meagre set meeting every Ramsey cone in both colours has the ordinary Baire property while failing to be a Ramsey set of infinite subsets.