Ellentuck topology (${\tau }^{\ast }$)

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.