# Topological games in Ramsey spaces

Julián C. Cano***Universidad de los Andes and Universidad Nacional de Colombia
Sede Bogotá<EMAIL_ADDRESS> and Carlos A. Di Prisco†††Universidad
de los Andes and Instituto Venezolano de Investigaciones Científicas.
<EMAIL_ADDRESS>
###### Abstract

Topological Ramsey theory studies a class of combinatorial topological spaces,
called topological Ramsey spaces, unifying the essential features of those
combinatorial frames where the Ramsey property is equivalent to Baire
property. In this article, we expose a general overview of the combinatorial
structure of topological Ramsey spaces and their main properties, and we
propose an alternative proof of abstract Ellentuck theorem for a large family
of topological Ramsey spaces. Additionally, we present the notion of selective
topological Ramsey space, and generalize Kastanas games in order to
characterize the Baire property for the family of selective topological Ramsey
spaces.

Key words and phrases: topological Ramsey space, Ellentuck topology, Ramsey
and Baire properties, topological game.

2020 Mathematics Subject Classification: 03E02, 03E05, 05D10.

##  1 Introduction

Infinite dimensional Ramsey theory has developed in different directions since
its beginning in the 1960’s. For instance, topological Ramsey spaces, first
introduced by Carlson and Simpson in [5], and later reformulated and extended
by Todorčević in [31], provide a general framework to study a wide variety of
combinatorial properties that have topological counterparts.

The prototype example of the topological characterization of an infinite
dimensional combinatorial property is Ellentuck’s result in [13], which
establishes an infinite dimensional version of Ramsey’s theorem restricted to
Baire measurable partitions of $[\mathbb{N}]^{\omega}$. On the Ramsey property
of subsets of $[\mathbb{N}]^{\omega}$, Erdős and Rado proved in [14], using
the axiom of choice, that there exist subsets of $[\mathbb{N}]^{\omega}$ which
are not Ramsey. Later, Galvin and Prikry showed in [16] that Borel subsets of
$[\mathbb{N}]^{\omega}$ in the metrizable topology are Ramsey, and Silver in
[29] extended this fact to analytic subsets. Finally, Ellentuck in [13]
characterized topologically Ramsey subsets of $[\mathbb{N}]^{\omega}$ as those
having the Baire property with respect to the exponential topology.
Ellentuck’s proof of this result uses a powerful combinatorial technique, due
to Nash-Williams in [27], now known as combinatorial forcing (see [31, 30,
18]). However, Matet in [23] gives a shorter proof of Ellentuck’s result
avoiding the use of combinatorial forcing.

Ellentuck’s theorem can be generalized using the combinatorial properties of
two types of families of infinite subsets of integers, namely selective and
semiselective coideals. This kind of coideals allow to construct local
versions of Ellentuck’s space, where homogeneity of Borel partitions is
delimited and located in a fixed coideal. The interaction between Ellenctuk’s
theorem and coideals on $\mathbb{N}$ has been studied by Mathias, Matet and
Todorčević in [24, 22, 30] for the notion of selectivity, and extended by
Farah and Todorčević in [15, 31] for the notion of semiselectivity.

Kastanas in [20] gave a characterization of the Ramsey property of subsets of
$[\mathbb{N}]^{\omega}$ in terms of determined topological games. For a fixed
partition of $[\mathbb{N}]^{\omega}$ into finite pieces, he presents an
infinite game such that either player has a winning strategy if and only if
some homogeneous set for the partition can be constructed. Local versions of
Kastanas’ game has been proposed by Matet in [22] for the class of selective
coideals, and extended by Di Prisco, Mijares and Uzcátegui in [7] for the
class of semiselective coideals.

Topological Ramsey spaces deal with the optimal conditions that must be
imposed to a combinatorial structure in order to guarantee the existence of
homogeneous substructures with respect to finite partitions whose parts are
topologically definable. In this regard, topological Ramsey spaces form a
general combinatorial framework in which the respective notions of Ramsey
property and Baire property are equivalent.

The Ellentuck’s space $[\mathbb{N}]^{\omega}$ ([13]) is the quintessential
example of a topological Ramsey space, although there are several
combinatorial contexts that correspond to specific examples of topological
Ramsey spaces, such as: the space of infinite block sequences of finite
subsets of $\mathbb{N}$ ([25]); the space of strong subtrees of a fixed rooted
pruned finitely branching tree of height $\omega$ ([26]); the space of
infinite partitions of $\mathbb{N}$ ([4]); the space of reduced echelon
infinite matrices over a fixed finite field ([2]); the space of rapidly
increasing infinite sequences of variable words over a fixed alphabet ([3]);
the space of infinite block sequences of vectors of finite subsets of
$\mathbb{N}$ ([17]); the space of infinite block sequences of vectors of
located variable words over a fixed alphabet ([31]); the hierarchy of Ramsey
spaces $\\{\mathcal{R}_{\alpha}\\}_{\alpha<\omega_{1}}$ corresponding to
recursive projections of trees ([10, 11]); the hierarchy of Ramsey spaces
$\\{\mathcal{E}_{k}\\}_{k<\omega}$ corresponding to high dimensional Ellentuck
spaces ([8]); construction of topological Ramsey spaces through uniform
barriers on $\mathbb{N}$ ([9]); construction of topological Ramsey spaces from
Fraïssé classes ([12]); spaces of alternating equivalence relations for finite
partitions of $\mathbb{N}$ ([21]); among others.

The main objective of this article is to propose an alternative proof of the
abstract Ellentuck theorem about the Ramsey property for subsets of a
topological Ramsey space given in [31], and also to characterize this property
in terms of infinite games through an abstract version of Kastanas’ theorem.

The article starts with a quick review of the Ramsey property of subsets of
the Ellentuck space of infinite subsets of $\mathbb{N}$ and the Kastanas game
characterization (Section 2). Then follows a presentation of Todorčević’s
formulation of topological Ramsey spaces (Section 3). After that, we define
strong topological Ramsey spaces, and give an alternative proof of the
abstract Ellentuck theorem for this class of Ramsey spaces that include a wide
family of topological Ramsey spaces (Section 4). Finally, we present an
abstract notion of selectivity, and generalize the Kastanas game
characterization of the Ramsey property to selective topological Ramsey spaces
(Section 5).

At this point, let us remember some important concepts related to any
topological space $\mathcal{R}$. First, a set
$\mathcal{N}\subseteq\mathcal{R}$ is nowhere dense if for each basic open set
$\mathcal{O}$ there is a basic open subset $\mathcal{Q}\subseteq\mathcal{O}$
such that $\mathcal{N}\cap\mathcal{Q}=\emptyset$. Also, a set
$\mathcal{M}\subseteq\mathcal{R}$ is meager if
$\mathcal{M}=\bigcup_{n\in\mathbb{N}}\mathcal{N}_{n}$ where each set
$\mathcal{N}_{n}$ is nowhere dense. In addition, a set
$\mathcal{A}\subseteq\mathcal{R}$ has the abstract Baire property if for every
basic open set $\mathcal{O}$ there exists some basic open subset
$\mathcal{Q}\subseteq\mathcal{O}$ such that either
$\mathcal{Q}\subseteq\mathcal{A}$ or $\mathcal{A}\cap\mathcal{Q}=\emptyset$.
Finally, a set $\mathcal{B}\subseteq\mathcal{R}$ has the Baire property if the
set $\mathcal{B}\triangle\mathcal{O}$ is meager for some open set
$\mathcal{O}$.

##  2 Ramsey property on Ellentuck space

In this section, we briefly present the Ramsey property for families of
infinite subsets of integers, and expose both Ellentuck theorem and Kastanas
theorem. To do this, let us first fix some notation:

Given any countably infinite set $X$ and $n\in\mathbb{N}$, we denote by
$[X]^{n}$ the set of subsets of $X$ with exactly $n$ elements; also, we denote
by $[X]^{<\omega}$ the set of finite subsets of $X$, hence
$[X]^{<\omega}=\bigcup_{n\in\mathbb{N}}[X]^{n}$; and analogously, we denote by
$[X]^{\omega}$ the set of infinite subsets of $X$.

Given $s\in[\mathbb{N}]^{<\omega}$ and $A\in[\mathbb{N}]^{\omega}$, the
notation $s\sqsubset A$ means that $s$ is initial segment of $A$. We denote by
$A/s$ the set $A/s=\\{n\in A\,|\,s<n\\}$, where we write $s<n$ whenever
$s=\emptyset$ or $\max s<n$. Lastly, for each $s\in[\mathbb{N}]^{<\omega}$ and
each $B\in[\mathbb{N}]^{\omega}$ we define the set $[s,B]$ as follows:

$[s,B]=\\{A\in[\mathbb{N}]^{\omega}\,|\,A\subseteq B\wedge s\sqsubset A\\}$.

The collection $\left\\{[s,B]\;|\;s\in[\mathbb{N}]^{<\omega}\wedge
B\in[\mathbb{N}]^{\omega}\right\\}$ forms an uncountable basis of open-closed
sets for a non-metrizable topology over $[\mathbb{N}]^{\omega}$, called the
Ellentuck topology of $[\mathbb{N}]^{\omega}$. It should be noted that the
Ellentuck topology is finer than the metrizable topology, which is obtained by
identifying $[\mathbb{N}]^{\omega}$ as a subspace of the Cantor space
$2^{\mathbb{N}}$.

Ramsey theorem is a deep generalization of the combinatorial result known as
the pigeonhole principle, which states that if any infinite set is partitioned
into a finite number of pieces, then at least one of these pieces is infinite.

######  Theorem 2.1.

(Ramsey, [28]). For all $n,k\in\mathbb{N}$ with $n,k\neq 0$, if
$[\mathbb{N}]^{n}=\bigcup_{i=1}^{k}G_{i}$ is any finite partition of
$[\mathbb{N}]^{n}$, then there is $H\in[\mathbb{N}]^{\omega}$ such that
$[H]^{n}\subseteq G_{i}$ for some $i\in\\{1,\ldots,k\\}$.

A set $\mathcal{A}\subseteq[\mathbb{N}]^{\omega}$ determines a partition of
$[\mathbb{N}]^{\omega}$ into two parts, namely $\mathcal{A}$ and
$\mathcal{A}^{\complement}$; so, one can ask when there exists some
$H\in[\mathbb{N}]^{\omega}$ such that $[H]^{\omega}$ is contained in one of
these parts. This question motivated the definition of the Ramsey property,
which is a combinatorial quality about subsets of $[\mathbb{N}]^{\omega}$ that
captures the optimal conditions to guarantee an ideal infinite dimensional
version of Ramsey theorem.

######  Definition 2.2.

A set $\mathcal{A}\subseteq[\mathbb{N}]^{\omega}$ is Ramsey if for all
$N\in[\mathbb{N}]^{\omega}$ and $s\in[N]^{<\omega}$, there is some
$M\in[N]^{\omega}$ such that $s\in[M]^{<\omega}$ and either
$[s,M]\subseteq\mathcal{A}$ or $[s,M]\cap\mathcal{A}=\emptyset$. Moreover, if
the second option always holds, then we say that $\mathcal{A}$ is Ramsey null.

The Ramsey property has two important set-theoretic characterizations. The
first one is given by the Ellentuck theorem, which states a topological
characterization of the Ramsey property through the Baire property with
respect to the Ellentuck topology on $[\mathbb{N}]^{\omega}$. The second one
is given by the Kastanas theorem, which states a game-theoretic
characterization of the Ramsey property through topological games on
$[\mathbb{N}]^{\omega}$.

######  Theorem 2.3.

(Ellentuck, [13]). Consider the space $[\mathbb{N}]^{\omega}$ endowed with the
Ellentuck topology.

  1. 1.

For any set $\mathcal{A}\subseteq[\mathbb{N}]^{\omega}$, the following
statements are equivalent:

    1. (a)

$\mathcal{A}$ is Ramsey null.

    2. (b)

$\mathcal{A}$ is nowhere dense.

    3. (c)

$\mathcal{A}$ is meager.

  2. 2.

For any set $\mathcal{A}\subseteq[\mathbb{N}]^{\omega}$, the following
statements are equivalent:

    1. (a)

$\mathcal{A}$ is Ramsey.

    2. (b)

$\mathcal{A}$ has the abstract Baire property.

    3. (c)

$\mathcal{A}$ has the Baire property.

  3. 3.

If $[\mathbb{N}]^{\omega}=\bigcup_{i=1}^{k}\mathcal{P}_{i}$ is any finite
partition of $[\mathbb{N}]^{\omega}$, where each piece $\mathcal{P}_{i}$ has
the Baire property in the Ellentuck topology, then there is
$H\in[\mathbb{N}]^{\omega}$ such that $[H]^{\omega}\subseteq\mathcal{P}_{i}$
for some $i\in\\{1,\ldots,k\\}$.

######  Definition 2.4.

(Kastanas, [20]). Let $\mathcal{A}\subseteq[\mathbb{N}]^{\omega}$,
$N\in[\mathbb{N}]^{\omega}$ and $s\in[N]^{<\omega}$ be given. The Kastanas
game $\mathcal{J}(\mathcal{A},N,s)$ is defined as follows:

$\setcounter{MaxMatrixCols}{12}\begin{matrix}\textrm{I}&&M_{0}&&M_{1}&&\cdots&&M_{k}&&\cdots&\\\
\textrm{II}&&&(n_{0},N_{0})&&(n_{1},N_{1})&&\cdots&&(n_{k},N_{k})&&\cdots\end{matrix}$

Two players I and II take turns in playing sequence of sets
$\\{M_{k}\\}_{k\in\mathbb{N}}\subseteq[\mathbb{N}]^{\omega}$ and
$\\{N_{k}\\}_{k\in\mathbb{N}}\subseteq[\mathbb{N}]^{\omega}$ respectively, but
additionally player II is required to play a sequence of integers
$\\{n_{k}\\}_{k\in\mathbb{N}}\subseteq\mathbb{N}$, such that the moves of both
players respect the following conditions:

Rules of the game. For every $k\in\mathbb{N}$:

$\textbf{(i) }M_{0}\subseteq N/s\;\;\;\;\;\;\;\;\textbf{(ii) }n_{k}\in
M_{k}\;\;\;\;\;\;\;\;\textbf{(iii) }N_{k}\subseteq
M_{k}/\\{n_{k}\\}\;\;\;\;\;\;\;\;\textbf{(iv) }M_{k+1}\subseteq N_{k}$

Let $N_{\infty}\in[s,N]$ be the set given by
$N_{\infty}=s\cup\\{n_{k}\,|\,k\in\mathbb{N}\\}$. Then, for the game
$\mathcal{J}(\mathcal{A},N,s)$ the following is decided:

  * $\bullet$

Player I wins the game if $N_{\infty}\in\mathcal{A}$.

  * $\bullet$

Player II wins the game if $N_{\infty}\notin\mathcal{A}$.

######  Theorem 2.5.

(Kastanas, [20]). For any set $\mathcal{A}\subseteq[\mathbb{N}]^{\omega}$, the
following statements are equivalent:

  1. (a)

$\mathcal{A}$ is Ramsey.

  2. (b)

For all $N\in[\mathbb{N}]^{\omega}$ and $s\in[N]^{<\omega}$, the Kastanas game
$\mathcal{J}(\mathcal{A},N,s)$ is determined, this means that one of the
players has a winning strategy in the game.

We notice that the Kastanas game $\mathcal{J}(\mathcal{A},N,s)$ on
$[\mathbb{N}]^{\omega}$ is equivalent to a similar topological game in which
the moves executed by the player II are nonempty basic open sets of the
Ellentuck topology, satisfying the rules that we state in definition 5.1 for
the abstract context. Thus, theorem 2.5 is just a particular case of our
theorem 5.7.

According to Todorčević in [31], the Ellentuck space is the combinatorial
structure $([\mathbb{N}]^{\omega},\subseteq,r)$, where $[\mathbb{N}]^{\omega}$
is ordered by the inclusion relation $\subseteq$, and $r$ is the function
$r:[\mathbb{N}]^{\omega}\times\mathbb{N}\longrightarrow[\mathbb{N}]^{<\omega}$
given by $r(B,n)=r_{n}(B)=s$ if $s\in[\mathbb{N}]^{n}$ and $s\sqsubset B$.
Notice that the basis for Ellentuck topology of $[\mathbb{N}]^{\omega}$ is the
collection of all sets $[s,B]=\\{A\in[\mathbb{N}]^{\omega}\,|\,A\subseteq
B\wedge(\exists\,n\in\mathbb{N})(r_{n}(A)=s)\\}$, with
$s\in[\mathbb{N}]^{<\omega}$ and $B\in[\mathbb{N}]^{\omega}$. This alternative
presentation of the Ellentuck space allow to study the Ramsey property in an
abstract sense through the combinatorial notion of topological Ramsey space,
such as will be discussed in the next section.

##  3 Topological Ramsey spaces

In this section, we formally present the combinatorial structure of
topological Ramsey spaces as well as the Ramsey property for this context, and
we conclude with the exposition of the abstract Ellentuck theorem. The
background on topological Ramsey spaces throughout this section is taken from
[31].

Consider a combinatorial structure of the form

$(\mathcal{R},\leq,r)$

where $\mathcal{R}$ is a nonempty set of objects, $\leq$ is a quasi-order
relation on $\mathcal{R}$, and $r$ is a mapping
$r:\mathcal{R}\times\mathbb{N}\rightarrow\mathcal{AR}$ giving us the sequence
of approximations mappings $r_{n}:\mathcal{R}\rightarrow\mathcal{AR}_{n}$ such
that $r_{n}(A)=r(A,n)$ for every $A\in\mathcal{R}$ and $n\in\mathbb{N}$, where
the range of $r$ is the set $\mathcal{AR}$ of all the finite approximations of
elements of $\mathcal{R}$, and the range of $r_{n}$ is the set
$\mathcal{AR}_{n}$ of all the $n$-approximations, hence
$\mathcal{AR}=\bigcup_{n\in\mathbb{N}}\mathcal{AR}_{n}$. Bearing this in mind,
the idea is to identify each object $A\in\mathcal{R}$ with the sequence of its
respective approximations
$\\{r_{n}(A)\\}_{n\in\mathbb{N}}\subseteq\mathcal{AR}$.

For every object $B\in\mathcal{R}$, the restriction of the structure
$\mathcal{R}$ to $B$ is the set $\mathcal{R}\\!\restriction\\!B$ defined by
all objects $A\in\mathcal{R}$ such that $A$ is below $B$ with respect to
quasi-order $\leq$, hence
$\mathcal{R}\\!\restriction\\!B=\\{A\in\mathcal{R}\,|\,A\leq B\\}$.

Following Todorčević in [31], will consider combinatorial structures
$(\mathcal{R},\leq,r)$ which satisfy the following list of combinatorial
properties which will be called axioms of metrization, finitization,
amalgamation and homogeneity (or abstract pigeonhole principle).

A.1. Axioms of Metrization.

  1. 1.

$r_{0}(A)=\emptyset$ for all $A\in\mathcal{R}$.

  2. 2.

For $A,B\in\mathcal{R}$, if $A\neq B$ then $r_{n}(A)\neq r_{n}(B)$ for some
$n\in\mathbb{N}$.

  3. 3.

For $A,B\in\mathcal{R}$, if $r_{n}(A)=r_{m}(B)$ then $n=m$ and
$r_{k}(A)=r_{k}(B)$ for each $k\leq n$.

Let $A\in\mathcal{R}$ and $a,b\in\mathcal{AR}$ be given. Then, we write
$a\sqsubset A$, and say that $a$ is an initial segment of $A$, if there is
some $n\in\mathbb{N}$ such that $a=r_{n}(A)$. Likewise, we write $a\sqsubseteq
b$, and say that $a$ is an initial segment of $b$, if there are
$B\in\mathcal{R}$ and $n,m\in\mathbb{N}$ with $n\leq m$ such that $a=r_{n}(B)$
and $b=r_{m}(B)$. Similarly, we write $a\sqsubset b$, and say that $a$ is a
proper initial segment of $b$, if $a\sqsubseteq b$ and $a\neq b$. Lastly, the
length of $a$, denoted by $|a|$, is the unique $n\in\mathbb{N}$ such that
$a=r_{n}(B)$ for some $B\in\mathcal{R}$.

A.2. Axioms of Finitization. There is a quasi-order $\leq_{\text{fin}}$ on
$\mathcal{AR}$ such that:

  1. 1.

For all $b\in\mathcal{AR}$, the set of approximations
$\\{a\in\mathcal{AR}\,|\,a\leq_{\text{fin}}b\\}$ is finite.

  2. 2.

Let $A,B\in\mathcal{R}$, then $A\leq B$ if and only if for all
$n\in\mathbb{N}$ there is $m\in\mathbb{N}$ such that
$r_{n}(A)\leq_{\text{fin}}r_{m}(B)$.

  3. 3.

For all $a,b,c\in\mathcal{AR}$, if $a\sqsubset b$ and $b\leq_{\text{fin}}c$
then there is $d\in\mathcal{AR}$ such that $d\sqsubset c$ and
$a\leq_{\text{fin}}d$.

For every approximation $a\in\mathcal{AR}$, the set $[a]$ is defined as the
family of all objects $A\in\mathcal{R}$ such that $a\sqsubset A$, that is,

$[a]=\\{A\in\mathcal{R}\,|\,r_{|a|}(A)=a\\}$.

The collection $\\{[a]\,|\,a\in\mathcal{AR}\\}$ forms a basis for a topology
over $\mathcal{R}$, called the metrizable topology of $\mathcal{R}$, whose
complete metric $\rho:\mathcal{R}\times\mathcal{R}\rightarrow\mathbb{R}$ is
described by $\rho(A,B)=1/2^{k}$, whenever $A\neq B$ and
$k=\min\\{n\in\mathbb{N}\,|\,r_{n}(A)\neq r_{n}(B)\\}$.

For every approximation $a\in\mathcal{AR}$ and every object $B\in\mathcal{R}$,
the set $[a,B]$ is defined as the family of all objects $A\in\mathcal{R}$ such
that $a\sqsubset A$ and $A\leq B$, so that

$[a,B]=\\{A\in[a]\,|\,A\leq B\\}=\\{A\in\mathcal{R}\,|\,A\leq B\wedge
r_{|a|}(A)=a\\}$.

The collection $\\{[a,B]\,|\,a\in\mathcal{AR}\wedge B\in\mathcal{R}\\}$ also
forms a basis for a topology over $\mathcal{R}$, called the Ellentuck topology
of $\mathcal{R}$, which extends and refines the metrizable topology. For
simplicity, for each $m\in\mathbb{N}$ we denote:
$[m,B]=\\{A\in\mathcal{R}\,|\,A\leq B\wedge
r_{m}(A)=r_{m}(B)\\}=[r_{m}(B),B]$.

The depth of an approximation in an object of $\mathcal{R}$ is defined as
follows: given $a\in\mathcal{AR}$ and $B\in\mathcal{R}$, the depth of $a$ in
$B$, denoted by $\mathtt{depth}_{B}(a)$, is

$\mathtt{depth}_{B}(a)=\begin{cases}\min\\{n\in\mathbb{N}\,|\,a\leq_{\text{fin}}r_{n}(B)\\}&\text{if
such minimum exists.}\\\ \infty&\text{otherwise.}\end{cases}$

For each basic open set $[a,B]\neq\emptyset$ in the Ellentuck topology of
$\mathcal{R}$, the restriction of the collection of approximations
$\mathcal{AR}$ to $[a,B]$ is the set
$\mathcal{AR}\\!\restriction\\![a,B]=\\{b\in\mathcal{AR}\,|\,a\sqsubseteq
b\wedge\mathtt{depth}_{B}(b)<\infty\\}$.

A.3. Axioms of Amalgamation.

  1. 1.

Let $a\in\mathcal{AR}$ and $B\in\mathcal{R}$, if
$\mathtt{depth}_{B}(a)<\infty$ then $[a,A]\neq\emptyset$ for all
$A\in[\mathtt{depth}_{B}(a),B]$, in particular $[a,B]\neq\emptyset$.

  2. 2.

Let $a\in\mathcal{AR}$ and $A,B\in\mathcal{R}$, if $A\leq B$ and
$\mathtt{depth}_{A}(a)\leq\mathtt{depth}_{B}(a)<\infty$ then there is some
$A^{\prime}\in[\mathtt{depth}_{B}(a),B]$ such that
$[a,A^{\prime}]\subseteq[a,A]$.

Let $[a,A]$ be a basic open set in the Ellentuck topology of $\mathcal{R}$;
then, for each $n\in\mathbb{N}$ with $n\geq|a|$, we denote by
$r_{n}\text{''}\,[a,A]$ the direct image of the set $[a,A]$ through the $n$-th
approximation function $r_{n}$, hence
$r_{n}\text{''}\,[a,A]=\\{d\in\mathcal{AR}_{n}\,|\,(\exists
D\in[a,A])(d=r_{n}(D))\\}=\\{d\in\mathcal{AR}\\!\restriction\\![a,A]\,|\,|d|=n\\}$.

A.4. Abstract Pigeonhole Principle. For all $a\in\mathcal{AR}$ and
$B\in\mathcal{R}$ such that $\mathtt{depth}_{B}(a)<\infty$, if
$\mathcal{O}\subseteq\mathcal{AR}_{|a|+1}$ then there exists some
$A\in[\mathtt{depth}_{B}(a),B]$ such that either
$r_{|a|+1}\text{''}\,[a,A]\subseteq\mathcal{O}$ or $\mathcal{O}\cap
r_{|a|+1}\text{''}\,[a,A]=\emptyset$.

The combinatorial structure $(\mathcal{R},\leq,r)$ is called metrically closed
if $\mathcal{R}$ is a closed subspace of $\mathcal{AR}^{\mathbb{N}}$ with the
product topology, where each object of $\mathcal{R}$ has been identified with
the sequence of its initial segments. Naturally, we consider the product
topology on $\mathcal{AR}^{\mathbb{N}}$ whenever $\mathcal{AR}$ has the
discrete topology, in which case the subspace topology on $\mathcal{R}$
inherited from $\mathcal{AR}^{\mathbb{N}}$ coincides exactly with the
metrizable topology of $\mathcal{R}$.

######  Definition 3.1.

(Todorčević, [31]). A metrically closed combinatorial structure
$(\mathcal{R},\leq,r)$ is a topological Ramsey space if $\mathcal{R}$ is
endowed with the Ellentuck topology and $(\mathcal{R},\leq,r)$ satisfies the
axioms A.1, A.2, A.3 and A.4.

A topological Ramsey space $(\mathcal{R},\leq,r)$ does not have isolated
points if each of its nonempty basic open sets is infinite. Notice that the
space $(\mathcal{R},\leq,r)$ does not have isolated points whenever the set of
approximations $r_{|a|+1}\text{''}\,[a,B]$ is infinite for every basic open
set $[a,B]\neq\emptyset$.

The Ellentuck space $([\mathbb{N}]^{\omega},\subseteq,r)$ is the
quintessential example of a topological Ramsey space without isolated points,
and this fact naturally leads us to extend definition 2.2 to the abstract
context. So, the Ramsey property related to topological Ramsey spaces is
defined as follows:

######  Definition 3.2.

Let $(\mathcal{R},\leq,r)$ be a topological Ramsey space. A set
$\mathcal{X}\subseteq\mathcal{R}$ is Ramsey if for every $[a,A]\neq\emptyset$
there is some $B\in[a,A]$ such that either $[a,B]\subseteq\mathcal{X}$ or
$[a,B]\cap\mathcal{X}=\emptyset$. Moreover, if the second option always holds,
then we say that $\mathcal{X}$ is Ramsey null.

Todorčević proved in [31] that for every topological Ramsey space
$(\mathcal{R},\leq,r)$ without isolated points there is a set
$\mathcal{X}\subseteq\mathcal{R}$ which is not Ramsey. In the proof the use of
the axiom of choice is indispensable.

An important tool in the study of topological Ramsey spaces is the concept of
fusion sequence, formed by basic open sets of a space, since through it we can
recursively construct new objects of the space satisfying specific and special
properties.

######  Definition 3.3.

Let $(\mathcal{R},\leq,r)$ be a topological Ramsey space. A sequence
$\\{[n_{k},Y_{k}]\\}_{k\in\mathbb{N}}$ of basic open sets of $\mathcal{R}$ is
a fusion sequence if:

  * $\bullet$

$\\{n_{k}\\}_{k\in\mathbb{N}}\subseteq\mathbb{N}$ is a strictly increasing
sequence.

  * $\bullet$

$Y_{k+1}\in[n_{k},Y_{k}]$ for all $k\in\mathbb{N}$.

The limit of the fusion sequence $\\{[n_{k},Y_{k}]\\}_{k\in\mathbb{N}}$ always
exists and it is the unique object $Y_{\infty}\in\mathcal{R}$, denoted by
$Y_{\infty}=\lim_{n_{k}}Y_{k}$, such that $Y_{\infty}\in[n_{k},Y_{k}]$ for all
$k\in\mathbb{N}$; hence, $r_{n_{k}}(Y_{\infty})=r_{n_{k}}(Y_{k})$ for all
$k\in\mathbb{N}$. In other words, $Y_{\infty}=\lim_{n_{k}}Y_{k}$ if and only
if
$\\{Y_{\infty}\\}=\bigcap_{k\in\mathbb{N}}[n_{k},Y_{k}]=\bigcap_{k\in\mathbb{N}}[r_{n_{k}}(Y_{k})]$.

At last, we present the main result of the topological Ramsey theory, which
corresponds to the natural generalization of theorem 2.3 for the abstract
context. In this regard, the abstract Ellentuck theorem states the equivalence
between the Ramsey property and the Baire property on the combinatorial
structure of topological Ramsey spaces.

######  Theorem 3.4.

[Abstract Ellentuck Theorem]. (Todorčević, [31]). Let $(\mathcal{R},\leq,r)$
be a topological Ramsey space. Then:

  1. 1.

For any $\mathcal{X}\subseteq\mathcal{R}$, the following statements are
equivalent:

    1. (a)

$\mathcal{X}$ is Ramsey null.

    2. (b)

$\mathcal{X}$ is nowhere dense.

    3. (c)

$\mathcal{X}$ is meager.

  2. 2.

For any $\mathcal{X}\subseteq\mathcal{R}$, the following statements are
equivalent:

    1. (a)

$\mathcal{X}$ is Ramsey.

    2. (b)

$\mathcal{X}$ has the abstract Baire property.

    3. (c)

$\mathcal{X}$ has the Baire property.

  3. 3.

If $\mathcal{R}=\bigcup_{i=1}^{k}\mathcal{P}_{i}$ is any finite partition of
the space $\mathcal{R}$, where each piece $\mathcal{P}_{i}$ has the Baire
property, then there is $H\in\mathcal{R}$ such that
$\mathcal{R}\\!\restriction\\!H\subseteq\mathcal{P}_{i}$ for some
$i\in\\{1,\ldots,k\\}$.

The original proof of the abstract Ellentuck theorem given by Todorčević in
[31] mainly applies a combinatorial technique known as abstract notion of
combinatorial forcing, whose essential idea is to force the recursive
construction of an object in a topological Ramsey space such that it turns out
to be homogeneous with respect to a fixed partition of the space that is
finite and Baire measurable. In the next section, we present a new proof of
abstract Ellentuck theorem for a large family of topological Ramsey spaces, in
which we do not use combinatorial forcing.

##  4 On an alternative proof of the abstract Ellentuck theorem

In this section, we introduce a generalization of axioms of amalgamation to
define the wide class of strong topological Ramsey spaces, with the aim of
proposing an alternative proof of the abstract Ellentuck theorem for this new
kind of Ramsey spaces.

Let $(\mathcal{R},\leq,r)$ be a topological Ramsey space, and let
$A,B\in\mathcal{R}$, $b\in\mathcal{AR}$ and $m\in\mathbb{N}$ be such that
$B\leq A$ and $\mathtt{depth}_{A}(b)\leq m$. We define the set $\langle
b,B\rangle_{m}^{A}$ as follows:

$\langle
b,B\rangle_{m}^{A}=\\{X\in[b,B]\;|\;\mathtt{depth}_{A}(r_{|b|+1}(X))>m\\}.$

Notice that $\langle b,B\rangle_{m}^{A}$ is an open set of $\mathcal{R}$,
since $\langle b,B\rangle_{m}^{A}=\bigcup\\{[c,B]\,|\,c\in
r_{|b|+1}\text{''}\,[b,B]\wedge\mathtt{depth}_{A}(c)>m\\}$. Furthermore,
notice that if $\mathtt{depth}_{A}(b)=m$ then $\langle
b,B\rangle_{m}^{A}=[b,B]$.

A.3*. Axioms of Strong Amalgamation.

  1. 1.

Let $b\in\mathcal{AR}$ and $A\in\mathcal{R}$, if $\mathtt{depth}_{A}(b)\leq
m<\infty$ then $\langle b,B\rangle_{m}^{A}\neq\emptyset$ for all $B\in[m,A]$,
in particular $\langle b,A\rangle_{m}^{A}\neq\emptyset$.

  2. 2.

Let $b\in\mathcal{AR}$ and $A,B\in\mathcal{R}$, if $B\leq A$ and
$\mathtt{depth}_{B}(b)\leq\mathtt{depth}_{A}(b)\leq m<\infty$ then there is
some $B^{\prime}\in[m,A]$ such that $\langle
b,B^{\prime}\rangle_{m}^{A}\subseteq\langle b,B\rangle_{m}^{A}$.

We point out that the axioms of strong amalgamation A.3* imply the axioms of
amalgamation A.3, since taking $m=\mathtt{depth}_{A}(b)$, we can see that A.3
is a particular case of A.3*.

######  Definition 4.1.

A metrically closed combinatorial structure $(\mathcal{R},\leq,r)$ is a strong
topological Ramsey space if $\mathcal{R}$ is endowed with the Ellentuck
topology and $(\mathcal{R},\leq,r)$ satisfies the axioms A.1, A.2, A.3* and
A.4.

It should be noted that for each topological Ramsey space
$(\mathcal{R},\leq,r)$, the axiom A.3*.1 holds if and only if for every basic
open set $[a,B]\neq\emptyset$ the set of approximations
$r_{|a|+1}\text{''}\,[a,B]$ is infinite. Therefore, if the space
$(\mathcal{R},\leq,r)$ is strong, then it does not have isolated points.

In addition, clearly every strong topological Ramsey space is itself a
topological Ramsey space, since A.3* imply A.3; however, we do not know the
answer to the following question: is there some topological Ramsey space
without isolated points that is not strong?

Next, we present an alternative proof of abstract Ellentuck theorem restricted
to the collection of strong topological Ramsey spaces. The new proof of
theorem 3.4 that we propose here is inspired on a short proof of theorem 2.3
given by Matet in [23].

Let $(\mathcal{R},\leq,r)$ be any topological Ramsey space. Given
$\mathcal{Q}\subseteq\mathcal{R}$ and $b\in\mathcal{AR}$, we define the set
$\mathcal{S}_{b}^{\mathcal{Q}}\subseteq\mathcal{R}$ as follows:

$\mathcal{S}_{b}^{\mathcal{Q}}=\\{D\in\mathcal{R}\,|\,\emptyset\neq[b,D]\subseteq\mathcal{Q}\\}$.

######  Lemma 4.2.

Let $(\mathcal{R},\leq,r)$ be a strong topological Ramsey space. For each
$\mathcal{Q}\subseteq\mathcal{R}$, $m\in\mathbb{N}$, $A\in\mathcal{R}$, and
$b,d\in\mathcal{AR}$, if $\mathtt{depth}_{A}(b)\leq\mathtt{depth}_{A}(d)\leq
m$ and $b\leq_{\text{fin}}d$, then there is some $B\in[m,A]$ such that either
$\langle d,B\rangle_{m}^{A}\subseteq\mathcal{S}_{b}^{\mathcal{Q}}$ or $\langle
d,B\rangle_{m}^{A}\cap\mathcal{S}_{b}^{\mathcal{Q}}=\emptyset$.

###### Proof.

If $\langle d,A\rangle_{m}^{A}\cap\mathcal{S}_{b}^{\mathcal{Q}}=\emptyset$
then we can simply take $B=A$. Otherwise, if $\langle
d,A\rangle_{m}^{A}\cap\mathcal{S}_{b}^{\mathcal{Q}}\neq\emptyset$ then we fix
$C\in\langle d,A\rangle_{m}^{A}\cap\mathcal{S}_{b}^{\mathcal{Q}}$, hence
$C\in[d,A]$ and $\emptyset\neq[b,C]\subseteq\mathcal{Q}$. Now, by A.3* there
exists some $B\in[m,A]$ such that $\emptyset\neq\langle
d,B\rangle_{m}^{A}\subseteq\langle d,C\rangle_{m}^{A}$. Finally, notice that
$\langle d,B\rangle_{m}^{A}\subseteq\mathcal{S}_{b}^{\mathcal{Q}}$, since
every $D\in\langle d,B\rangle_{m}^{A}\subseteq\langle
d,C\rangle_{m}^{A}\subseteq[d,C]$ satisfies that
$\emptyset\neq[b,D]\subseteq[b,C]\subseteq\mathcal{Q}$, therefore
$D\in\mathcal{S}_{b}^{\mathcal{Q}}$. ∎

######  Lemma 4.3.

Let $(\mathcal{R},\leq,r)$ be a strong topological Ramsey space, and let
$[a,A]\neq\emptyset$ be a basic open set of $\mathcal{R}$. If
$\\{\mathcal{Q}_{i}\\}_{i\in\mathbb{N}}\subseteq\wp(\mathcal{R})$ is any
sequence of subsets of $\mathcal{R}$ for which
$[a,B]\not\subseteq\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}$ for all
$B\in[a,A]$, then there are $i\in\mathbb{N}$ and $[c,C]\neq\emptyset$, with
$C\in[a,A]$, $[c,C]\subseteq[a,A]$, and $|a|\leq|c|\leq|a|+i$, such that
$[d,D]\not\subseteq\mathcal{Q}_{i}$ for each $[d,D]\neq\emptyset$ with
$[d,D]\subseteq[c,C]$.

###### Proof.

Let $[a,A]\neq\emptyset$ and
$\\{\mathcal{Q}_{i}\\}_{i\in\mathbb{N}}\subseteq\wp(\mathcal{R})$ be such that
$[a,B]\not\subseteq\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}$ for all
$B\in[a,A]$. We may suppose without loss of generality that $a\sqsubset A$.

First of all, we recursively construct a fusion sequence
$\\{[n_{k},A_{k}]\\}_{k\in\mathbb{N}}$ such that for all $k\in\mathbb{N}$ we
have:

  1. (i)

$A_{0}=A$.

  2. (ii)

$n_{k}=\mathtt{depth}_{A}(a)+k=|a|+k$.

  3. (iii)

For all $i\leq k$ and for each $b,d\in\mathcal{AR}\\!\restriction\\![a,A_{k}]$
with $\mathtt{depth}_{A_{k}}(b)\leq\mathtt{depth}_{A_{k}}(d)\leq n_{k}$ and
$b\leq_{\text{fin}}d$, the object $A_{k+1}\in[n_{k},A_{k}]$ is such that
either $\langle
d,A_{k+1}\rangle_{n_{k}}^{A_{k}}\subseteq\mathcal{S}_{b}^{\mathcal{Q}_{i}}$ or
$\langle
d,A_{k+1}\rangle_{n_{k}}^{A_{k}}\cap\mathcal{S}_{b}^{\mathcal{Q}_{i}}=\emptyset$.

Suppose inductively that the basic open set $[n_{k},A_{k}]$ has been
constructed. Applying A.2, we consider the finite set
$\\{(b,d)\in\mathcal{AR}\\!\restriction\\![a,A_{k}]\times\mathcal{AR}\\!\restriction\\![a,A_{k}]\,|\,\mathtt{depth}_{A_{k}}(b)\leq\mathtt{depth}_{A_{k}}(d)\leq
n_{k}\wedge b\leq_{\text{fin}}d\\}$, for which we fix an arbitrary enumeration
$\\{(b_{0},d_{0}),\ldots,(b_{p},d_{p})\\}$ with $p\in\mathbb{N}$. Now, we
construct a finite sequence $X_{0}\geq\cdots\geq X_{p+1}$ of elements of
$[n_{k},A_{k}]$ such that for all $j\leq p$ we have:

  1. (iv)

$X_{0}=A_{k}$.

  2. (v)

$X_{j+1}\in[n_{k},X_{j}]$.

  3. (vi)

For all $i\leq k$, either $\langle
d_{j},X_{j+1}\rangle_{n_{k}}^{A_{k}}\subseteq\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}$
or $\langle
d_{j},X_{j+1}\rangle_{n_{k}}^{A_{k}}\cap\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}=\emptyset$.

Suppose that the object $X_{j}\in[n_{k},A_{k}]$ has been defined, thus
$r_{n_{k}}(X_{j})=r_{n_{k}}(A_{k})$, which implies that
$\mathtt{depth}_{X_{j}}(b_{j})=\mathtt{depth}_{A_{k}}(b_{j})$ and
$\mathtt{depth}_{X_{j}}(d_{j})=\mathtt{depth}_{A_{k}}(d_{j})$. Then, applying
iteratively lemma 4.2, we construct a finite sequence $X_{j}^{0}\geq\cdots\geq
X_{j}^{k+1}$ of elements of $[n_{k},X_{j}]\subseteq[n_{k},A_{k}]$ such that
for all $i\leq k$ we have:

  1. (vii)

$X_{j}^{0}=X_{j}$.

  2. (viii)

$X_{j}^{i+1}\in[n_{k},X_{j}^{i}]$.

  3. (ix)

Either $\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{A_{k}}\subseteq\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}$
or $\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{A_{k}}\cap\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}=\emptyset$.

Suppose that the object $X_{j}^{i}\in[n_{k},X_{j}]$ has been defined, in which
case
$\mathtt{depth}_{X_{j}^{i}}(b_{j})\leq\mathtt{depth}_{X_{j}^{i}}(d_{j})\leq
n_{k}$, so by lemma 4.2, we deduce that there is some
$X_{j}^{i+1}\in[n_{k},X_{j}^{i}]\subseteq[n_{k},X_{j}]\subseteq[n_{k},A_{k}]$
such that either $\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{X_{j}^{i}}\subseteq\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}$
or $\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{X_{j}^{i}}\cap\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}=\emptyset$;
however, since $r_{n_{k}}(X_{j}^{i})=r_{n_{k}}(X_{j})=r_{n_{k}}(A_{k})$, then
we infer that $\langle d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{X_{j}^{i}}=\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{A_{k}}$, whereby we conclude that either
$\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{A_{k}}\subseteq\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}$
or $\langle
d_{j},X_{j}^{i+1}\rangle_{n_{k}}^{A_{k}}\cap\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}=\emptyset$.

Let $X_{j+1}\in[n_{k},X_{j}]$ be the object $X_{j+1}=X_{j}^{k+1}$, and so we
now define the object $A_{k+1}\in[n_{k},A_{k}]$ by $A_{k+1}=X_{p+1}$. Thus,
for each $i\leq k$ and $j\leq p$ we have that either $\langle
d_{j},A_{k+1}\rangle_{n_{k}}^{A_{k}}\subseteq\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}$
or $\langle
d_{j},A_{k+1}\rangle_{n_{k}}^{A_{k}}\cap\mathcal{S}_{b_{j}}^{\mathcal{Q}_{i}}=\emptyset$,
completing the induction on $k\in\mathbb{N}$.

Let $B\in[a,A]$ be the limit $B=\lim_{n_{k}}A_{k}$ of the fusion sequence
$\\{[n_{k},A_{k}]\\}_{k\in\mathbb{N}}$, so that $B\in[n_{k},A_{k}]$ and
$r_{n_{k}}(B)=r_{n_{k}}(A_{k})$ for all $k\in\mathbb{N}$. Since
$[a,B]\not\subseteq\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}$, then there is
some $i\in\mathbb{N}$ such that $[a,B]\not\subseteq\mathcal{Q}_{i}$; so, we
choose $E\in[a,B]\setminus\mathcal{Q}_{i}$ and we take $c=r_{\ell}(E)$ with
$\ell=\max\\{m\in\mathbb{N}\,|\,\mathtt{depth}_{B}(r_{m}(E))\leq n_{i}\\}$,
obtaining $|a|\leq|c|\leq|a|+i$. Thus, $a\sqsubseteq c\sqsubset E$ and
$E\in[c,B]$, consequently we have that $[c,B]\not\subseteq\mathcal{Q}_{i}$; in
fact, it is also true that $[c,E]\not\subseteq\mathcal{Q}_{i}$.

For each $b\in\mathcal{AR}\\!\restriction\\![c,E]$, we proceed to consider the
family of approximations $T_{b}\subseteq r_{|b|+1}\text{''}\,[b,E]$ as well as
the collection of objects $\mathcal{K}_{b}\subseteq[\mathtt{depth}_{E}(b),E]$
defined as follows:

$\displaystyle T_{b}$ $\displaystyle=\\{q\in
r_{|b|+1}\text{''}\,[b,E]\;|\;[q,E]\subseteq\mathcal{Q}_{i}\\},\text{ and }$
$\displaystyle\mathcal{K}_{b}$
$\displaystyle=\\{D\in[\mathtt{depth}_{E}(b),E]\;|\;[b,D]\subseteq\mathcal{Q}_{i}\\}=\mathcal{S}_{b}^{\mathcal{Q}_{i}}\cap[\mathtt{depth}_{E}(b),E].$

Claim 1. If $b\in\mathcal{AR}\\!\restriction\\![c,E]$ and
$D_{1},D_{2}\in[\mathtt{depth}_{E}(b),E]$ be such that
$D_{1}\in\mathcal{K}_{b}$ and $D_{2}\leq D_{1}$ then
$D_{2}\in\mathcal{K}_{b}$. Indeed, notice that
$[b,D_{2}]\subseteq[b,D_{1}]\subseteq\mathcal{Q}_{i}$ whenever
$D_{1}\in\mathcal{K}_{b}$ and $D_{2}\leq D_{1}$, so that
$D_{2}\in\mathcal{K}_{b}$.

Claim 2. For all $b\in\mathcal{AR}\\!\restriction\\![c,E]$, either
$\mathcal{K}_{b}=[\mathtt{depth}_{E}(b),E]$ or $\mathcal{K}_{b}=\emptyset$.
Indeed, let $b\in\mathcal{AR}\\!\restriction\\![c,E]$ be fixed and put
$d=r_{\mathtt{depth}_{E}(b)}(E)$, hence $b\leq_{\text{fin}}d$ and
$[\mathtt{depth}_{E}(b),E]=[d,E]$. Now, suppose that
$\mathcal{K}_{b}\neq\emptyset$ and let $D\in[d,E]$ be such that
$D\in\mathcal{K}_{b}$, which implies that
$D\in\mathcal{S}_{b}^{\mathcal{Q}_{i}}$. We know that $B=\lim_{n_{k}}A_{k}$,
so we can take $k\in\mathbb{N}$ such that
$n_{k}=\max\\{\mathtt{depth}_{B}(d),n_{i}\\}$. Thus $i\leq k$ and
$\mathtt{depth}_{B}(b)\leq\mathtt{depth}_{B}(d)\leq n_{k}$, and since
$B\in[n_{k+1},A_{k+1}]\subseteq[n_{k},A_{k}]$, we also have that
$\mathtt{depth}_{B}(b)=\mathtt{depth}_{A_{k}}(b)$ and
$\mathtt{depth}_{B}(d)=\mathtt{depth}_{A_{k}}(d)$. Therefore, we deduce that
either $\langle
d,A_{k+1}\rangle_{n_{k}}^{A_{k}}\subseteq\mathcal{S}_{b}^{\mathcal{Q}_{i}}$ or
$\langle
d,A_{k+1}\rangle_{n_{k}}^{A_{k}}\cap\mathcal{S}_{b}^{\mathcal{Q}_{i}}=\emptyset$,
and since $r_{n_{k}}(B)=r_{n_{k}}(A_{k+1})=r_{n_{k}}(A_{k})$ and $B\leq
A_{k+1}\leq A_{k}$, then we also obtain that either $\langle
d,B\rangle_{n_{k}}^{B}\subseteq\mathcal{S}_{b}^{\mathcal{Q}_{i}}$ or $\langle
d,B\rangle_{n_{k}}^{B}\cap\mathcal{S}_{b}^{\mathcal{Q}_{i}}=\emptyset$.
However, notice that $D\in\langle
d,B\rangle_{n_{k}}^{B}\cap\mathcal{S}_{b}^{\mathcal{Q}_{i}}$, so it must
necessarily be true that $\langle
d,B\rangle_{n_{k}}^{B}\subseteq\mathcal{S}_{b}^{\mathcal{Q}_{i}}$, and we
infer that $E\in\mathcal{S}_{b}^{\mathcal{Q}_{i}}$ because $E\in\langle
d,B\rangle_{n_{k}}^{B}$. We then conclude that $E\in\mathcal{K}_{b}$, and
consequently $\mathcal{K}_{b}=[\mathtt{depth}_{E}(b),E]$.

Claim 3. For all $b\in\mathcal{AR}\\!\restriction\\![c,E]$, it is true that
$\mathcal{K}_{b}\neq\emptyset$ if and only if
$T_{b}=r_{|b|+1}\text{''}\,[b,E]$. Indeed, if $\mathcal{K}_{b}\neq\emptyset$
then $\mathcal{K}_{b}=[\mathtt{depth}_{E}(b),E]$, in which case
$E\in\mathcal{K}_{b}$ and hence $[b,E]\subseteq\mathcal{Q}_{i}$, thus
$[q,E]\subseteq[b,E]\subseteq\mathcal{Q}_{i}$ for each $q\in
r_{|b|+1}\text{''}\,[b,E]$, as a result we obtain that
$T_{b}=r_{|b|+1}\text{''}\,[b,E]$. Conversely, if
$T_{b}=r_{|b|+1}\text{''}\,[b,E]$, then $[b,E]\subseteq\bigcup_{q\in
T_{b}}[q,E]\subseteq\mathcal{Q}_{i}$, therefore $E\in\mathcal{K}_{b}$ and in
particular $\mathcal{K}_{b}\neq\emptyset$.

Claim 4. For every $b\in\mathcal{AR}\\!\restriction\\![c,E]$, if
$\mathcal{K}_{b}\neq\emptyset$ then $b\neq c$ and $b\in T_{b^{-}}$, where
$c\sqsubseteq b^{-}\sqsubset b$ with $|b^{-}|=|b|-1$. Indeed, since
$[c,E]\not\subseteq\mathcal{Q}_{i}$ then $E\notin\mathcal{K}_{c}$, in which
case $\mathcal{K}_{c}\neq[\mathtt{depth}_{E}(c),E]=[c,E]$ and hence
$\mathcal{K}_{c}=\emptyset$. Consequently, if $\mathcal{K}_{b}\neq\emptyset$
then $b\neq c$ and also $\mathcal{K}_{b}=[\mathtt{depth}_{E}(b),E]$, so that
$E\in\mathcal{K}_{b}$ and hence $[b,E]\subseteq\mathcal{Q}_{i}$, therefore
$b\in T_{b^{-}}$ where $c\sqsubseteq b^{-}\sqsubset b$ with $|b^{-}|=|b|-1$.

After that stage, we recursively construct a new fusion sequence
$\\{[m_{k},M_{k}]\\}_{k\in\mathbb{N}}$ such that for all $k\in\mathbb{N}$ we
have:

  1. (i)

$M_{0}=E$.

  2. (ii)

$m_{k}=\mathtt{depth}_{E}(c)+k=|c|+k$.

  3. (iii)

For all $b\in\mathcal{AR}\\!\restriction\\![c,E]$ with
$\mathtt{depth}_{M_{k}}(b)=m_{k}$, the object $M_{k+1}\in[m_{k},M_{k}]$ is
such that $T_{b}\cap r_{|b|+1}\text{''}\,[b,M_{k+1}]=\emptyset$.

By induction, suppose that the basic open set $[m_{k},M_{k}]$ has been
constructed. Applying A.2, we consider any enumeration of the finite set
$\\{b\in\mathcal{AR}\\!\restriction\\![c,E]\,|\,\mathtt{depth}_{M_{k}}(b)=m_{k}\\}=\\{b_{0},\ldots,b_{p}\\}$
with $p\in\mathbb{N}$. Now, we construct a finite sequence
$Z_{0}\geq\cdots\geq Z_{p+1}$ of elements of $[m_{k},M_{k}]$ such that for
each $j\leq p$ we have:

  1. (iv)

$Z_{0}=M_{k}$.

  2. (v)

$Z_{j+1}\in[m_{k},Z_{j}]$.

  3. (vi)

$T_{b_{j}}\cap r_{|b_{j}|+1}\text{''}\,[b_{j},Z_{j+1}]=\emptyset$.

Suppose that the object $Z_{j}\in[m_{k},M_{j}]$ has been defined, and notice
that $\mathtt{depth}_{Z_{j}}(b_{j})=\mathtt{depth}_{M_{k}}(b_{j})=m_{k}$, then
applying the abstract pigeonhole principle A.4 to the family $T_{b_{j}}$, we
deduce that there exists some $Z_{j+1}\in[m_{k},Z_{j}]$ such that either
$r_{|b_{j}|+1}\text{''}\,[b_{j},Z_{j+1}]\subseteq T_{b_{j}}$ or $T_{b_{j}}\cap
r_{|b_{j}|+1}\text{''}\,[b_{j},Z_{j+1}]=\emptyset$. However, the first
alternative is impossible, since
$r_{|b_{j}|+1}\text{''}\,[b_{j},Z_{j+1}]\subseteq T_{b_{j}}$ implies that
$[b_{j},Z_{j+1}]\subseteq\mathcal{Q}_{i}$, hence
$\mathcal{K}_{b_{j}}\neq\emptyset$ because $Z_{j+1}\in\mathcal{K}_{b_{j}}$, so
that $b_{j}\neq c$ and $b_{j}\in T_{b_{j}^{-}}$; in fact, notice that
$b_{j}\in T_{b_{j}^{-}}\cap r_{|b_{j}|}\text{''}\,[b_{j}^{-},Z_{j}]$, which is
contradictory. Therefore, we must have $T_{b_{j}}\cap
r_{|b_{j}|+1}\text{''}\,[b_{j},Z_{j+1}]=\emptyset$.

We now define the object $M_{k+1}\in[m_{k},M_{j}]$ by $M_{k+1}=Z_{p+1}$; thus,
for all $j\leq p$ we have $T_{b_{j}}\cap
r_{|b_{j}|+1}\text{''}\,[b_{j},M_{k+1}]=\emptyset$. In this way, we have
completed the induction on $k\in\mathbb{N}$.

Finally, let $C\in[c,E]\subseteq[c,B]\subseteq[a,A]$ be the object given by
the limit of the fusion sequence $\\{[m_{k},M_{k}]\\}_{k\in\mathbb{N}}$, that
is $C=\lim_{m_{k}}M_{k}$, so that $C\in[m_{k},M_{k}]$ and
$r_{m_{k}}(C)=r_{m_{k}}(M_{k})$ for all $k\in\mathbb{N}$. Now, if
$[d,D]\neq\emptyset$ is any basic open set of $\mathcal{R}$ such that
$[d,D]\subseteq[c,C]\subseteq[c,E]$, then we have that
$d\in\mathcal{AR}\\!\restriction\\![c,C]\subseteq\mathcal{AR}\\!\restriction\\![c,E]$.
Furthermore, we can assume that $D\leq E$, in which case there is some
$D^{\prime}\in[\mathtt{depth}_{E}(d),E]$ such that
$\emptyset\neq[d,D^{\prime}]\subseteq[d,D]$, in view of A.3. In closing, let
$k\in\mathbb{N}$ be such that
$m_{k}=\mathtt{depth}_{C}(d)=\mathtt{depth}_{M_{k}}(d)$, thus we infer that
$T_{d}\cap r_{|d|+1}\text{''}\,[d,M_{k+1}]=\emptyset$, and since $M_{k+1}\leq
E$, then we deduce that $T_{d}\neq r_{|d|+1}\text{''}\,[d,E]$, for this reason
$\mathcal{K}_{d}=\emptyset$, so that $D^{\prime}\notin\mathcal{K}_{d}$ and
hence $[d,D^{\prime}]\not\subseteq\mathcal{Q}_{i}$, due to
$D^{\prime}\notin\mathcal{S}_{d}^{\mathcal{Q}_{i}}$. Therefore, we ultimately
conclude that $[d,D]\not\subseteq\mathcal{Q}_{i}$ whenever
$\emptyset\neq[d,D]\subseteq[c,C]$. ∎

Applying previous lemma, we finally conclude with our alternative proof of the
abstract Ellentuck theorem for the class of strong topological Ramsey spaces.
With reference to the statement of theorem 3.4, its first item will be proved
in proposition 4.4, its second item will be shown in proposition 4.5, and its
third item will be verified in corollary 4.6.

######  Proposition 4.4.

Let $(\mathcal{R},\leq,r)$ be a strong topological Ramsey space. Then, for any
$\mathcal{X}\subseteq\mathcal{R}$, the following statements are equivalent:

  1. (a)

$\mathcal{X}$ is Ramsey null.

  2. (b)

$\mathcal{X}$ is nowhere dense.

  3. (c)

$\mathcal{X}$ is meager.

###### Proof.

Notice that it is immediate that every Ramsey null set is also a nowhere dense
set and hence is itself a meager set. So, let us verify that every meager set
is Ramsey null. Indeed, if $\mathcal{X}\subseteq\mathcal{R}$ is a meager set,
then there is a sequence
$\\{\mathcal{Q}_{i}\\}_{i\in\mathbb{N}}\subseteq\wp(\mathcal{R})$ of nowhere
dense sets such that $\mathcal{X}=\bigcup_{i\in\mathbb{N}}\mathcal{Q}_{i}$, so
by considering the sequence
$\\{\mathcal{Q}_{i}^{\complement}\\}_{i\in\mathbb{N}}$, we have that
$\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}^{\complement}=(\bigcup_{i\in\mathbb{N}}\mathcal{Q}_{i})^{\complement}=\mathcal{X}^{\complement}$.
Now, by reduction to absurd, suppose that $\mathcal{X}$ is not a Ramsey null
set, then there is $[a,A]\neq\emptyset$ such that for every $B\in[a,A]$ it is
true that $[a,B]\cap\mathcal{X}\neq\emptyset$, so
$[a,B]\not\subseteq\mathcal{X}^{\complement}$ and hence
$[a,B]\not\subseteq\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}^{\complement}$.
Thus, by virtue of lemma 4.3, we deduce that there are $i\in\mathbb{N}$ and
$[c,C]\neq\emptyset$, with $C\in[a,A]$ and $[c,C]\subseteq[a,A]$, such that
for each $[d,D]\neq\emptyset$, if $[d,D]\subseteq[c,C]$ then
$[d,D]\not\subseteq\mathcal{Q}_{i}^{\complement}$ and hence
$[d,D]\cap\mathcal{Q}_{i}\neq\emptyset$; therefore, we conclude that
$\mathcal{Q}_{i}$ is not a nowhere dense set, which is contradictory. ∎

######  Proposition 4.5.

Let $(\mathcal{R},\leq,r)$ be a strong topological Ramsey space. Then, for any
$\mathcal{X}\subseteq\mathcal{R}$, the following statements are equivalent:

  1. (a)

$\mathcal{X}$ is Ramsey.

  2. (b)

$\mathcal{X}$ has the abstract Baire property.

  3. (c)

$\mathcal{X}$ has the Baire property.

###### Proof.

As in the proof of proposition 4.4 it is immediate that every Ramsey set has
the abstract Baire property and hence also has the Baire property.

Conversely, if $\mathcal{X}\subseteq\mathcal{R}$ has the Baire property, then
$\mathcal{X}\triangle\mathcal{O}$ is meager for some open set $\mathcal{O}$,
hence $\mathcal{X}\triangle\mathcal{O}$ is nowhere dense by proposition 4.4;
thus, for each $[a,A]\neq\emptyset$ there is
$\emptyset\neq[b,B]\subseteq[a,A]$ such that
$[b,B]\cap(\mathcal{X}\triangle\mathcal{O})=\emptyset$. Now, since
$\mathcal{O}$ has the Baire property, there is
$\emptyset\neq[c,C]\subseteq[b,B]\subseteq[a,A]$ such that either
$[c,C]\subseteq\mathcal{O}$ or $[c,C]\cap\mathcal{O}=\emptyset$. Also, since
$[c,C]\subseteq[b,B]$, we have
$[c,C]\cap(\mathcal{X}\triangle\mathcal{O})=\emptyset$. Therefore, when
$[c,C]\subseteq\mathcal{O}$ we deduce that $[c,C]\subseteq\mathcal{X}$, and
when $[c,C]\cap\mathcal{O}=\emptyset$ we infer that
$[c,C]\cap\mathcal{X}=\emptyset$; as a result, we conclude that $\mathcal{X}$
has the abstract Baire property.

Finally, let us check that any set with the abstract Baire property is in turn
a Ramsey set. Indeed, suppose that $\mathcal{X}\subseteq\mathcal{R}$ is not a
Ramsey set, then there exists $[a,A]\neq\emptyset$ such that for every
$B\in[a,A]$ it is true that $[a,B]\not\subseteq\mathcal{X}$ and
$[a,B]\not\subseteq\mathcal{X}^{\complement}$. Now, we consider the eventually
constant sequence
$\\{\mathcal{Q}_{i}\\}_{i\in\mathbb{N}}\subseteq\wp(\mathcal{R})$ defined by
$\mathcal{Q}_{0}=\mathcal{X}$ and $\mathcal{Q}_{i}=\mathcal{R}$ for each
$i>0$, so that
$\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}=\mathcal{Q}_{0}=\mathcal{X}$ and
hence $[a,B]\not\subseteq\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}$ for all
$B\in[a,A]$. Then, applying lemma 4.3 to the sequence
$\\{\mathcal{Q}_{i}\\}_{i\in\mathbb{N}}$, we necessarily conclude that there
exists $C\in[a,A]$ such that $[d,D]\not\subseteq\mathcal{X}$ whenever
$\emptyset\neq[d,D]\subseteq[a,C]$. Next, we consider the constant sequence
$\\{\mathcal{Q}_{i}^{*}\\}_{i\in\mathbb{N}}\subseteq\wp(\mathcal{R})$ defined
by $\mathcal{Q}_{i}^{*}=\mathcal{X}^{\complement}$ for each $i\in\mathbb{N}$,
so that
$\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}^{*}=\mathcal{Q}_{i}^{*}=\mathcal{X}^{\complement}$
and hence $[a,B]\not\subseteq\bigcap_{i\in\mathbb{N}}\mathcal{Q}_{i}^{*}$ for
all $B\in[a,C]\subseteq[a,A]$. Then, applying lemma 4.3 to the sequence
$\\{\mathcal{Q}_{i}^{*}\\}_{i\in\mathbb{N}}$, we deduce that there is some
$[e,E]\neq\emptyset$, with $E\in[a,C]$ and $[e,E]\subseteq[a,C]$, such that
$[d,D]\not\subseteq\mathcal{X}^{\complement}$ whenever
$\emptyset\neq[d,D]\subseteq[e,E]$. Therefore, the basic open set
$[e,E]\neq\emptyset$ satisfies both $[d,D]\not\subseteq\mathcal{X}$ and
$[d,D]\not\subseteq\mathcal{X}^{\complement}$ for every $[d,D]\neq\emptyset$
such that $[d,D]\subseteq[e,E]$, and we can conclude that $\mathcal{X}$ does
not have the abstract Baire property. ∎

######  Corollary 4.6.

Let $(\mathcal{R},\leq,r)$ be a strong topological Ramsey space. If
$\mathcal{R}=\bigcup_{i=1}^{k}\mathcal{P}_{i}$ is any finite partition of the
space $\mathcal{R}$, where each piece $\mathcal{P}_{i}$ has the Baire
property, then there is $H\in\mathcal{R}$ such that
$\mathcal{R}\\!\restriction\\!H\subseteq\mathcal{P}_{i}$ for some
$i\in\\{1,\ldots,k\\}$.

###### Proof.

For $k=2$, let $\mathcal{R}=\mathcal{P}_{1}\cup\mathcal{P}_{2}$ be any
partition of $\mathcal{R}$ into two pieces such that $\mathcal{P}_{1}$ and
$\mathcal{P}_{2}$ have the Baire property, being
$\mathcal{P}_{2}=\mathcal{R}\setminus\mathcal{P}_{1}$. By virtue of the
proposition 4.5 we deduce that $\mathcal{P}_{1}$ is a Ramsey set, then for any
basic open set $[\emptyset,A]$ there is $H\in[\emptyset,A]$ such that either
$[\emptyset,H]\subseteq\mathcal{P}_{1}$ or
$[\emptyset,H]\cap\mathcal{P}_{1}=\emptyset$, where
$[\emptyset,H]=\mathcal{R}\\!\restriction\\!H$. Therefore, the object
$H\in\mathcal{R}$ satisfies that either
$\mathcal{R}\\!\restriction\\!H\subseteq\mathcal{P}_{1}$ or
$\mathcal{R}\\!\restriction\\!H\subseteq\mathcal{P}_{2}$. Now, for $k>2$ a
completely analogous argument is used and the result is similarly proved by
applying induction on $k$. ∎

##  5 On an abstract version of Kastanas theorem

In this section, we present an abstract version of Kastanas game in order to
characterize the Ramsey property for the class of selective topological Ramsey
spaces via topological games, so we conclude with the abstract Kastanas
theorem, which is the main result of our article.

######  Definition 5.1.

Let $(\mathcal{R},\leq,r)$ be a topological Ramsey space, and let
$\mathcal{X}\subseteq\mathcal{R}$ and $[a_{0},B_{0}]\neq\emptyset$ be fixed
but arbitrary. Then, the abstract Kastanas game
$\mathcal{G}_{[a_{0},B_{0}]}(\mathcal{X})$ is defined as follows:

$\setcounter{MaxMatrixCols}{12}\begin{matrix}\textrm{I}&&A_{1}&&A_{2}&&\cdots&&A_{k}&&\cdots&\\\
\textrm{II}&&&[a_{1},B_{1}]&&[a_{2},B_{2}]&&\cdots&&[a_{k},B_{k}]&&\cdots\end{matrix}$

Two players I and II take turns to play alternately both objects in
$\mathcal{R}$ and basic open sets of $\mathcal{R}$, such that the moves of
both players respect the following conditions:

Rules of the game. For every $k\in\mathbb{N}$, both the sequence of objects
$\\{A_{k}\\}_{k\in\mathbb{N}}\subseteq\mathcal{R}$ and
$\\{B_{k}\\}_{k\in\mathbb{N}}\subseteq\mathcal{R}$, as well as the sequence of
approximations $\\{a_{k}\\}_{k\in\mathbb{N}}\subseteq\mathcal{AR}$, must
satisfy the following rules:

$\textbf{(i) }A_{k+1}\in[a_{k},B_{k}]\;\;\;\;\;\;\;\;\textbf{(ii)
}a_{k}\sqsubset a_{k+1}\;\;\;\;\;\;\;\;\textbf{(iii)
}|a_{k+1}|=|a_{k}|+1\;\;\;\;\;\;\;\;\textbf{(iv) }B_{k}\in[a_{k},A_{k}]$

Let $B_{\infty}\in[a_{0},B_{0}]$ be the object given by
$\\{B_{\infty}\\}=\bigcap_{k\in\mathbb{N}}[a_{k}]$. Then, for the game
$\mathcal{G}_{[a_{0},B_{0}]}(\mathcal{X})$ the following is decided:

  * $\bullet$

Player I wins the game if $B_{\infty}\in\mathcal{X}$.

  * $\bullet$

Player II wins the game if $B_{\infty}\notin\mathcal{X}$.

It should be noted that in the definition of the abstract Kastanas game
$\mathcal{G}_{[a_{0},B_{0}]}(\mathcal{X})$, we can take
$[a_{0},B_{0}]\neq\emptyset$ with $a_{0}\sqsubset B_{0}$, in which case we
have that the object $B_{\infty}\in[a_{0},B_{0}]$ is such that
$\\{B_{\infty}\\}=\bigcap_{k\in\mathbb{N}}[a_{k}]$ if and only if
$B_{\infty}=\lim_{n_{k}}B_{k}$ where $n_{k}=\mathtt{depth}_{B_{k}}(a_{k})$.

A topological game, such as the abstract Kastanas game given in the above
definition, is said to be determined if one of the players has a winning
strategy for the game. The concept of strategy is well known (see for instance
[19, chapter 33]), so we do not give a formal definition, though we can say
that a strategy for one of the players is just a function that tells the
player what to play at each step depending on the previous moves of both
players. Furthermore, a strategy in the game is a winning strategy for one of
the players if, using this strategy, the outcome of the game is a win for this
player independently of the moves of the other player.

We start with the following result about topological Ramsey spaces, which
characterizes when player I has a winning strategy in the abstract Kastanas
game.

######  Proposition 5.2.

Let $(\mathcal{R},\leq,r)$ be a topological Ramsey space, and let
$\mathcal{X}\subseteq\mathcal{R}$ and $[a,A]\neq\emptyset$ be given. Then, the
player I has a winning strategy in the game $\mathcal{G}_{[a,A]}(\mathcal{X})$
if and only if there is $H\in[a,A]$ such that $[a,H]\subseteq\mathcal{X}$.

###### Proof.

Given a topological Ramsey space $(\mathcal{R},\leq,r)$, we fix a set of
objects $\mathcal{X}\subseteq\mathcal{R}$ and a basic open set
$[a,M]\neq\emptyset$, for which we assume without loss of generality that
$a\sqsubset M$. So, we proceed to consider the abstract Kastanas game
$\mathcal{G}_{[a,M]}(\mathcal{X})$.

First of all, suppose that there is an object $H\in[a,N]$ such that
$[a,H]\subseteq\mathcal{X}$, in which case it is natural to define a winning
strategy for the player I in the game $\mathcal{G}_{[a,M]}(\mathcal{X})$,
namely the player I must select the object $H$ as the first move. Indeed,
consider any complete run in the game $\mathcal{G}_{[a,M]}(\mathcal{X})$ of
the form

$\mathcal{G}_{[a,M]}(\mathcal{X})$

$\begin{matrix}\textrm{I}&&H&&A_{2}&&A_{3}&&\cdots&\\\
\textrm{II}&&&[a_{1},B_{1}]&&[a_{2},B_{2}]&&[a_{3},B_{3}]&&\cdots\end{matrix}$

So, notice that the object $B_{\infty}\in[a,M]$ given by
$\\{B_{\infty}\\}=\bigcap_{k=1}^{\infty}[a_{k}]$ is such that
$B_{\infty}\in[a_{k},B_{k}]$ for every $k\in\mathbb{N}-\\{0\\}$, consequently
$B_{\infty}\in[a,H]$ and hence $B_{\infty}\in\mathcal{X}$; therefore, the
player I wins the game $\mathcal{G}_{[a,M]}(\mathcal{X})$.

Conversely, suppose that the player I has a winning strategy $\sigma$ in the
game $\mathcal{G}_{[a,M]}(\mathcal{X})$, and we consider the strictly
increasing sequence $\\{n_{k}\\}_{n\in\mathbb{N}}\subseteq\mathbb{N}$
described by $n_{k}=\mathtt{depth}_{M}(a)+k=|a|+k$ for each $k\in\mathbb{N}$.
Now, our goal is to recursively construct a fusion sequence
$\\{[n_{k},A_{k}]\\}_{k\in\mathbb{N}}$ such that its limit
$H=\lim_{n_{k}}A_{k}$ will satisfy that $[a,H]\subseteq\mathcal{X}$.

For the first stage of the process, we put $A_{0}\in\mathcal{R}$ and
$a_{0}\in\mathcal{AR}$ as $A_{0}=M$ and $a_{0}=r_{n_{0}}(A_{0})=a$. Next, let
$A_{1}\in[a_{0},A_{0}]$ be the object determined by
$A_{1}=\sigma\langle\emptyset\rangle$, so that $A_{1}$ is the first move of
the player I in the game $\mathcal{G}_{[a,M]}(\mathcal{X})$ using the winning
strategy $\sigma$, in which case we take the approximation
$a_{1}\in\mathcal{AR}$ given by $a_{1}=r_{n_{1}}(A_{1})$, then $a_{0}\sqsubset
a_{1}$ and $|a_{1}|=|a_{0}|+1$.

A detailed description of the next stage of the construction will help the
reader to go through the general inductive step. In the next stage of the
process, by A.2.1 we can take the finite set
$\\{b\in\mathcal{AR}\,|\,a\sqsubset
b\wedge\mathtt{depth}_{A_{1}}(b)=n_{1}\\}\subseteq\\{b\in\mathcal{AR}\,|\,b\leq_{\text{fin}}a_{1}\\}$,
for which we fix an enumeration
$\\{b_{1}^{0},b_{1}^{1},\ldots,b_{1}^{p_{1}}\\}$ with $p_{1}\in\mathbb{N}$,
where $b_{1}^{0}=b_{1}^{p_{1}}=a_{1}$ and $b_{1}^{i}\neq b_{1}^{j}$ for every
$1\leq i<j\leq p_{1}$. Next, we recursively construct a finite sequence of
objects
$C_{1}^{0},B_{1}^{0},A_{1}^{0},\ldots,C_{1}^{p_{1}},B_{1}^{p_{1}},A_{1}^{p_{1}}$
in $[a,A_{1}]$, putting initially $C_{1}^{0}=B_{1}^{0}=A_{1}^{0}=A_{1}$.
Afterwards, inductively suppose that $C_{1}^{m-1}\geq B_{1}^{m-1}\geq
A_{1}^{m-1}$ have been defined for some $1\leq m\leq p_{1}$, where
$C_{1}^{m-1}\in[a_{1}]$ and $B_{1}^{m-1},A_{1}^{m-1}\in[b_{1}^{m-1}]$. Notice
that $r_{n_{1}}(C_{1}^{m-1})=a_{1}$, hence
$\mathtt{depth}_{A_{1}^{m-1}}(b_{1}^{m-1})\leq\mathtt{depth}_{C_{1}^{m-1}}(b_{1}^{m-1})=n_{1}$,
thus by A.3.2 there is $C_{1}^{m}\in[a_{1},C_{1}^{m-1}]$ such that
$[b_{1}^{m-1},C_{1}^{m}]\subseteq[b_{1}^{m-1},A_{1}^{m-1}]$. Now, since also
$r_{n_{1}}(C_{1}^{m})=a_{1}$ then
$\mathtt{depth}_{C_{1}^{m}}(b_{1}^{m})=n_{1}$, thus by A.3.1 we have that
$[b_{1}^{m},C_{1}^{m}]\neq\emptyset$, so we take
$B_{1}^{m}\in[b_{1}^{m},C_{1}^{m}]$. Later, we consider
$A_{1}^{m}\in[b_{1}^{m},B_{1}^{m}]$ determined by $A_{1}^{m}=\sigma\langle
A_{1};[b_{1}^{m},B_{1}^{m}]\rangle$, so that $A_{1}^{m}$ is the second move of
the player I in the game $\mathcal{G}_{[a,M]}(\mathcal{X})$ using the winning
strategy $\sigma$, in response to the throw $[b_{1}^{m},B_{1}^{m}]$ executed
by the player II. As a result, we obtain the desired objects $C_{1}^{m}\geq
B_{1}^{m}\geq A_{1}^{m}$ for all $m\leq p_{1}$. Finally, for the last case
$b_{1}^{p_{1}}=a_{1}$, we have that $A_{1}^{p_{1}}\in[a_{1}]$ and hence
$r_{n_{1}}(A_{1}^{p_{1}})=a_{1}$; consequently, we define
$A_{2}\in[a_{1},A_{1}]$ as the object $A_{2}=A_{1}^{p_{1}}$, and we take
$a_{2}\in\mathcal{AR}$ given by $a_{2}=r_{n_{2}}(A_{2})$, then $a_{1}\sqsubset
a_{2}$ and $|a_{2}|=|a_{1}|+1$.

In this order of ideas, by induction we fix $k\in\mathbb{N}-\\{0\\}$ and
suppose that the $j$-th stage of the process has been constructed for each
$j\in\\{0,\ldots,k-1\\}$, where we have defined a finite sequence of basic
open sets $\\{[n_{j},A_{j}]\\}_{j=0}^{k}$ such that
$[n_{0},A_{0}]\supseteq\cdots\supseteq[n_{k},A_{k}]$, which it has associated
a finite sequence of approximations $\\{a_{j}\\}_{j=0}^{k}$ described by
$a_{j}=r_{n_{j}}(A_{j})$, hence $a_{0}\sqsubset\cdots\sqsubset a_{k}$ and
$|a_{j+1}|=|a_{j}|+1$.

Now, for the $k$-th stage of the process, we consider the set of
approximations $\\{b\in\mathcal{AR}\,|\,a\sqsubset
b\wedge\mathtt{depth}_{A_{k}}(b)=n_{k}\\}$, which corresponds to a finite set
by virtue of A.2.1, since $\\{b\in\mathcal{AR}\,|\,a\sqsubset
b\wedge\mathtt{depth}_{A_{k}}(b)=n_{k}\\}\subseteq\\{b\in\mathcal{AR}\,|\,b\leq_{\text{fin}}a_{k}\\}$;
thus, we fix an enumeration $\\{b\in\mathcal{AR}\,|\,a\sqsubset
b\wedge\mathtt{depth}_{A_{k}}(b)=n_{k}\\}=\\{b_{k}^{0},b_{k}^{1},\ldots,b_{k}^{p_{k}}\\}$
with $p_{k}\in\mathbb{N}$, such that $b_{k}^{0}=b_{k}^{p_{k}}=a_{k}$ and
$b_{k}^{i}\neq b_{k}^{j}$ for every $1\leq i<j\leq p_{k}$.

Next, we recursively define a finite sequence of objects
$C_{k}^{0},B_{k}^{0},A_{k}^{0},C_{k}^{1},B_{k}^{1},A_{k}^{1},\ldots,C_{k}^{p_{k}},B_{k}^{p_{k}},A_{k}^{p_{k}}$
in $[a,A_{k}]$, such that it satisfies the following properties:

  1. (i)

$C_{k}^{0}=B_{k}^{0}=A_{k}^{0}=A_{k}$.

  2. (ii)

$C_{k}^{m}\geq B_{k}^{m}\geq A_{k}^{m}$ for every $m\in\\{0,\ldots,p_{k}\\}$.

  3. (iii)

$r_{n_{k}}(C_{k}^{m})=r_{n_{k}}(A_{k})=a_{k}$ for every
$m\in\\{0,\ldots,p_{k}\\}$.

  4. (iv)

$C_{k}^{m+1}\in[a_{k},C_{k}^{m}]$ for every $m\in\\{0,\ldots,p_{k}-1\\}$.

  5. (v)

$[b_{k}^{m},C_{k}^{m+1}]\subseteq[b_{k}^{m},A_{k}^{m}]$ for every
$m\in\\{0,\ldots,p_{k}-1\\}$.

  6. (vi)

$B_{k}^{m}\in[b_{k}^{m},C_{k}^{m}]$ for every $m\in\\{0,\ldots,p_{k}\\}$.

  7. (vii)

$A_{k}^{m}=\sigma\langle\overset{\longrightarrow}{\rho}[b_{j}^{n_{m}}];A_{j}^{n_{m}};[b_{k}^{m},B_{k}^{m}]\rangle$
for every $m\in\\{1,\ldots,p_{k}\\}$, where $j\in\\{0,\ldots,k-1\\}$ and
$n_{m}\in\\{1,\ldots,p_{j}\\}$ are such that
$\mathtt{depth}_{A_{j}}(b_{j}^{n_{m}})=n_{j}$ and $a\sqsubseteq
b_{j}^{n_{m}}\sqsubset b_{k}^{m}$ with $|b_{j}^{n_{m}}|=|b_{k}^{m}|-1$, and
being $\overset{\longrightarrow}{\rho}[b_{j}^{n_{m}}]$ the partial run in the
game $\mathcal{G}_{[a,M]}(\mathcal{X})$ coded by $b_{j}^{n_{m}}$ in some
previous stage of the construction, specifically in the $j$-th stage, for
which
$\sigma\langle\overset{\longrightarrow}{\rho}[b_{j}^{n_{m}}]\rangle=A_{j}^{n_{m}}$.

By induction, suppose that the objects $C_{k}^{m-1}\geq B_{k}^{m-1}\geq
A_{k}^{m-1}$ in $[a,A_{k}]$ have been defined for some $1\leq m\leq p_{k}$,
where $C_{k}^{m-1}\in[a_{k}]$ and $B_{k}^{m-1},A_{k}^{m-1}\in[b_{k}^{m-1}]$.
Since $r_{n_{k}}(C_{k}^{m-1})=r_{n_{k}}(A_{k})=a_{k}$, then
$\mathtt{depth}_{C_{k}^{m-1}}(b_{k}^{m-1})=\mathtt{depth}_{A_{k}}(b_{k}^{m-1})=n_{k}$,
hence
$\mathtt{depth}_{A_{k}^{m-1}}(b_{k}^{m-1})\leq\mathtt{depth}_{C_{k}^{m-1}}(b_{k}^{m-1})<\infty$,
so by applying A.3.2 we infer that there is an object
$C_{k}^{m}\in[a_{k},C_{k}^{m-1}]$ such that
$[b_{k}^{m-1},C_{k}^{m}]\subseteq[b_{k}^{m-1},A_{k}^{m-1}]$. Also, notice that
$\mathtt{depth}_{C_{k}^{m}}(b_{k}^{m})=\mathtt{depth}_{A_{k}}(b_{k}^{m})=n_{k}$
because $C_{k}^{m}\in[a_{k},A_{k}]$ with $a_{k}=r_{n_{k}}(A_{k})$, thus by
A.3.1 we have that $\mathtt{depth}_{C_{k}^{m}}(b_{k}^{m})<\infty$ implies
$[b_{k}^{m},C_{k}^{m}]\neq\emptyset$, then we take any object
$B_{k}^{m}\in[b_{k}^{m},C_{k}^{m}]$.

Moreover, we know that $b_{k}^{m}\leq_{\text{fin}}a_{k}$; thus, if
$d\in\mathcal{AR}$ is the approximation such that $a\sqsubseteq d\sqsubset
b_{k}^{m}$ and $|d|=|b_{k}^{m}|-1$, then by A.2.3 we deduce that there is an
approximation $a_{j}\sqsubset a_{k}$ with $j\in\\{0,\ldots,k-1\\}$ such that
$d\leq_{\text{fin}}a_{j}$; in fact, we can assume that $j$ is the smallest
satisfying this property, in which case $\mathtt{depth}_{A_{j}}(d)=n_{j}$
because $a_{j}=r_{n_{j}}(A_{j})$, consequently there is some
$n_{m}\in\\{1,\ldots,p_{j}\\}$ such that $d=b_{j}^{n_{m}}$, whereby
$\mathtt{depth}_{A_{j}}(b_{j}^{n_{m}})=n_{j}$ and $a\sqsubseteq
b_{j}^{n_{m}}\sqsubset b_{k}^{m}$ with $|b_{j}^{n_{m}}|=|b_{k}^{m}|-1$.

Now, in the $j$-th stage previously constructed, we consider the partial run
$\overset{\longrightarrow}{\rho}[b_{j}^{n_{m}}]$ in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$ coded by the approximation $b_{j}^{n_{m}}$,
for which we have that
$\sigma\langle\overset{\longrightarrow}{\rho}[b_{j}^{n_{m}}]\rangle=A_{j}^{n_{m}}$,
so at some point $A_{j}^{n_{m}}$ was the last move made by the player I
through the winning strategy $\sigma$. To continue, notice that
$B_{k}^{m}\in[b_{k}^{m},A_{j}^{n_{m}}]$, since
$B_{k}^{m}\in[b_{k}^{m},C_{k}^{m}]\subseteq[b_{k}^{m},A_{k}]\subseteq[b_{j}^{n_{m}},C_{j}^{n_{m}+1}]\subseteq[b_{j}^{n_{m}},A_{j}^{n_{m}}]$
whenever $1\leq n_{m}<p_{j}$, and also
$B_{k}^{m}\in[b_{k}^{m},C_{k}^{m}]\subseteq[b_{k}^{m},A_{k}]\subseteq[a_{j},A_{j+1}]$
if $n_{m}=p_{j}$. Therefore, we define the object
$A_{k}^{m}\in[b_{k}^{m},B_{k}^{m}]$ given by
$A_{k}^{m}=\sigma\langle\overset{\longrightarrow}{\rho}[b_{j}^{n_{m}}];A_{j}^{n_{m}};[b_{k}^{m},B_{k}^{m}]\rangle$,
so that $A_{k}^{m}$ is the next move of the player I in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$ using the winning strategy $\sigma$, in
response to the throw $[b_{k}^{m},B_{k}^{m}]$ executed by the player II. To
conclude, in this way we have constructed the desired objects $C_{k}^{m}\geq
B_{k}^{m}\geq A_{k}^{m}$ for all $m\leq p_{k}$.

In the end, for the last case corresponding to $b_{k}^{p_{k}}=a_{k}$, we
obtain that
$A_{k}^{p_{k}}=\sigma\langle\overset{\longrightarrow}{\rho}[a_{k-1}];A_{k};[a_{k},B_{k}^{p_{k}}]\rangle$,
then $A_{k}^{p_{k}}\in[a_{k},B_{k}^{p_{k}}]$ and hence
$r_{n_{k}}(A_{k}^{p_{k}})=a_{k}$. Therefore, we finally define
$A_{k+1}\in[a_{k},A_{k}]$ as the object $A_{k+1}=A_{k}^{p_{k}}$, and we take
the approximation $a_{k+1}\in\mathcal{AR}$ given by
$a_{k+1}=r_{n_{k+1}}(A_{k+1})$, so that $a_{k}\sqsubset a_{k+1}$ and
$|a_{k+1}|=|a_{k}|+1$.

In this way, we have successfully constructed, among other considerations, the
fusion sequence $\\{[n_{k},A_{k}]\\}_{k\in\mathbb{N}}\subseteq\wp([a,M])$,
where $n_{k}=\mathtt{depth}_{M}(a)+k=|a|+k$ and $r_{n_{k}}(A_{k})=a_{k}$ for
each $k\in\mathbb{N}$. Thus, if the object $H\in[a,M]$ is the limit
$H=\lim_{n_{k}}A_{k}$ of the fusion sequence
$\\{[n_{k},A_{k}]\\}_{k\in\mathbb{N}}$, then it is true that
$H\in[a_{k},A_{k}]$ for every $k\in\mathbb{N}$ and consequently
$\\{H\\}=\bigcap_{n\in\mathbb{N}}[a_{k}]$.

Next, we proceed to consider the following complete run in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$:

$\mathcal{G}_{[a,M]}(\mathcal{X})$

$\begin{matrix}\textrm{I}&&A_{1}&&A_{2}&&A_{3}&&\cdots&\\\
\textrm{II}&&&[a_{1},B_{1}^{p_{1}}]&&[a_{2},B_{2}^{p_{2}}]&&[a_{3},B_{3}^{p_{3}}]&&\cdots\end{matrix}$

In view of the fact that this complete run is obtained by the player I
applying the winning strategy $\sigma$ in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$, due to
$A_{1}=\sigma\langle\emptyset\rangle$ and
$A_{k+1}=\sigma\langle\overset{\longrightarrow}{\rho}[a_{k-1}];A_{k};[a_{k},B_{k}^{p_{k}}]\rangle$
for every $k\in\mathbb{N}-\\{0\\}$, then we deduce that the player I wins the
game $\mathcal{G}_{[a,M]}(\mathcal{X})$; therefore, we infer that
$H\in\mathcal{X}$ because $\\{H\\}=\bigcap_{n\in\mathbb{N}}[a_{k}]$.

Finally, let us check that the object $H\in[a,M]\cap\mathcal{X}$ satisfies
that $[a,H]\subseteq\mathcal{X}$. Indeed, if $B\in[a,H]$ then $B\leq H$ with
$r_{|a|}(B)=a=r_{|a|}(H)$; so, by virtue of A.2.2 we infer that for each
$m\in\mathbb{N}$ there is some $j_{m}\in\mathbb{N}$ such that
$r_{|a|+m}(B)\leq_{\text{fin}}r_{|a|+j_{m}}(H)$, hence
$r_{|a|+m}(B)\leq_{\text{fin}}a_{j_{m}}$, and without loss of generality we
can suppose that $j_{m}$ is the smallest satisfying this property, in which
case $\mathtt{depth}_{A_{j_{m}}}(r_{|a|+j_{m}}(B))=n_{j_{m}}$, whereby
$r_{|a|+j_{m}}(B)=b_{j_{m}}^{i_{m}}$ for some
$i_{m}\in\\{1,\ldots,p_{j_{m}}\\}$. Therefore, the object $B\in[a,H]$
satisfies that $\\{B\\}=\bigcap_{m\in\mathbb{N}}[b_{j_{m}}^{i_{m}}]$; in view
of the foregoing, we proceed to consider the following complete run in the
game $\mathcal{G}_{[a,M]}(\mathcal{X})$:

$\mathcal{G}_{[a,M]}(\mathcal{X})$

$\begin{matrix}\textrm{I}&&A_{1}&&A_{j_{2}}^{i_{2}}&&A_{j_{3}}^{i_{3}}&&\cdots&\\\
\textrm{II}&&&[b_{j_{1}}^{i_{1}},B_{j_{1}}^{i_{1}}]&&[b_{j_{2}}^{i_{2}},B_{j_{2}}^{i_{2}}]&&[b_{j_{3}}^{i_{3}},B_{j_{3}}^{i_{3}}]&&\cdots\end{matrix}$

It should be noted that this complete run is obtained by applying the winning
strategy $\sigma$ of the player I in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$, hence the player I wins the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$, as a result we deduce that
$B\in\mathcal{X}$ because
$\\{B\\}=\bigcap_{m\in\mathbb{N}}[b_{j_{m}}^{i_{m}}]$. Consequently, we
finally conclude that $[a,H]\subseteq\mathcal{X}$. ∎

Motivated by the concept of selective coideal on the Ellentuck space, given by
Mathias in [24], as well as by the abstract approximation of local Ramsey
theory in the context of topological Ramsey spaces, proposed by Di Prisco,
Mijares and Nieto in [6], we now introduce the wide class of selective
topological Ramsey spaces.

######  Definition 5.3.

Let $(\mathcal{R},\leq,r)$ be a topological Ramsey space. Given any
$[a,B]\neq\emptyset$ and a sequence
$\\{D_{n}\\}_{n\in\mathbb{N}}\subseteq[a,B]$, we say that an object
$D_{\infty}\in[a,B]$ is a diagonalization of $\\{D_{n}\\}_{n\in\mathbb{N}}$
whenever $\emptyset\neq[b,D_{\infty}]\subseteq[b,D_{n_{b}}]$ for every
$b\in\mathcal{AR}\\!\restriction\\![a,D_{\infty}]$, where
$n_{b}=\mathtt{depth}_{B}(b)-\mathtt{depth}_{B}(a)$.

######  Definition 5.4.

A topological Ramsey space $(\mathcal{R},\leq,r)$ is selective if for any
$[a,B]\neq\emptyset$ and for all sequence
$\\{D_{n}\\}_{n\in\mathbb{N}}\subseteq[a,B]$ such that $D_{n+1}\leq D_{n}$ for
each $n\in\mathbb{N}$, there exists some $D_{\infty}\in[a,B]$ such that
$D_{\infty}$ is a diagonalization of $\\{D_{n}\\}_{n\in\mathbb{N}}$.

We point out that many well known examples of topological Ramsey spaces are
selective; however, there are exceptions, for instance, spaces of strong
subtrees (see [26, 31]).

We turn now to the case when player II has a winning strategy in the abstract
Kastanas game. To do this, first we show a lemma about selectivity that we
will need later, which is inspired by the development of local version of
Kastanas game given by Matet in [22].

######  Lemma 5.5.

Let $(\mathcal{R},\leq,r)$ be a selective topological Ramsey space. Let
$[a,B]\neq\emptyset$ with $a\sqsubset B$, and let
$f:[a,B]\rightarrow\mathcal{AR}_{|a|+1}$ and $g:[a,B]\rightarrow[a,B]$ be
functions such that $a\sqsubset f(A)$ and $g(A)\in[f(A),A]$ for each
$A\in[a,B]$. Then, there exists some $E_{f,g}\in[a,B]$ which satisfies that
for every $q\in r_{|a|+1}\text{''}\,[a,E_{f,g}]$ there is $A\in[a,B]$ such
that $f(A)=q$ and $\emptyset\neq[q,E_{f,g}]\subseteq[q,g(A)]$.

###### Proof.

Let $(\mathcal{R},\leq,r)$ be a selective topological Ramsey space. Given a
basic open set $[a,B]\neq\emptyset$ with $a\sqsubset B$, we fix functions
$f:[a,B]\rightarrow\mathcal{AR}_{|a|+1}$ and $g:[a,B]\rightarrow[a,B]$ such
that $a\sqsubset f(A)$ and $g(A)\in[f(A),A]$ for each $A\in[a,B]$.

First of all, for every $p\in r_{|a|+1}\text{''}\,[a,B]$ we consider the set
$\mathcal{W}_{p}\subseteq[a,B]$ defined by

$\mathcal{W}_{p}=\\{D\in[a,B]\,|\,(\exists\,A\in[a,B])(g(A)=D\wedge
f(A)=p)\\}$.

We proceed to recursively construct a sequence of objects
$\\{D_{n}\\}_{n\in\mathbb{N}}\subseteq[a,B]$, with $D_{0}=B$ and $D_{n+1}\leq
D_{n}$ for all $n\in\mathbb{N}$, as follows:

By induction, let $n\in\mathbb{N}$ and suppose that the objects $D_{0}\geq
D_{1}\geq\cdots\geq D_{n-1}$ in $[a,B]$ has been constructed. Now, let
$\Gamma_{n}=\\{p\in
r_{|a|+1}\text{''}\,[a,B]\,|\,\mathtt{depth}_{B}(p)=|a|+n\\}$, then the set
$\Gamma_{n}$ is finite by virtue of A.2.1, since
$\Gamma_{n}\subseteq\\{p\in\mathcal{AR}\,|\,p\leq_{\text{fin}}r_{|a|+n}(B)\\}$;
so, we fix any enumeration $\Gamma_{n}=\\{p_{n}^{1},\ldots,p_{n}^{k_{n}}\\}$
with $k_{n}\in\mathbb{N}$. Next, we recursively define a finite sequence of
objects $D_{n}^{0},D_{n}^{1},\ldots,D_{n}^{k_{n}}$ in $[a,D_{n-1}]$. To do
this, we put $D_{n}^{0}=D_{n-1}$, and assuming that the object $D_{n}^{i-1}$
has been determined for $1\leq i\leq k_{n}$, we consider the set
$\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]$ and we analyze the following
cases: on the one hand, if
$\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]=\emptyset$, then we take
$D_{n}^{i}=D_{n}^{i-1}$; on the other hand, if
$\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]\neq\emptyset$, then we take
$D_{n}^{i}$ as any fixed object
$D_{n}^{i}\in\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]$. In this way, we get
finitely many objects $D_{n}^{0}\geq D_{n}^{1}\geq\cdots\geq D_{n}^{k_{n}}$ in
$[a,D_{n-1}]$; thus, we define $D_{n}\in[a,D_{n-1}]$ as the object
$D_{n}=D_{n}^{k_{n}}$.

As an additional remark to the construction just explained, it should be noted
that the family of approximations $r_{|a|+1}\text{''}\,[a,B]$ can be described
in terms of the sequence of finite sets $\\{\Gamma_{n}\\}_{n\in\mathbb{N}}$ as
follows:

$r_{|a|+1}\text{''}\,[a,B]=\bigcup_{n\in\mathbb{N}}\Gamma_{n}=\\{p_{n}^{i}\,|\,n\in\mathbb{N}\wedge
1\leq i\leq k_{n}\\}$.

Now, we have that the sequence of objects
$\\{D_{n}\\}_{n\in\mathbb{N}}\subseteq[a,B]$ is decreasing with respect to
quasi-order $\leq$, then using the fact that our topological Ramsey space
$(\mathcal{R},\leq,r)$ is selective, we fix an object $D_{\infty}\in[a,B]$
such that $D_{\infty}$ corresponds to a diagonalization of sequence
$\\{D_{n}\\}_{n\in\mathbb{N}}$; therefore, for every
$b\in\mathcal{AR}\\!\restriction\\![a,D_{\infty}]$, if
$\mathtt{depth}_{B}(b)=|a|+n$ then
$\emptyset\neq[b,D_{\infty}]\subseteq[b,D_{n}]$.

Next, let $\mathcal{O}\subseteq\mathcal{AR}_{|a|+1}$ be the collection of
approximations defined by

$\mathcal{O}=\\{p_{n}^{i}\in
r_{|a|+1}\text{''}\,[a,B]\;|\;\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]\neq\emptyset\\}.$

Applying the abstract pigeonhole principle A.4 to the family $\mathcal{O}$, we
deduce that there exists some object $E_{f,g}\in[a,D_{\infty}]\subseteq[a,B]$
such that $r_{|a|+1}\text{''}\,[a,E_{f,g}]$ is either contained in
$\mathcal{O}$ or disjoint from $\mathcal{O}$, so

$r_{|a|+1}\text{''}\,[a,E_{f,g}]\subseteq\mathcal{O}\;\;\;\;\;\vee\;\;\;\;\;\mathcal{O}\cap
r_{|a|+1}\text{''}\,[a,E_{f,g}]=\emptyset$.

However, the second option is impossible, this due to
$f(E_{f,g})\in\mathcal{O}\cap r_{|a|+1}\text{''}\,[a,E_{f,g}]$. Indeed, on the
one hand $a\sqsubset f(E_{f,g})$ with $|f(E_{f,g})|=|a|+1$ and also
$g(E_{f,g})\in[f(E_{f,g}),E_{f,g}]$, so $r_{|a|+1}(g(E_{f,g}))=f(E_{f,g})$ and
hence $f(E_{f,g})\in r_{|a|+1}\text{''}\,[a,E_{f,g}]$. On the other hand,
$f(E_{f,g})=p_{n}^{i}$ for some $n\in\mathbb{N}$ and $1\leq i\leq k_{n}$, so
that $\mathtt{depth}_{B}(f(E_{f,g}))=|a|+n$ and also
$g(E_{f,g})\in\mathcal{W}_{p_{n}^{i}}$, then
$[f(E_{f,g}),E_{f,g}]\subseteq[f(E_{f,g}),D_{\infty}]\subseteq[f(E_{f,g}),D_{n}]\subseteq[a,D_{n}]\subseteq[a,D_{n}^{i-1}]$,
thus $g(E_{f,g})\in[a,D_{n}^{i-1}]$; therefore, we deduce that
$\mathcal{W}_{f(E_{f,g})}\cap[a,D_{n}^{i-1}]=\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]\neq\emptyset$,
hence $f(E_{f,g})\in\mathcal{O}$. Consequently, it must necessarily be
satisfied that $r_{|a|+1}\text{''}\,[a,E_{f,g}]\subseteq\mathcal{O}$.

Finally, let us verify that for every $q\in r_{|a|+1}\text{''}\,[a,E_{f,g}]$
there is some $A\in[a,B]$ such that $f(A)=q$ and
$\emptyset\neq[q,E_{f,g}]\subseteq[q,g(A)]$. Indeed, if $q\in
r_{|a|+1}\text{''}\,[a,E_{f,g}]$ then $q=p_{n}^{i}$ for some $n\in\mathbb{N}$
and $1\leq i\leq k_{n}$; additionally, since
$r_{|a|+1}\text{''}\,[a,E_{f,g}]\subseteq\mathcal{O}$ then $q\in\mathcal{O}$,
so that $\mathcal{W}_{p_{n}^{i}}\cap[a,D_{n}^{i-1}]\neq\emptyset$ and in
particular $D_{n}^{i}\in\mathcal{W}_{p_{n}^{i}}$, that is
$D_{n}^{i}\in\mathcal{W}_{q}$, which means that there exists some $A\in[a,B]$
such that $f(A)=q$ y $g(A)=D_{n}^{i}$, hence $[q,g(A)]=[q,D_{n}^{i}]$;
moreover, since $\mathtt{depth}_{B}(q)=|a|+n$ then we have that
$[q,E_{f,g}]\subseteq[q,D_{\infty}]\subseteq[q,D_{n}]\subseteq[q,D_{n}^{i}]$,
reason for which we conclude that $\emptyset\neq[q,E_{f,g}]\subseteq[q,g(A)]$.
∎

Applying the above lemma, we consider selective topological Ramsey spaces to
obtain the following result that characterizes when player II has a winning
strategy in the abstract Kastanas game.

######  Proposition 5.6.

Let $(\mathcal{R},\leq,r)$ be a selective topological Ramsey space, and let
$\mathcal{X}\subseteq\mathcal{R}$ and $[a,M]\neq\emptyset$ be given. Then, the
player II has a winning strategy in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$ if and only if for every $N\in[a,M]$ there
is $H\in[a,N]$ such that $[a,H]\cap\mathcal{X}=\emptyset$.

###### Proof.

Given a selective topological Ramsey space $(\mathcal{R},\leq,r)$, we fix a
set of objects $\mathcal{X}\subseteq\mathcal{R}$ and a basic open set
$[a,M]\neq\emptyset$, for which we assume without loss of generality that
$a\sqsubset M$. So, we proceed to consider the abstract Kastanas game
$\mathcal{G}_{[a,M]}(\mathcal{X})$.

First of all, suppose that for all $N\in[a,M]$ there exists an object
$H\in[a,N]$ such that $[a,H]\cap\mathcal{X}=\emptyset$, in this case a winning
strategy for the player II in the game $\mathcal{G}_{[a,M]}(\mathcal{X})$, is
given as follows: player II must be the basic open set $[r_{|a|+1}(H),H]$ as
the first move, in response to the first throw $N$ executed by player I.
Indeed, consider any complete run in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$ of the form

$\mathcal{G}_{[a,M]}(\mathcal{X})$

$\begin{matrix}\textrm{I}&&N&&A_{2}&&A_{3}&&\cdots&\\\
\textrm{II}&&&[r_{|a|+1}(H),H]&&[a_{2},B_{2}]&&[a_{3},B_{3}]&&\cdots\end{matrix}$

So, notice that the object $B_{\infty}\in[a,M]$ given by
$\\{B_{\infty}\\}=\bigcap_{k=2}^{\infty}[a_{k}]$ is such that
$B_{\infty}\in[a,H]$, thus $B_{\infty}\notin\mathcal{X}$ and consequently the
player II wins the game $\mathcal{G}_{[a,M]}(\mathcal{X})$.

Conversely, suppose that the player II has a winning strategy $\tau$ in the
game $\mathcal{G}_{[a,M]}(\mathcal{X})$, and let $N\in[a,M]$ be a fixed but
arbitrary object. Now, our goal is to construct a winning strategy $\sigma$
for the player I in the game
$\mathcal{G}_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$.

In order to successfully accomplish our goal, we must iteratively apply lemma
5.5 to recursively construct a complete run in the game
$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$ with consecutive moves of the
player II given by a sequence of basic open sets
$\\{[a_{n},B_{n}]\\}_{n=1}^{\infty}$, for which we define sequences of
functions $\\{f_{n}\\}_{n=1}^{\infty}$ and $\\{g_{n}\\}_{n=1}^{\infty}$ as
well as sequences of objects $\\{A_{n}\\}_{n=1}^{\infty}$ and
$\\{E_{f_{n},g_{n}}\\}_{n=1}^{\infty}$ in the space $\mathcal{R}$, such that:

  1. (1)

$[f_{1}(A),g_{1}(A)]=\tau\langle A\rangle$ for every $A\in[a,N]$.

  2. (2)

$[f_{n+1}(A),g_{n+1}(A)]=\tau\left\langle
A_{1};[f_{1}(A_{1}),g_{1}(A_{1})];\ldots;A_{n};[f_{n}(A_{n}),g_{n}(A_{n})];A\right\rangle$
for every $A\in[a_{n},B_{n}]$.

  3. (3)

$A_{1},E_{f_{1},g_{1}}\in[a,N]$ and
$A_{n+1},E_{f_{n+1},g_{n+1}}\in[a_{n},B_{n}]$.

  4. (4)

$f_{n}(A_{n})=a_{n}$ and
$B_{n}\in[a_{n},E_{f_{n},g_{n}}]\subseteq[a_{n},g_{n}(A_{n})]$.

For the first stage of the process, we take functions
$f_{1}:[a,N]\rightarrow\mathcal{AR}_{|a|+1}$ and $g_{1}:[a,N]\rightarrow[a,N]$
defined by $[f_{1}(A),g_{1}(A)]=\tau\langle A\rangle$ for each $A\in[a,N]$, so
the basic open set $[f_{1}(A),g_{1}(A)]$ is the first move of the player II
using the winning strategy $\tau$ in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$, in response to any throw $A$ in $[a,N]$
given by the player I; then, we have $a\sqsubset f_{1}(A)$ and
$g_{1}(A)\in[f_{1}(A),A]$ for each $A\in[a,N]$. Therefore, by lemma 5.5, we
infer that there exists an object $E_{f_{1},g_{1}}\in[a,N]$ with the property
that for every $q\in r_{|a|+1}\text{''}\,[a,E_{f_{1},g_{1}}]$ there is
$A\in[a,N]$ such that $f_{1}(A)=q$ and
$\emptyset\neq[q,E_{f_{1},g_{1}}]\subseteq[q,g_{1}(A)]$. Thus, we define
$\sigma\langle\emptyset\rangle=E_{f_{1},g_{1}}$.

Again, we give a detailed description of the next stage of the construction
for the benefit of the reader. In the next stage of the process, suppose that
the short partial run given by $E_{f_{1},g_{1}}$ followed by $[a_{1},B_{1}]$,
as the first moves of the players I and II, respectively, has been developed
in the game $G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$, so that
$B_{1}\in[a_{1},E_{f_{1},g_{1}}]$ and $a\sqsubset a_{1}$ with $|a_{1}|=|a|+1$.
Since $a_{1}\in r_{|a|+1}\text{''}\,[a,E_{f_{1},g_{1}}]$, then there is
$A_{1}\in[a,N]$ such that $f_{1}(A_{1})=a_{1}$ and
$\emptyset\neq[a_{1},E_{f_{1},g_{1}}]\subseteq[a_{1},g_{1}(A_{1})]$, so
$B_{1}\in[a_{1},g_{1}(A_{1})]$ and hence
$\emptyset\neq[a_{1},B_{1}]\subseteq[a_{1},g_{1}(A_{1})]$. Let
$f_{2}:[a_{1},B_{1}]\rightarrow\mathcal{AR}_{|a_{1}|+1}$ and
$g_{2}:[a_{1},B_{1}]\rightarrow[a_{1},B_{1}]$ be functions defined by
$[f_{2}(A),g_{2}(A)]=\tau\langle A_{1};[a_{1},g_{1}(A_{1})];A\rangle$ for each
$A\in[a_{1},B_{1}]$, so the second move of the player II using the winning
strategy $\tau$ in the game $\mathcal{G}_{[a,M]}(\mathcal{X})$ is
$[f_{2}(A),g_{2}(A)]$, in response to any throw $A$ in $[a_{1},B_{1}]$ given
by the player I, who had selected $A_{1}$ as previous throw; then,
$a_{1}\sqsubset f_{2}(A)$ and $g_{2}(A)\in[f_{2}(A),A]$ for each
$A\in[a_{1},B_{1}]$. Therefore, by the lemma 5.5, there exists
$E_{f_{2},g_{2}}\in[a_{1},B_{1}]$ with the property that for every $q\in
r_{|a_{1}|+1}\text{''}\,[a_{1},E_{f_{2},g_{2}}]$ there is $A\in[a_{1},B_{1}]$
such that $f_{2}(A)=q$ and
$\emptyset\neq[q,E_{f_{2},g_{2}}]\subseteq[q,g_{2}(A)]$. Thus, we define
$\sigma\langle E_{f_{1},g_{1}};[a_{1},B_{1}]\rangle=E_{f_{2},g_{2}}$.

In this order of ideas, by induction we fix $n\in\mathbb{N}-\\{0\\}$ and
suppose that the $j$-th stage of the process has been constructed for each
$j\in\\{1,\ldots,n\\}$, that is, we are assuming defined the first $n$ stages
of the strategy $\sigma$ for the player I in the game
$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$, where the finite sequence of
basic open sets $\\{[a_{j},B_{j}]\\}_{j=1}^{n}$ corresponds to the first $n$
consecutive moves of the player II, which it has associated both functions
$\\{f_{n}\\}_{j=1}^{n}$ and $\\{g_{n}\\}_{j=1}^{n}$ as well as objects
$\\{A_{j}\\}_{j=1}^{n}$ and $\\{E_{f_{j},g_{j}}\\}_{j=1}^{n}$ in the space
$\mathcal{R}$, satisfying the conditions established in items (1), (2), (3),
(4) above, such that $\sigma\langle\emptyset\rangle=E_{f_{1},g_{1}}$ and
$\sigma\langle
E_{f_{1},g_{1}};[a_{1},B_{1}];\ldots;E_{f_{j-1},g_{j-1}};[a_{j-1},B_{j-1}]\rangle=E_{f_{j},g_{j}}$
for every $1\leq j\leq n$.

Now, in the game $G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$ we consider the
partial run described by the consecutive moves
$E_{f_{1},g_{1}},\ldots,E_{f_{n},g_{n}}$ executed by the player I, which are
alternated together with the successive throws
$[a_{1},B_{1}],\ldots,[a_{n},B_{n}]$ of the player II, so in particular we
have $B_{n}\in[a_{n},E_{f_{n},g_{n}}]$ and $a_{n-1}\sqsubset a_{n}$ with
$|a_{n}|=|a_{n-1}|+1$, hence $a_{n}\in
r_{|a_{n-1}|+1}\text{''}\,[a_{n-1},E_{f_{n},g_{n}}]$; then, there exists some
object $A_{n}\in[a_{n-1},B_{n-1}]$ such that $f_{n}(A_{n})=a_{n}$ and
$\emptyset\neq[a_{n},E_{f_{n},g_{n}}]\subseteq[a_{n},g_{n}(A_{n})]$, this by
virtue of the fact that the object $E_{f_{n},g_{n}}$ was previously
constructed using lemma 5.5; therefore, $B_{n}\in[a_{n},g_{n}(A_{n})]$ and
hence $\emptyset\neq[a_{n},B_{n}]\subseteq[a_{n},g_{n}(A_{n})]$.

Next, we consider functions
$f_{n+1}:[a_{n},B_{n}]\rightarrow\mathcal{AR}_{|a_{n}|+1}$ and
$g_{n+1}:[a_{n},B_{n}]\rightarrow[a_{n},B_{n}]$ defined by
$[f_{n+1}(A),g_{n+1}(A)]=\tau\langle
A_{1};[a_{1},g_{1}(A_{1})];\ldots;A_{n};[a_{n},g_{n}(A_{n})];A\rangle$ for
each $A\in[a_{n},B_{n}]$, that is, the basic open set
$[f_{n+1}(A),g_{n+1}(A)]$ corresponds to the $(n+1)$-th move of the player II
using the winning strategy $\tau$ in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$, in response to any throw $A$ in
$[a_{n},B_{n}]$ given by the player I, who had successively selected
$A_{1},\ldots,A_{n}$ as previous throws; then, it is clear that
$a_{n}\sqsubset f_{n+1}(A)$ and $g_{n+1}(A)\in[f_{n+1}(A),A]$ for each
$A\in[a_{n},B_{n}]$. Therefore, by applying lemma 5.5, we deduce the existence
of some object $E_{f_{n+1},g_{n+1}}\in[a_{n},B_{n}]$ with the special property
that for every $q\in r_{|a_{n}|+1}\text{''}\,[a_{n},E_{f_{n+1},g_{n+1}}]$
there is $A\in[a_{n},B_{n}]$ such that $f_{n+1}(A)=q$ and
$\emptyset\neq[q,E_{f_{n+1},g_{n+1}}]\subseteq[q,g_{n+1}(A)]$. On merit of the
above, we define $\sigma\langle
E_{f_{1},g_{1}};[a_{1},B{1}];\ldots;E_{f_{n},g_{n}};[a_{n},B_{n}]\rangle=E_{f_{n+1},g_{n+1}}$.

In this way, we have successfully constructed, among other considerations, the
strategy $\sigma$ for the player I in the game
$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$, which consists in consecutively
selecting the objects of the sequence $\\{E_{f_{n},g_{n}}\\}_{n=1}^{\infty}$
in each of the moves, thus:

$\sigma\langle\emptyset\rangle=E_{f_{1},g_{1}}\,\text{ and
}\,\sigma\left\langle
E_{f_{1},g_{1}};[a_{1},B_{1}];\ldots;E_{f_{n},g_{n}};[a_{n},B_{n}]\right\rangle=E_{f_{n+1},g_{n+1}}\,\text{
for every }n\in\mathbb{N}-\\{0\\}.$

Now, let $B_{\infty}\in[a,N]$ be the object determined by
$\\{B_{\infty}\\}=\bigcap_{n=1}^{\infty}[a_{n}]$. To continue, we proceed to
consider the following complete run in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$:

$\mathcal{G}_{[a,M]}(\mathcal{X})$

$\begin{matrix}\textrm{I}&&A_{1}&&A_{2}&&A_{3}&&\cdots&\\\
\textrm{II}&&&[a_{1},g_{1}(A_{1})]&&[a_{2},g_{2}(A_{2})]&&[a_{3},g_{3}(A_{3})]&&\cdots\end{matrix}$

In view of the fact that this complete run is obtained by applying the winning
strategy $\tau$ for the player II in the game
$\mathcal{G}_{[a,M]}(\mathcal{X})$, since $\tau\left\langle
A_{1};[a_{1},g_{1}(A_{1})];\ldots;A_{n};[a_{n},g_{n}(A_{n})];A_{n+1}\right\rangle=[a_{n+1},g_{n+1}(A_{n+1})]$
for each $n\in\mathbb{N}-\\{0\\}$, then we deduce that the player II wins the
game $\mathcal{G}_{[a,M]}(\mathcal{X})$, thus $B_{\infty}\notin\mathcal{X}$
and hence $B_{\infty}\in\mathcal{R}\setminus\mathcal{X}$.

On the other hand, by applying the strategy $\sigma$ that has been recursively
constructed for the player I in the game
$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$, where
$\\{E_{f_{n},g_{n}}\\}_{n=1}^{\infty}$ corresponds to the sequence of
consecutive moves of the player I, with $\\{[a_{n},B_{n}]\\}_{n=1}^{\infty}$
being the respective sequence of successive moves of the player II, then we
obtain the following complete run in the game
$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$:

$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$

$\begin{matrix}\textrm{I}&&E_{f_{1},g_{1}}&&E_{f_{2},g_{2}}&&E_{f_{3},g_{3}}&&\cdots&\\\
\textrm{II}&&&[a_{1},B_{1}]&&[a_{2},B_{2}]&&[a_{3},B_{3}]&&\cdots\end{matrix}$

Consequently, by virtue of the fact that
$B_{\infty}\in\mathcal{R}\setminus\mathcal{X}$, then it follows that the
player I wins the game $G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$; thus, we
conclude that the strategy $\sigma$ is actually a winning strategy for the
player I in the game $G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$.

Finally, since the player I has a winning strategy in th game
$G_{[a,N]}(\mathcal{R}\setminus\mathcal{X})$, then by proposition 5.2 we
deduce that there exists some object $H\in[a,N]$ such that
$[a,H]\subseteq\mathcal{R}\setminus\mathcal{X}$, and thus
$[a,H]\cap\mathcal{X}=\emptyset$. ∎

At last, we are ready to present the main result of our article, which
corresponds to the generalization of theorem 2.5 for the abstract context.
Thus, our abstract version of Kastanas theorem characterizes the Ramsey
property through the existence of winning strategies in the abstract Kastanas
game on the combinatorial structure of selective topological Ramsey spaces.

######  Theorem 5.7.

[Abstract Kastanas Theorem]. Let $(\mathcal{R},\leq,r)$ be a selective
topological Ramsey space. Then, for every $\mathcal{X}\subseteq\mathcal{R}$,
the following statements are equivalent:

  1. (a)

$\mathcal{X}$ is Ramsey.

  2. (b)

For all $[a,A]\neq\emptyset$, the abstract Kastanas game
$\mathcal{G}_{[a,A]}(\mathcal{X})$ is determined.

###### Proof.

Let $(\mathcal{R},\leq,r)$ be a selective topological Ramsey space, and fix
both a set of objects $\mathcal{X}\subseteq\mathcal{R}$ and a basic open set
$[a,A]\neq\emptyset$.

If the game $\mathcal{G}_{[a,A]}(\mathcal{X})$ is determined, then
propositions 5.2 and 5.6 clearly imply that there exists some $H\in[a,A]$ such
that either $[a,H]\subseteq\mathcal{X}$ or $[a,H]\cap\mathcal{X}=\emptyset$,
so that $\mathcal{X}$ is a Ramsey set.

Conversely, suppose that the set $\mathcal{X}$ is Ramsey and that the player
II does not have a winning strategy in the game
$\mathcal{G}_{[a,A]}(\mathcal{X})$; thus, by proposition 5.6, there exists
some $B\in[a,A]$ such that $[a,H]\cap\mathcal{X}\neq\emptyset$ for all
$H\in[a,B]$. Since $\mathcal{X}$ is Ramsey, then for some
$H\in[a,B]\subseteq[a,A]$ we have $[a,H]\subseteq\mathcal{X}$, so that the
player I has a winning strategy in the game
$\mathcal{G}_{[a,A]}(\mathcal{X})$, by virtue of proposition 5.2. Therefore,
we conclude that the game $\mathcal{G}_{[a,A]}(\mathcal{X})$ is determined. ∎

As a final comment we mention that in view of these results it might be
interesting to explore the Ramsey property on topological Ramsey spaces under
determinacy axioms.

## Acknowledgments

The authors thank Natasha Dobrinen and José Mijares for several invaluable
conversations on the topics and results of this article.

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