A directed graph $R^{\circ}$ on a set $X$ is a set of ordered pairs of
distinct points called _arcs_. It is a tournament when every pair of distinct
points is connected by an arc in one direction or the other (and not both). We
can describe a tournament $R\subset X\times X$ as a total, antisymmetric
relation, i.e. $R\cup R^{-1}=X\times X$ and $R\cap R^{-1}$ is the diagonal
$1_{X}=\\{(x,x):x\in X\\}$. The set of arcs is $R^{\circ}=R\setminus
1_{X}=(X\times X)\setminus R^{-1}$. A topological tournament on a compact
Hausdorff space $X$ is a tournament $R$ which is a closed subset of $X\times
X$. We construct uncountably many non-isomorphic examples on the Cantor set
$X$ as well as examples of arbitrarily large cardinality. We also describe
compact Hausdorff spaces which do not admit any topological tournament.

###### Key words and phrases:

tournaments, topological tournaments, lexicographic products, inverse limits
of tournaments,prime topological tournaments, arc cyclic tournaments,
classification of arc cyclic topological tournaments

2010 Mathematical Subject Classification 05C20, 05C25, 05C38, 05C63, 06A05,
54D30

###### Contents

  1. 1  Introduction
  2. 2  Topological Tournaments
  3. 3  Lexicographic Products
  4. 4  Inverse Limits
  5. 5  Connectedness and Compactness
  6. 6  Group Tournaments
  7. 7  Cycle Points
  8. 8  LOTS Constructions
  9. 9  Big Examples
  10. 10 WAC Tournaments and Prime Quotients
  11. 11  Classification of WAC Tournaments
  12. 12  Prime Tournament Constructions
    1. 12.1  Doubles and Reduced Doubles
    2. 12.2  Compact Countably Infinite Tournaments
    3. 12.3  Adjusting Lexicographic Products
    4. 12.4  The Attachment Construction
  13. 13  Prime Tournament Examples
  14. 14 Semi-Prime Tournaments
  15. 15 Appendix
    1. 15.1 Alternative Game Subsets for the 2-adics
    2. 15.2 Sections Over the Cantor Set

##  1\.  Introduction

A directed graph (or just _digraph_) consists of a non-empty finite set $X$ of
elements called _vertices_ and a finite set $R^{\circ}$ of ordered pairs of
distinct vertices called _arcs_. In addition, we assume that $(x,y)\in
R^{\circ}$ implies $(y,x)\not\in R^{\circ}$. That is, for any pair of distinct
vertices there is at most one arc between them. A digraph is called a
_tournament_ if for any pair of distinct vertices there is exactly one arc
between them. That is, either $(x,y)\in R^{\circ}$ or $(y,x)\in R^{\circ}$ but
not both. Digraphs have been the object of considerable study, see e.g. [5].
For the special case of tournaments, see [9] and [12]. When the tournament $R$
is understood, we will write $x\rightharpoonup y$ when $(x,y)\in R^{\circ}$.

In considering tournaments on infinite set $X$, it will be convenient to
attach the diagonal set $1_{X}=\\{(x,x)\in X\\}$. So we will call $R$ a
tournament on $X$ when it is an anti-symmetric, total relation on $X$. That
is, $R\subset X\times X$ with $R\cap R^{-1}=1_{X}$ and $R\cup R^{-1}=X\times
X$, where $R^{-1}=\\{(x,y):(y,x)\in R\\}$. The set of arcs is

$R^{\circ}\ =\ R\setminus 1_{X}\ =\ (X\times X)\setminus R^{-1}.$

A tournament $(X,R)$ is a _trivial tournament_ when $X$ is a singleton set,
and is an _arc tournament_ when $X$ is a two point set. If $A\subset X$, then
$(A,R|A)$ is called the _restriction_ to $A$ where $R|A=R\cap(A\times A)$.

We will call $R$ a topological tournament on a topological space $X$ when it
is a tournament, closed as a subset of $X\times X$. Of course, when $X$ is
given the discrete topology, any tournament on $X$ is a topological
tournament. We will be primarily interested in the case when $X$ is compact.

All our spaces are assumed to be Hausdorff, but as they need not be
metrizable, we will use the convergence theory of nets. These are analogues of
sequences, indexed by directed sets instead of by the natural numbers
$\mathbb{N}$. For the theory of nets, see [11] Chapter 2 on Moore-Smith
Convergence.

For a topological tournament $R$ and a point $x\in X$, the _outset_ is
$R^{\circ}(x)=R(x)\setminus\\{x\\}$, and the _inset_ is $R^{\circ-1}(x)$. We
call a point _right balanced_ if it is in the closure of its outset and _left
balanced_ if it is in the closure of its inset. It is _balanced_ if it is both
left and right balanced. A point is neither left nor right balanced if and
only if it is an isolated point. We call a point $x$ a _cycle point_ when
every neighborhood of $x$ contains a $3-$cycle which includes $x$. Clearly a
cycle point is balanced and we will see that in the compact case a cycle point
is a $G_{\delta}$ point and so has a countable neighborhood base. Of course,
balanced points of any sort only occur when the space $X$ is infinite.

######  Proposition 1.1.

If $x$ is a cycle point for a tournament $R$ on a compact space $X$, then $x$
is a $G_{\delta}$ point and so has a countable neighborhood base in $X$.

A tournament is _arc cyclic_ when every arc is contained in a $3-$cycle. We
call a topological tournament _weakly arc cyclic_ or just _wac_ when every
non-isolated point is a cycle point. As the name suggests, an arc cyclic
topological tournament is wac. A finite tournament is always wac and although
not all finite tournaments are arc cyclic, many are.

For compact topological tournaments $(X_{1},R_{1})$ and $(X_{2},R_{2})$ a
_quotient map_ $h:(X_{2},R_{2})\to(X_{1},R_{1})$ is a surjective continuous
map $h:X_{2}\to X_{1}$ such that $(h\times h)(R_{2})\subset R_{1}$. When $h$
is injective, it is an _isomorphism_ with inverse
$h^{-1}:(X_{1},R_{1})\to(X_{2},R_{2})$. It is an _automorphism_ when, in
addition, $(X_{2},R_{2})=(X_{1},R_{1})$.

If $(X_{1},R_{1})$ and $\\{(Y_{x},S_{x}):x\in X_{1}\\}$ are tournaments, then
the _lexicographic product_

$(X_{2},R_{2})\ =\ (X_{1},R_{1})\ltimes\\{(Y_{x},S_{x}):x\in X_{1}\\}$

is the tournament with $X_{2}=\bigcup\\{\\{x\\}\times Y_{x}:x\in X_{1}\\}$ and
$((x,y),(x^{\prime},y^{\prime}))\in R_{2}$ when $(x,x^{\prime})\in
R_{1}^{\circ}$ or $x=x^{\prime}$ and $(y,y^{\prime})\in S_{x}$. The map
$\pi:X_{2}\to X_{1}$ is the projection to the first coordinate. The product is
called a _compact topological lexicographic product_ when the following
conditions are satisfied.

  * (i)

The space $X_{1}$ and each $Y_{x}$ is compact.

  * (ii)

If $x\in X_{1}$ is non-isolated point of $X_{1}$, then $(Y_{x},S_{x})$ is a
trivial tournament.

  * (iii)

The space $X_{2}$ is given the topology with basis $\mathcal{B}$ where
$U\in\mathcal{B}$ when either $U=\pi^{-1}(V)$ for $V$ some open subset of
$X_{1}$, or $U=\\{x\\}\times V$ for $x$ isolated in $X_{1}$ and $V$ some open
subset of $Y_{x}$.

In that case, $(X_{2},R_{2})$ is a compact topological tournament and
$\pi:(X_{2},R_{2})\to(X_{1},R_{1})$ is an open quotient map.

A sequence $\\{(X_{i},R_{i},f_{i}):i\in\mathbb{N}\\}$ is a _compact inverse
sequence_ when each $(X_{i},R_{i})$ is a compact topological tournament and
each $f_{i}:(X_{i+1},R_{i+1})\to(X_{i},R_{i})$ is a quotient map. The _inverse
limit_ is the compact tournament $(X,R)\ =\
\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ with

$\begin{split}X\ =\ \\{x\in\prod_{i}X_{i}:x_{i}=f_{i}(x_{i+1})\ \ \text{ for
all }\ i\in\mathbb{N}\\},\hskip 28.45274pt\\\ R\ =\ \\{(x,x^{\prime})\in
X\times X:(x_{i},x^{\prime}_{i})\in R_{i}\ \ \text{ for all }\
i\in\mathbb{N}\\}.\end{split}$

The projection $\pi_{i}:(X,R)\to(X_{i},R_{i})$ given by $\pi_{i}(x)=x_{i}$ is
a quotient map for each $i\in\mathbb{N}$.

######  Proposition 1.2.

Let $h:(X_{2},R_{2})\to(X_{1},R_{1})$ be a quotient map.

(a) If $y\in X_{1}$ is a cycle point for $R_{1}$, then $h^{-1}(y)$ is a
singleton $\\{x\\}\subset X_{2}$ and $x$ is a cycle point for $R_{2}$. In
particular, if every point of $X_{1}$ is a cycle point, then $h$ is an
isomorphism.

(b) If $(X_{2},R_{2})$ is an arc cyclic (or wac) tournament, then
$(X_{1},R_{1})$ is arc cyclic (resp. wac) and there is an isomorphism

$q:(X_{1},R_{1})\ltimes\\{(h^{-1}(y),R_{2}|h^{-1}(y)):y\in X_{1}\\}\ \to\
(X_{2},R_{2})$

such that $\pi=h\circ q$. In particular, $h$ is an open map.

(c) The inverse limit $\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ is an arc
cyclic (or wac) tournament, if and only if each $(X_{i},R_{i})$ is arc cyclic
(resp. wac). In particular, if each $X_{i}$ is finite, then
$\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ is wac.

If $X$ is a group with identity $e$, a _game subset_ $A$ is a subset of $X$
such that $A\cap A^{-1}=\\{e\\}$ and $A\cup A^{-1}=X$, where
$A^{-1}=\\{x^{-1}:x\in A\\}$. That is, for every $x\not=e$ in $X$, exactly one
member of the pair $\\{x,x^{-1}\\}$ lies in $A$. Thus, $X$ admits a game
subset if and only if it contains no elements of order two. When $X$ is
finite, this means $X$ has odd order. Associated with a game subset $A$ is the
tournament $\widehat{A}=\\{(x,y):x^{-1}y\in A\\}$. If $X$ is a topological
group and $A$ is a closed game subset, then $\widehat{A}$ is a topological
tournament on $X$.

######  Theorem 1.3.

Let $X$ be an infinite, compact topological group with no elements of order
two.

(a) There exists a closed game subset $A$ for $X$ if and only if the space $X$
is totally disconnected and metrizable, i.e. it is a Cantor set.

(b) If $A$ is a closed game subset for $X$, then the tournament $\widehat{A}$
is arc cyclic and so every point of $X$ is a cycle point.

Notice that a topological group contains an isolated point if and only if it
is discrete. Hence, any infinite, compact topological group has no isolated
points.

On the one hand, there are many different topological group structures on the
Cantor set. For example, the $p-$adic integers for any prime $p$ including $2$
admits closed game subsets.

On the other hand, this illustrates that the existence of a topological
tournament is a demanding condition. For example, if $X$ is an uncountable
product of finite groups of odd order, then because the product topology is
not metrizable, the product group does not admit a closed game subset.

A transitive tournament is just a linear order. A linear order on $X$ is a
topological tournament when $X$ is given the order topology. Conversely, if
$L$ is a transitive topological tournament on a compact space $X$, then, as we
will see, the topology on $X$ is the order topology associated with $L$, i.e.
$X$ is a compact _LOTS_ (= linearly ordered topological space).

A topological tournament $R$ on $X$ is called _nowhere locally transitive_ if
no restriction of $R$ to a nonempty open set is transitive. This is equivalent
to the condition that every nonempty open subset contains a $3-$cycle.

Clearly, if $X$ contains a dense set of cycle points, then $R$ is nowhere
locally transitive. Conversely, we have

######  Theorem 1.4.

Let $(X,R)$ be a compact topological tournament.

(a) If $R$ is nowhere locally transitive, then $X$ is totally disconnected and
contains a dense set of cycle points.

(b) If $R$ is balanced and the space $X$ is totally disconnected, then $R$ is
nowhere locally transitive.

Despite these limitations, it is possible to construct big examples by using
lexicographic products of LOTS and inverse limits.

######  Theorem 1.5.

Let $\aleph$ be an arbitrary uncountable cardinal. There exists a compact,
totally disconnected LOTS $X$ such that every nonempty open subset has
cardinality at least that of $\aleph$ and $X$ admits a balanced topological
tournament $R$. The set of cycle points for $R$ is a dense $G_{\delta}$ subset
of $X$, while the set of non $G_{\delta}$ points is also a dense subset of
$X$. Furthermore, no open subset is separable.

A nontrivial, compact topological tournament $(Y,P)$ is a _prime tournament_
when the only quotient maps $h:(Y,P)\to(Y_{1},P_{1})$ with $(Y_{1},P_{1})$
nontrivial are isomorphisms. For example, an arc tournament is prime. A
quotient map $(X,R)\to(Y,P)$ with $(Y,P)$ prime is called a _prime quotient
map_ and $(Y,P)$ is called a _prime quotient_ of $(X,R)$.

######  Theorem 1.6.

If $(X,R)$ is a wac tournament, then it has a prime quotient which is unique
up to isomorphism. If the prime quotient $(Y,P)$ is not an arc, then the prime
quotient map is unique up to isomorphism. That is, If $h:(X,R)\to(Y,P)$ and
$h_{1}:(X,R)\to(Y_{1},P_{1})$ are prime quotient maps and $(Y,P)$ is not an
arc, then there exists an isomorphism $q:(Y,P)\to(Y_{1},P_{1})$ such that
$q\circ h=h_{1}$.

If $(Y,L)$ is an order on a nontrivial finite set, then a _maximum order
quotient map_ $h:(X,R)\to(Y,L)$ is a quotient map such that for all $y\in Y$,
the restriction $(h^{-1}(y),R|h^{-1}(y))$ does not have an arc quotient.

######  Addendum 1.7.

If $(X,R)$ is a wac tournament with an arc quotient, then it has a maximum
order quotient map unique up to isomorphism.

Thus, every wac tournament $(X,R)$ has a base quotient map $h:(X,R)\to(Y,P)$,
unique up to isomorphism, as follows

  * •

If $(X,R)$ is nontrivial and does not have an arc quotient, then $h$ is a
prime quotient map.

  * •

If $(X,R)$ is nontrivial and has an arc quotient, then $h$ is a maximum order
quotient map.

  * •

If $(X,R)$ is trivial, then $(Y,P)$ is trivial.

######  Definition 1.8.

For a wac tournament $(X,R)$ the _classifier system_ is an inverse system
$\\{(X_{i},R_{i},f_{i})\\}$ of topological tournaments, together with quotient
maps $h_{i}:(X,R)\to(X_{i},R_{i})$ which satisfy the following properties.

  * (i)

$h_{i}=f_{i}\circ h_{i+1}$ for all $i\in\mathbb{N}$.

  * (ii)

$h_{1}:(X,R)\to(X_{1},R_{1})$ is a base quotient map.

  * (iii)

For each $x_{i}\in X_{i}$, the restriction
$(h_{i}^{-1}(x_{i}),R|h^{-1}_{i}(x_{i}))$ is a wac tournament and the map

$h_{i+1}:(h_{i}^{-1}(x_{i}),R|h^{-1}_{i}(x_{i}))\to(f_{i}^{-1}(x_{i}),R_{i+1}|f^{-1}_{i}(x_{i}))$

is a base quotient map.

######  Theorem 1.9.

If $(X,R)$ is a wac tournament, then it has a classifier system which is
unique up to isomorphism and the product map
$\prod_{i}h_{i}:(X,R)\to\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ is an
isomorphism. Furthermore, the classifier can be constructed so that
$(X_{i+1},R_{i+1})=(X_{i},R_{i})\ltimes\\{(Y_{ix_{i}},S_{ix_{i}}:x_{i}\in
X_{i}\\}$ with each $(Y_{ix_{i}},S_{ix_{i}})$ either prime, a nontrivial
finite order or trivial.

Using the uniqueness of the classifier system, we are able to construct an
uncountable number of arc cyclic tournaments on the Cantor set with each
$(X_{i},R_{i})$ a finite arc cyclic tournament, such that no two are
isomorphic.

On the compact group $\mathbb{Z}[2]$ of $2-$adic integers, there exists a
closed game subset $A$ such that $\widehat{A}$ is a prime tournament. Using
it, we are able to construct an uncountable number of prime, arc cyclic
tournaments on the Cantor set, as well as arc cyclic, prime tournaments with
countable sets of isolated points and with Cantor subsets.

######  Background 1.10.

We briefly review some standard results about compact spaces which we will be
using. All of our spaces, compact or not, are assumed to be Hausdorff.

  1. (1)

If $\\{A_{n}\\}$ is a decreasing sequence of compact sets in a space $X$ with
intersection $A$ and $U$ is an open set with $A\subset U$, then for
sufficiently large $n$, $A_{n}\subset U$, because $\\{U\\}\cup\\{X\setminus
A_{n}\\}$ is an open cover of $A_{1}$ and so has a finite subcover. In
particular, if $A$ is clopen (= open as well as closed), then, using $U=A$,
$A_{n}=A$ for sufficiently large $n$.

  2. (2)

A component $A$ in a compact space is the intersection of the clopen subsets
which meet and therefore contain it. Hence, if $X$ is totally disconnected,
i.e. the only connected subsets are singletons, then the clopen subsets form a
base for the topology.

  3. (3)

If $X$ is compact and metrizable, and so has a countable base, then there are
only countably many clopen subsets because each clopen set is a finite union
of members of the base.

  4. (4)

A Cantor set is a compact, metrizable, totally disconnected space with no
isolated points. Any Cantor set is homeomorphic to the product
$\\{0,1\\}^{\mathbb{N}}$ and so to any other Cantor set. In particular, it is
homeomorphic to the classical Cantor Set $C$ contained in the unit interval of
$\mathbb{R}$.

  5. (5)

For any compact, metrizable, totally disconnected space $X$, the product
$X\times C$ is a Cantor set, homeomorphic to $C$ itself and so $X$ can be
embedded in $C$.

  6. (6)

A countable compact space $X$ is totally disconnected since any non-trivial
connected compact space maps onto the unit interval and so is uncountable. The
diagonal $1_{X}=\\{(x,x):x\in X\\}$ is clearly a $G_{\delta}$ subset. In a
compact space, a closed subset is a $G_{\delta}$ set if and only if it has a
countable base of neighborhoods. A compact space is metrizable if and only if
the diagonal is $G_{\delta}$ because then the uniformity of neighborhoods of
the diagonal has a countable base, see, e.g. [11] Theorem 6.13. Hence, the
countable compact space $X$ can be embedded in a Cantor set.

  7. (7)

The countably infinite product of non-trivial, compact, metrizable, totally
disconnected spaces is a compact, metrizable, totally disconnected space with
no isolated points, i.e. a Cantor set.

  8. (8)

On $\\{0,1\\}^{\mathbb{N}}$ the metric $u$ defined to by
$u(x,x^{\prime})=\max_{i}2^{-i}|x_{i}-x^{\prime}_{i}|$ is compatible with the
product topology and is an _ultra-metric_. That is, it satisfies the
strengthening of the triangle inequality: $u(x,x^{\prime\prime})\ \leq\
\max(u(x,x^{\prime}),u(x^{\prime},x^{\prime\prime}))$. For any ultra-metric
$u$ and $\epsilon>0$ the set
$V_{\epsilon}=\\{(x,x^{\prime}):u(x,x^{\prime})<\epsilon\\}$ is an equivalence
relation with finitely many clopen equivalence classes, namely, the $\epsilon$
balls $V_{\epsilon}(x)$. Hence,
$V_{\epsilon}=\bigcup_{x}\\{V_{\epsilon}(x)\times V_{\epsilon}(x)\\}$ is
clopen. Any compact, metrizable, totally disconnected space admits such an
ultrametric.

  9. (9)

If $X$ is a general compact, totally disconnected space and $V$ is a
neighborhood of the diagonal $1_{X}$, then there exists a clopen equivalence
relation $E$ on $X$ such that $E\subset V$. We can choose a finite cover
$\\{U_{1},\dots,U_{n}\\}$ of $X$ by clopen sets such that $U_{i}\times
U_{i}\subset V$. With $U_{0}=\emptyset$ we let
$U^{\prime}_{i}=U_{i}\setminus\bigcup_{j<i}U_{j}$ to get a clopen partition
and then let $E=\bigcup_{i}U_{i}^{\prime}\times U_{i}^{\prime}$.

  10. (10)

Let $X$ be a compact metric space with metric $d$. If $G$ is a compact
topological group with a continuous action $(g,x)\mapsto gx$ on $X$, then
$d_{G}(x,x^{\prime})=\max\\{d(gx,gx^{\prime}):g\in G\\}$ is a $G$ invariant
metric compatible with the topology on $X$. Notice that if $\epsilon>0$, then
$\min\\{d(g^{-1}x,g^{-1}x^{\prime}):g\in
G,d(x,x^{\prime})\geq\epsilon\\}=\delta>0$ and so $d(x,x^{\prime})<\delta$
implies $d_{G}(x,x^{\prime})<\epsilon$. If $d$ is an ultra-metric, then
$d_{G}$ is a $G$ invariant ultra-metric.

##  2\.  Topological Tournaments

Following [1] we will use the language of relations. For sets $X,Y$ (not
necessarily finite) a _relation_ $F$ from $X$ to $Y$ is just a subset of the
product set set $X\times Y$ of ordered pairs. We let $\pi_{1}$ and $\pi_{2}$
denote the coordinate projections.

We define for a relation $F$ from $X$ to $Y$ and $x\in X,A\subset X,B\subset
Y$:

(2.1) $\begin{split}F(x)\ =\ \\{y\in Y:(x,y)\in F\\},\hskip 56.9055pt\\\ F(A)\
=\ \bigcup\\{F(x):x\in A\\}\ =\ \pi_{2}((A\times Y)\cap F)\\\ F^{-1}\ =\
\\{(y,x)\in Y\times X:(x,y)\in F\\},\hskip 42.67912pt\\\ F^{*}(B)\ =\
X\setminus F^{-1}(Y\setminus B).\hskip 56.9055pt\end{split}$

The _reverse relation_ $F^{-1}$ is a relation from $Y$ to $X$.

Notice that

(2.2) $\begin{split}F^{-1}(B)\ =\ \\{x\in X:F(x)\cap B\not=\emptyset\\},\hskip
42.67912pt\\\ F^{*}(B)\ =\ \\{x\in X:F(x)\subset B\\}.\hskip
42.67912pt\end{split}$

We think of a relation as a generalization of a mapping. The relation $F$ is a
function from $X$ to $Y$ when for every $x\in X$ the set $F(x)$ is a
singleton, i.e. $|F(x)|=1$ where we use $|A|$ to denote the cardinality of a
finite set $A$. For example, the identity map $1_{X}$ is the relation
$\\{(x,x):x\in X\\}$ on $X$. If $F$ is a mapping, then $F^{-1}(B)=F^{*}(B)$ is
the usual pre-image of $B$.

The relation $F$ is called _surjective_ when for all $x\in X,y\in Y$,
$F(x)\not=\emptyset$ and $F^{-1}(y)\not=\emptyset$, or, equivalently, for
every $x\in X$ there exists $y\in Y$ such that $y\in F(x)$ and for every $y\in
Y$ there exists $x\in X$ such that $y\in F(x)$. When $F$ is a function, it is
a surjective relation exactly when it is a surjective function.

If $F$ is a relation from $X$ to $Y$ and $G$ is a relation from $Y$ to $Z$,
the _composition_ is the relation $G\circ F$ from $X$ to $Z$ defined by

(2.3) $\begin{split}G\circ F\ =\ \pi_{13}((F\times Z)\cap(X\times G))\ =\
\hskip 56.9055pt\\\ \\{(x,z)\in X\times Z:\text{there exists }\ y\in Y\
\text{such that}\ (x,y)\in F,(y,z)\in G\\},\end{split}$

where $\pi_{13}$ is the coordinate projection from $X\times Y\times Z$ to
$X\times Z$.

Thus, for any subset $A$ of $X$, $(G\circ F)(A)=G(F(A))$. Clearly, $(G\circ
F)^{-1}=F^{-1}\circ G^{-1}$. As with functions, composition of relations is
associative.

When $X=Y$ $F$ is called a _relation on_ $X$.

If $R_{1}$ is a relation on $X_{1}$ and $R_{2}$ is a relation on $X_{2}$, then
a function $h:X_{2}\to X_{1}$ maps $R_{2}$ to $R_{1}$ when $(x,x^{\prime})\in
R_{2}$ implies $(h(x),h(x^{\prime}))\in R_{1}$. That is,

(2.4) $(h\times h)(R_{2})\subset R_{1},\quad\text{or, equivalently,}\quad
R_{2}\subset(h\times h)^{-1}(R_{1}),$

where $h\times h:X_{2}\times X_{2}\to X_{1}\times X_{1}$ is the product map
induced by $h$. It clearly follows that $h$ maps $R_{2}^{-1}$ to $R_{1}^{-1}$.

We define the _product relation_ $R_{1}\times R_{2}$ on $X_{1}\times X_{2}$ by

(2.5) $R_{1}\times R_{2}\ =\ \\{(x_{1},x_{2}),(y_{1},y_{2})):(x_{1},y_{1})\in
R_{1}\ \ \text{and}\ \ (x_{2},y_{2})\in R_{2}\\}.$

That is, we identify $(X_{1}\times X_{1})\times(X_{2}\times X_{2})$ with
$(X_{1}\times X_{2})\times(X_{1}\times X_{2})$.

If $Y\subset X$ and $R$ is a relation on $X$, then the _restriction_ of $R$ to
$Y$ is $R|Y=R\cap(Y\times Y)$.

A relation $R$ on $X$ is _reflexive_ when $1_{X}\subset R$, _symmetric_ when
$R=R^{-1}$ and _transitive_ when $R\circ R\subset R$.

For $n>1$ an _$n-$ cycle_ for the relation $R$ on $X$ is a sequence
$\\{x_{1},\dots,x_{n}\\}$ of distinct points of $X$ such that
$(x_{i},x_{i+1})\in R$ for $i=1,\dots,n$ (with addition mod $n$).

A _closed relation_ (or an _open relation_) is a relation $F$ between
Hausdorff topological spaces $X$ and $Y$ with $F$ a closed subset (resp. an
open subset) of $X\times Y$. Clearly, for a closed relation $F$, the reverse
relation $F^{-1}$ is closed and for each $x\in X,y\in Y$ the sets $F(x)$ and
$F^{-1}(y)$ are closed. The product and restriction of closed relations are
closed relations. Similarly, for an open relation $F$ the sets $F^{-1}$,
$F(x),F^{-1}(y)$ are open and the product and restriction of open relations
are open relations.

As all of our spaces are assumed to be Hausdorff, any continuous map between
spaces is a closed relation. If the spaces are compact, then the converse
holds. That is, if a mapping between compact spaces is a closed relation, then
it is a continuous map. For compact spaces the composition of closed relations
is closed and the image of a closed subset by a closed relation is closed.
Furthermore, in the compact case, if $B$ is open, then $F^{*}(B)$ is open.

Thus, a digraph is a relation $R^{\circ}$ on a finite set $X$ such that
$R^{\circ}\cap(R^{\circ})^{-1}=\emptyset$. It is a tournament when, in
addition, $R^{\circ}\cup(R^{\circ})^{-1}=X\times X\setminus 1_{X}$. A
tournament is said to be _regular_ when for every $x\in X$ the _inset_
$(R^{\circ})^{-1}(x)$ and the _outset_ $R^{\circ}(x)$ have the same
cardinality, i.e. $|(R^{\circ})^{-1}(x)|=|R^{\circ}(x)|$ for all $x\in X$. A
regular tournament exists on a finite set $X$ only when the cardinality $|X|$
is odd. Conversely, as we will see from the group examples below, a finite set
of odd cardinality admits regular tournaments.

Notice that an $n-$cycle for a digraph has length $n$ greater than $2$.

It will be convenient for our purposes to attach the identity $1_{X}$ to
$R^{\circ}$.

######  Definition 2.1.

For an arbitrary nonempty set $X$, a _tournament_ on $X$ is a relation $R$ on
$X$ which is anti-symmetric and total, i.e.

(2.6) $R\cap R^{-1}\ =\ 1_{X},\quad\text{and}\quad R\cup R^{-1}\ =\ X\times
X.$

We denote by $R^{\circ}$ the _arc-set_ $R\setminus 1_{X}$.

The tournament $1_{X}$ on a singleton set $X$ is called a _trivial
tournament_. A tournament on a two point set is called an _arc tournament_ or
simply an arc.

A _topological tournament_ is a tournament on a topological space which is a
closed relation.

We will call a pair $(X,R)$ a tournament when $R$ is a tournament on the set
$X$. The pair is a topological tournament when $X$ is a topological space and
$R$ is closed, and it is a compact topological tournament when, in addition,
the space $X$ is compact. A finite tournament is a tournament on a finite set,
always a compact topological tournament with the discrete topology on $X$.

If $R$ is a topological tournament on $X$, then

  * •

The reverse relation $R^{-1}$ is a topological tournament on $X$ with
$(R^{-1})^{\circ}=(R^{\circ})^{-1}$ which we will therefore write as
$R^{\circ-1}$.

  * •

The arc-set relation $R^{\circ}=R\setminus 1_{X}=(X\times X)\setminus R^{-1}$
is an open relation.

  * •

For each $x\in X$, the sets $R(x)$ and $R^{-1}(x)$ are closed subsets and the
_outset_ $R^{\circ}(x)=R(x)\setminus\\{x\\}$ and the _inset_
$(R^{-1})^{\circ}(x)=R^{-1}(x)\setminus\\{x\\}$ are open subsets of $X$.

When the tournament $R$ is understood we will write $x\rightharpoonup y$ or
$y\leftharpoonup x$ when $(x,y)\in R^{\circ}$ and we write $x\
\underline{\rightharpoonup}\ y$ when $(x,y)\in R$. For subsets $A,B$ of $X$,
we will write $A\rightharpoonup B$, when $x\rightharpoonup y$ for all $x\in
A,y\in B$, or, equivalently, when $A\times B\subset R^{\circ}$.

We will call a tournament $(X,R)$ _arc cyclic_ (or _point cyclic_ ) when every
arc $(x_{1},x_{2})\in R^{\circ}$ (respectively, every point $x_{1}\in X$) is
contained in a $3$-cycle $\\{x_{1},x_{2},x_{3}\\}$ for $R$.

Every finite regular tournament is arc cyclic, see, e.g. [2] Proposition 2.2
or [6] Proposition 5.1. However, there exist finite tournaments which are arc
cyclic but not regular. A non-trivial arc cyclic tournament is clearly point
cyclic. On the other hand, a trivial tournament is not point cyclic but is
vacuously arc cyclic.

For a tournament $(X,R)$ we will call $A\subset X$ an _arc cyclic subset_ when
every arc $(x_{1},x_{2})\in R^{\circ}$ with $x_{1},x_{2}\in A$ is contained in
a $3$-cycle $\\{x_{1},x_{2},x_{3}\\}$ for $R$. Note that $x_{3}$ need not be
in $A$. Thus, if the restriction $R|A$ is an arc cyclic tournament, then $A$
is an arc cyclic subset, but the converse need not be true. Clearly, $(X,R)$
is an arc cyclic tournament when $X$ is an arc cyclic subset.

We will call a topological tournament $(X,R)$ _locally arc cyclic_ when every
point $x\in X$ has a neighborhood $U$ which is an arc cyclic subset. Since a
trivial tournament is arc cyclic, the singleton set containing an isolated
point is an arc cyclic neighborhood of the point. In particular, every finite
tournament is locally arc cyclic. Of course, an arc cyclic topological
tournament is locally arc cyclic.

For a tournament $(X,R)$ we call $x\in X$ a _terminal point_ (or a _initial
point_) for $R$ when $R^{\circ}(x)=\emptyset$ (resp.
$R^{\circ-1}(x)=\emptyset$). A tournament has at most one terminal point since
$R^{\circ}(x)=\emptyset$ and $x\not=y$ implies $x\in R^{\circ}(y)$ because $R$
is total. Similarly, there is at most one initial point. A tournament $R$ is a
surjective relation if and only if it has neither a terminal point nor an
initial point.

For a topological tournament $R$ on $X$ we define for a point $x\in X$

(2.7) $\displaystyle\begin{split}x\ \ \text{is right
balanced}\quad&\Longleftrightarrow\quad\overline{R^{\circ}(x)}=R(x)\\\ x\ \
\text{is left
balanced}\quad&\Longleftrightarrow\quad\overline{R^{\circ-1}(x)}=R^{-1}(x)\\\
x\ \ \text{is balanced}\quad&\Longleftrightarrow\quad x\ \ \text{is both left
and right balanced}.\end{split}$

Note that if $x$ is not right balanced, if and only if
$\overline{R^{\circ}(x)}=R^{\circ}(x)$ and so $R^{\circ}(x)$ is closed as well
as open. Hence, $x$ is neither left nor right balanced, if and only if
$\\{x\\}$ is clopen and so $x$ is an isolated point.

We will call a topological tournament $(X,R)$ _balanced_ when every point of
$X$ is balanced.

We will call a topological tournament $(X,R)$ _regular_ when for every $x\in
X$ there is a homeomorphism $h_{x}$ from $X$ onto $X$ such that $h_{x}(x)=x$
and $h_{x}(R(x))=R^{-1}(x)$. Note that when $X$ is finite any bijection on $X$
is a homeomorphism and so this concept agrees with the usual notion of
regularity when $X$ is finite.

If $(X_{1},R_{1})$ and $(X_{2},R_{2})$ are tournaments, and $h$ is a function
from $X_{2}$ to $X_{1}$, then we call $h:(X_{2},R_{2})\to(X_{1},R_{1})$ a
_tournament map_ (or, equivalently, $h$ is a tournament map from $R_{2}$ to
$R_{1}$) when $h$ maps the relation $R_{2}$ to the relation $R_{1}$. Since $h$
then maps $R_{2}^{-1}$ to $R_{1}^{-1}$ it follows that
$(h(x),h(x^{\prime}))\in R_{1}^{\circ}$ implies $(x,x^{\prime})\in
R_{2}^{\circ}$ or, equivalently

(2.8) $R_{2}^{\circ}\supset(h\times h)^{-1}(R_{1}^{\circ}).$

Thus, the preimage of $R_{1}^{\circ}$ is contained in $R_{2}^{\circ}$ and
$R_{2}$ is contained in the union of the preimage of $R_{1}^{\circ}$ and that
of $1_{X_{1}}$.

If $h$ is a bijection, then the inverse map $h^{-1}:X_{1}\to X_{2}$ maps
$R_{1}$ to $R_{2}$ and we call $h$ a _tournament isomorphism_. When
$(X_{1},R_{1})$ and $(X_{2},R_{2})$ are topological tournaments and $h$ is a
homeomorphism we call it a _topological tournament isomorphism_. It is a
_topological tournament automorphism_ when the domain and the range are the
same.

If $Y\subset X$ and $R$ is a tournament on $X$, then the restriction $R|Y$ is
a tournament on $Y$ and the inclusion map from $Y$ to $X$ is a tournament map
from $R|Y$ to $R$. Conversely, if $h$ is a tournament map with $h:X_{2}\to
X_{1}$ is injective, then since $1_{X_{2}}=(h\times h)^{-1}(1_{X_{1}})$ it
follows that $h$ is an isomorphism from $R_{2}$ onto the restriction of
$R_{1}$ to the image of $h$.

We will call a topological tournament _rigid_ when the identity is the only
automorphism.

######  Proposition 2.2.

Let $h:(X_{2},R_{2})\to(X_{1},R_{1})$ be a tournament map and let
$x_{1},x_{2},x_{3}$ be distinct points of $X_{2}$ with
$y_{1}=h(x_{1}),y_{2}=h(x_{2}),y_{3}=h(x_{3})$.

If $\\{y_{1},y_{2},y_{3}\\}$ is a $3-$cycle in $X_{1}$, then
$\\{x_{1},x_{2},x_{3}\\}$ is a $3-$cycle in $X_{2}$.

Conversely, if $\\{x_{1},x_{2},x_{3}\\}$ is a $3-$cycle in $X_{2}$, then
either $\\{y_{1},y_{2},y_{3}\\}$ is a $3-$cycle in $X_{1}$ or else
$y_{1}=y_{2}=y_{3}$.

###### Proof.

That a cycle lifts to a cycle follows from (2.8). If $\\{x_{1},x_{2},x_{3}\\}$
is a $3-$cycle in $X_{2}$, then $y_{1}\ \underline{\rightharpoonup}\ y_{2}\
\underline{\rightharpoonup}\ y_{3}\ \underline{\rightharpoonup}\ y_{1}$. If
two of the points are equal, then all three are. For example, if $y_{1}=y_{2}$
then $y_{1}\ \underline{\rightharpoonup}\ y_{3}$ and $y_{3}\
\underline{\rightharpoonup}\ y_{1}$ and so $y_{1}=y_{3}$ by anti-symmetry.

∎

######  Corollary 2.3.

Let $h:(X_{2},R_{2})\to(X_{1},R_{1})$ be a surjective tournament map and let
$A$ be a subset of $X_{1}$.

The subset $h^{-1}(A)$ is an arc cyclic subset of $X_{2}$ if and only if

  * (i)

$A$ is an arc cyclic subset, and

  * (ii)

the restriction $R_{2}|h^{-1}(y)$ is arc cyclic for every $y\in A$.

In particular, the tournament $(X_{2},R_{2})$ is arc cyclic, if and only if
$(X_{1},R_{1})$ is arc cyclic and, in addition, the restriction
$R_{2}|h^{-1}(y)$ is arc cyclic for every $y\in X_{1}$.

The tournament $(X_{2},R_{2})$ is point cyclic, if either $(X_{1},R_{1})$ is
point cyclic, or the restriction $R_{2}|h^{-1}(y)$ is point cyclic for every
$y\in X_{1}$.

###### Proof.

Assume $h^{-1}(A)$ is an arc cyclic subset.

If $y_{1}\rightharpoonup y_{2}$ with $y_{1},y_{2}\in A$, then because $h$ is
surjective there exist $x_{1},x_{2}\in h^{-1}(A)$ such that
$y_{1}=h(x_{1}),y_{2}=h(x_{2})$. Because $h$ is a tournament map
$x_{1}\rightharpoonup x_{2}$. Because $h^{-1}(A)$ is an arc cyclic subset,
there exists $x_{3}$ such that $\\{x_{1},x_{2},x_{3}\\}$ is a $3-$cycle in
$X_{2}$. Since $y_{1}\not=y_{2}$, it follows from Proposition 2.2 that with
$y_{3}=h(x_{3})$ $\\{y_{1},y_{2},y_{3}\\}$ is a $3-$cycle in $X_{1}$.

If $x_{1}\rightharpoonup x_{2}$ and $h(x_{1})=y=h(x_{2})$ with $y\in A$, then
any $3-$cycle $\\{x_{1},x_{2},x_{3}\\}$ in $X_{2}$ is contained in $h^{-1}(y)$
by Proposition 2.2. Since $h^{-1}(A)$ is an arc cyclic subset, it follows that
$R_{2}|h^{-1}(y)$ is arc cyclic.

For the converse, suppose $x_{1}\rightharpoonup x_{2}$ with $x_{1},x_{2}\in
h^{-1}(A)$. If $h(x_{1})=y=h(x_{2})$, then there exists a $3-$cycle
$\\{x_{1},x_{2},x_{3}\\}$ in $h^{-1}(y)$ by assumption. If $h(x_{1})=y_{1}$
and $h(x_{2})=y_{2}$ are distinct, then $y_{1}\rightharpoonup y_{2}$ with
$y_{1},y_{2}\in A$. Because $A$ is an arc cyclic subset, there exists
$\\{y_{1},y_{2},y_{3}\\}$ a $3-$cycle in $X_{2}$ and so there exists a
$3-$cycle lift $\\{x_{1},x_{2},x_{3}\\}$ by Proposition 2.2again.

The point cyclicity result is obvious from Proposition 2.2. Notice that if
$(X_{1},R_{1})$ is trivial, then it is not point cyclic even when
$(X_{2},R_{2})$ is.

∎

The condition that $h$ be a continuous surjective tournament map between
topological tournaments is rather restrictive.

######  Theorem 2.4.

With $(X_{1},R_{1})$ and $(X_{2},R_{2})$ topological tournaments, assume that
$h$ is a continuous tournament map from $(X_{2},R_{2})$ to $(X_{1},R_{1})$.
Let $y\in X_{1}$ and define
$h^{*}(y)=X_{2}\setminus\overline{h^{-1}(R_{1}^{\circ}(y)\cup
R_{1}^{\circ-1}(y))}$.

  * (i)

$h^{*}(y)$ is an open subset of $X_{2}$ with $h^{*}(y)\subset h^{-1}(y)$.

  * (ii)

If $y$ is right balanced with respect to $R_{1}$, then there exists at most
one point $M\in\overline{h^{-1}(R_{1}^{\circ}(y))}\cap h^{-1}(y)$. If the
point $M$ exists, then it is a terminal point for the restriction
$R_{2}|h^{-1}(y)$. If, in addition, $y$ is not left balanced, then the open
set $h^{*}(y)$ is $h^{-1}(y)\setminus\\{M\\}$ or $h^{-1}(y)$ if $M$ does not
exist.

  * (iii)

If $y$ is left balanced with respect to $R_{1}$, then there exists at most one
point $m\in\overline{h^{-1}(R_{1}^{\circ-1}(y))}\cap h^{-1}(y)$. If the point
$m$ exists, then it is a initial point for the restriction $R_{2}|h^{-1}(y)$.
If, in addition, $y$ is not right balanced, then the open set $h^{*}(y)$ is
$h^{-1}(y)\setminus\\{m\\}$ or $h^{-1}(y)$ if $m$ does not exist.

  * (iv)

If $y$ is balanced, then the open set $h^{*}(y)$ is $h^{-1}(y)$ with $m$ and
$M$ removed when either exists.

  * (v)

If $y$ is isolated, then $h^{*}(y)=h^{-1}(y)$ is a clopen subset of $X_{2}$.

###### Proof.

(i) is clear since $\\{y\\}=X_{1}\setminus(R_{1}^{\circ}(y)\cup
R_{1}^{\circ-1}(y))$.

(ii) If $M\in\overline{h^{-1}(R_{1}^{\circ}(y))}\cap h^{-1}(y)$ and $x\in
h^{-1}(y)$, then for any $z\in h^{-1}(R_{1}^{\circ}(y))$, $x\rightharpoonup
z$. Since $R_{2}$ is closed, $x\ \underline{\rightharpoonup}\ M$. Hence, $M$
is the terminal point of $h^{-1}(y)$. If $y$ is not left balanced, then
$R_{1}^{\circ-1}(y)$ is closed and so
$h^{*}(y)=h^{-1}(y)\setminus[\overline{h^{-1}(R_{1}^{\circ}(y))}\cap
h^{-1}(y)]$.

(iii) Since $h$ maps $R_{2}^{-1}$ to $R_{1}^{-1}$, this follows from (ii).

(iv) and (v) are obvious.

∎

######  Addendum 2.5.

Let $h$ be a surjective continuous tournament map from the topological
tournament $(X_{2},R_{2})$ to $(X_{1},R_{1})$, with $X_{2}$ compact.

If $y\in X_{1}$ is right balanced (or left balanced) with respect to $R_{1}$,
then $\overline{h^{-1}(R_{1}^{\circ}(y))}\cap h^{-1}(y)$ (resp.
$\overline{h^{-1}(R_{1}^{\circ-1}(y))}\cap h^{-1}(y)$) is nonempty and so is a
singleton.

###### Proof.

By continuity and compactness, the surjective map $h$ sends
$\overline{h^{-1}(R_{1}^{\circ}(y))}$ onto a closed set which contains
$R_{1}^{\circ}(y)$. If $y$ is right balanced, then
$\overline{R_{1}^{\circ}(y)}=R_{1}(y)$ which contains $y$. Thus,
$\overline{h^{-1}(R_{1}^{\circ}(y))}$ meets $h^{-1}(y)$ and from Theorem 2.4
(ii) we see that the intersection is a singleton.

∎

######  Theorem 2.6.

If $h:(X_{2},R_{2})\to(X_{1},R_{1})$ is a continuous tournament map with
$X_{1},X_{2}$ compact metric spaces, then for every $\epsilon>0$, the set
$\\{y\in X_{1}:diam\ h^{-1}(y)\geq\epsilon\\}$ is finite.

###### Proof.

Suppose there is is a sequence of triples $\\{(y_{n},x_{n},z_{n})\in
X_{1}\times X_{2}\times X_{2}\\}$ with $\\{y_{n}\\}$ distinct points,
$h(z_{n})=y_{n}=h(x_{n})$ and $d(z_{n},x_{n})\geq\epsilon$. By going to a
subsequence, we may assume that the sequence converges to $(y,x,z)$ so that
$h(z)=y=h(x)$ and $d(z,x)\geq\epsilon$. By going to a further subsequence, we
may assume $y_{n}\in R_{1}^{\circ}(y)$ or $y_{n}\in R_{1}^{\circ-1}(y)$ for
all $n$. Suppose the latter holds. Then $z_{n}\rightharpoonup x$ and
$x_{n}\rightharpoonup z$ for all $n$ and so in the limit $z\rightharpoonup x$
and $x\rightharpoonup z$ contradicting anti-symmetry.

∎

##  3\.  Lexicographic Products

Let $(X_{1},R_{1})$ be a tournament. Assume that for each $x\in X_{1}$,
$(Y_{x},S_{x})$ is a tournament. The _lexicographic product_ is defined by:

(3.1) $\displaystyle\begin{split}X_{2}\ =\ X_{1}\times\\{Y_{x}\\}\ =\ &\
\bigcup_{x\in X}\ \\{x\\}\times Y_{x},\\\ R_{2}\ =\
R_{1}\ltimes\\{S_{x}\\}\quad\text{where}\ &\text{for}\
(x_{1},y_{1}),(x_{2},y_{2})\in X_{2},\\\ ((x_{1},y_{1}),(x_{2},y_{2}))\in
R_{2}\quad\Longleftrightarrow&\quad\begin{cases}(x_{1},x_{2})\in
R_{1}^{\circ}\quad\text{or}\\\ x_{1}=x_{2}\ \text{and}\ (y_{1},y_{2})\in
S_{x_{1}}.\end{cases}\end{split}$

It is clear that $(X_{2},R_{2})$ is a tournament and the first coordinate
projection $\pi:X_{2}\to X_{1}$ is a surjective tournament map from
$(X_{2},R_{2})$ to $(X_{1},R_{1})$. We call $X_{2}$ the _total space_ ,
$X_{1}$ the _base space_ and the $Y_{x}$’s the _fibers_ of the product.

######  Proposition 3.1.

If $R_{1}$ and each $S_{x}$ is transitive, then $R_{2}$ is transitive.

###### Proof.

Assume $((x_{1},y_{1}),(x_{2},y_{2})),((x_{2},y_{2}),(x_{3},y_{2}))\in R_{2}$.
If either
$(x_{1},x_{2})\in R_{1}^{\circ}$ or $(x_{2},x_{3})\in R_{1}^{\circ}$ then by
transitivity of $R_{1}$, $(x_{1},x_{3})\in R_{1}^{\circ}$ and so
$((x_{1},y_{1}),(x_{3},y_{3}))\in R_{2}$. Otherwise, $x_{1}=x_{2}=x_{3}$ and
$(y_{1},y_{2}),(y_{2},y_{3})\in S_{x_{1}}$. By transitivity of $S_{x_{1}}$,
$(y_{1},y_{3})\in S_{x_{1}}$ and so
$((x_{1},y_{1}),(x_{3},y_{3}))=((x_{1},y_{1}),(x_{1},y_{3}))\in R_{2}$.

∎

We will write $(X_{2},R_{2})$ as $(X_{1},R_{1})\ltimes\\{(Y_{x},S_{x})\\}$.

For the special case when $(Y_{x},S_{x})=(Y,S)$ for all $x$, we have $X_{2}\
=\ X_{1}\times Y$ and we write $R_{2}\ =\ R_{1}\ltimes S$ and
$(X_{2},R_{2})=(X_{1},R_{1})\ltimes(Y,S)$ is called the _lexicographic
product_ of $(X_{1},R_{1})$ and $(Y,S)$, see [7]. In that case,

(3.2) $R_{1}\ltimes S\ =\ [R_{1}^{\circ}\times(Y\times Y)]\ \cup\ [1_{X}\times
S],\hskip 28.45274pt$

i.e. the union of two product relations as in (2.5). If $R_{1}$ and $S$ are
topological tournaments with $Y$ non-trivial, then $R_{1}\ltimes S$ is closed
if and only if $R_{1}^{\circ}$ is closed and so $1_{X}$ is clopen, which means
that $X$ is discrete, i.e. every point of $X$ is isolated.

In particular, if $X_{1}$ is finite and $(Y,S)$ is a topological tournament,
then $(X_{1},R_{1})\ltimes(Y,S)$ is a topological tournament where
$X_{1}\times Y$ has the product topology. I emphasize the topology on
$X_{1}\times Y$ because we will deal with the problem of obtaining a closed
relation for the lexicographic product by adjusting the topology on the total
space.

######  Definition 3.2.

Let $(X_{1},R_{1})$ and the members of $\\{(Y_{x},S_{x}):x\in X\\}$ be
topological tournaments. The product tournament
$(X_{2},R_{2})=(X_{1},R_{1})\ltimes\\{(Y_{x},S_{x})\\}$ is called the
_topological lexicographic product_ when the following conditions hold:

  * (i)

For each $x\in X_{1}$, either $x$ is an isolated point of $X_{1}$ or else
$(Y_{x},S_{x})$ is a trivial tournament.

  * (ii)

The total space $X_{2}$ is given the topology with basis $\mathcal{B}$ where
$U\in\mathcal{B}$ when either $U=\pi^{-1}(V)$ for $V$ some open subset of
$X_{1}$, or $U=\\{x\\}\times V$ for $x$ isolated in $X_{1}$ and $V$ some open
subset of $Y_{x}$.

In particular, if $x\in X_{1}$ is non-isolated, then $\pi^{-1}(x)$ is a
singleton subset of $X_{2}$ which we will identify with $\\{x\\}$.

######  Theorem 3.3.

The topological lexicographic product $(X_{2},R_{2})=$
$(X_{1},R_{1})\ltimes\\{(Y_{x},S_{x})\\}$ is a topological tournament which
satisfies the following properties.

  1. (a)

The projection map $\pi$ is a continuous, open surjection mapping $R_{2}$ to
$R_{1}$.

  2. (b)

A point $(x,y)\in X_{2}$ is isolated if and only if $x$ is isolated in $X_{1}$
and $y$ is isolated in $Y_{x}$.

  3. (c)

Assume $x\in X_{1}$ is an isolated point. The map $y\mapsto(x,y)$ is a
homeomorphism from $Y_{x}$ onto the clopen subset $\\{x\\}\times Y_{x}$ of
$X_{2}$, mapping $S_{x}$ isomorphically to the restriction
$R_{2}|\\{x\\}\times Y_{x}$. In particular, a point $y\in Y_{x}$ is left (or
right) balanced for $S_{x}$ if and only if $(x,y)$ is left (resp. right)
balanced for $R_{2}$.

  4. (d)

If $X_{1}$ and each $Y_{x}$ is compact, then $X_{2}$ is compact.

  5. (e)

If $X_{1}$ and each $Y_{x}$ is countable, then $X_{2}$ is countable.

  6. (f)

If $R_{1}$ and each $S_{x}$ is transitive, then $R_{2}$ is transitive.

  7. (g)

If $R_{1}$ and each $S_{x}$ is arc cyclic, then $R_{2}$ is arc cyclic. If
$R_{1}$ is point cyclic, then $R_{2}$ is point cyclic.

###### Proof.

(a): It is clear that the collection $\mathcal{B}$ is closed under
intersection and so forms a basis for a topology with $\pi_{X}$ continuous.
Since $\pi(U)$ is open in $X_{1}$ for each $U\in\mathcal{B}$, it follows that
$\pi$ is an open map. From (i) and (ii) it easily follows that the topology on
$X_{2}$ is Hausdorff.

Now suppose that $\\{((x_{k},y_{k}),(u_{k},v_{k}))\\}$ is a net in $R_{2}$
converging to $((x,y),(u,v))\in X_{2}\times X_{2}$. Since $\pi$ is continuous,
$\\{(x_{k},u_{k})\\}$ converges to $(x,u)$ in $X_{1}\times X_{1}$.

Case 1: ($x\not=u$) If $(x,u)\in R_{1}^{\circ-1}$ then eventually
$(x_{k},u_{k})\in R_{1}^{\circ-1}$ and so $((x_{k},y_{k}),(u_{k},v_{k}))\in
R_{2}^{\circ-1}$ contrary to hypothesis. Hence, $(x,u)\in R_{1}^{\circ}$ and
so $((x,y),(u,v))\in R_{2}^{\circ}$.

Case 2a: ($x=u$ is isolated) In that case, eventually, $x_{k}=x$ and
$u_{k}=u=x$ and so $\\{(y_{k},v_{k})\\}$ is eventually a net in $S_{x}$.
Hence, the limit point $(y,v)\in S_{x}$ which implies
$((x,y),(u,v))=((x,y),(x,v))\in R_{2}$.

Case 2b: ($x=u$ is not isolated) In that case, $Y_{x}$ is a singleton and so
$y=v$. That is, $(x,y)=(u,v)$ and so $((x,y),(u,v))\in R_{2}$.

Thus, $R_{2}$ is a closed relation and so $(X_{2},R_{2})$ is a topological
tournament.

(b): Clearly, if $x$ is isolated in $X_{1}$ and $y$ is isolated in $Y_{x}$,
then $\\{x\\}\times\\{y\\}$ is a basic open set in $X_{2}$ and so $(x,y)$ is
isolated.

If $x$ is isolated and $\\{y_{k}\\}$ is a net in $Y_{x}\setminus\\{y\\}$
converging to $y$, then $\\{(x,y_{k})\\}$ converges to $(x,y)$ and so $(x,y)$
is not isolated.

If $\\{x_{k}\\}$ is a net in $X_{1}\setminus\\{x\\}$ converging to $x$ and
$Y_{x}=\\{y\\}$, then for any $y_{k}\in Y_{x_{k}}$, the net
$\\{(x_{k},y_{k})\\}$ in $X_{2}$ converges to $(x,y)$ and so $(x,y)$ is not
isolated.

(c): That the injection from $Y_{x}$ onto $\\{x\\}\times Y_{x}\subset X_{2}$
is a homeomorphism onto a clopen subset follows using the basis $\mathcal{B}$
in (ii).

(d): Now assume that $X_{1}$ and the $Y_{x}$’s are compact and that
$\mathcal{U}$ is an open cover of $X_{2}$. Let $\mathcal{U}_{1}$ be the open
cover of $X_{1}$ consisting of the singleton isolated points together with
open sets $V$ such that $\pi^{-1}(V)$ is contained in some member of
$\mathcal{U}$. Because $X_{1}$ is compact, $\mathcal{U}_{1}$ has a finite
subcover consisting of finitely many isolated point singletons $\\{x_{j}\\}$
together with finitely many open sets $V_{i}$ with $\pi^{-1}(V_{i})\subset
U_{i}$. For each $\\{x_{j}\\}$ there is a finite cover $\\{V_{jk}\\}$ of
$Y_{x_{j}}$ consisting of open sets with $\\{x_{j}\\}\times V_{jk}$ contained
in some member $U_{jk}$ of $\mathcal{U}$. Then, $\\{U_{i}\\}$ together with
$\\{U_{jk}\\}$ for each $x_{j}$ is a finite cover of $X_{2}$ by elements of
$\mathcal{U}$. It follows that $X_{2}$ is compact.

(e): The countability result is obvious.

(f): The transitivity result follows from Proposition 3.1.

(g): The cyclicity results follow from Corollary 2.3

∎

######  Addendum 3.4.

If $\\{z_{k}\\}$ is a net in $X_{2}$ and $z\in X_{2}$ with $\pi(z)$ non-
isolated, then $\\{z_{k}\\}$ converges to $z$ if and only if the net
$\\{\pi(z_{k})\\}$ in $X_{1}$ converges to $\pi(z)\in X_{1}$. In particular,
$x=\pi(z)$ is left (or right) balanced for $R_{1}$ if and only if, identified
with the point in $\pi^{-1}(x)$ it is left (resp. right) balanced for $R_{2}$.

###### Proof.

Suppose $\\{\pi(z_{k})\\}$ converges to $\pi(z)$. If $U\subset X$ is an open
set with $\pi(z)\in U$, then eventually $\pi(z_{k})\in U$ and so
$z_{k}\in\pi^{-1}(U)$. From the definition of the topology on $X_{2}$, it
follows that $\\{z_{k}\\}$ converges to $z$.

The converse is obvious from the continuity of $\pi$.

The balance results follow because $(z_{k},z)\in R_{2}^{\circ}$ if and only if
$(\pi(z_{k}),\pi(z))\in R_{1}^{\circ}$.

∎

For a topological lexicographic product
$(X_{2},R_{2})=(X_{1},R_{1})\ltimes\\{(Y_{x},S_{x})\\}$ a _section_ is a
function $\xi:X_{1}\to X_{2}$ such that $\pi\circ\xi\ =\ 1_{X_{1}}$. That is,
$\xi$ is essentially a choice function $\tilde{\xi}$ for the family
$\\{Y_{x}\\}$ with $\xi(x)=(x,\tilde{\xi}(x))$.

######  Lemma 3.5.

For a topological lexicographic product $(X_{2},R_{2})=(X_{1},R_{1})$
$\ltimes\\{(Y_{x},S_{x})\\}$ any section is continuous. Furthermore any
section $\xi$ is a topological tournament isomorphism from $(X_{1},R_{1})$
onto the restriction of $R_{2}$ to the image $\xi(X_{1})$.

###### Proof.

Continuity at $x$ when $x$ is isolated is obvious. When $x$ is non-isolated,
continuity follows from Addendum 3.4. It is clear that the injection $\xi$
maps $R_{1}$ to $R_{2}$.

∎

######  Definition 3.6.

We call a topological tournament $(X,R)$ a _brick_ when it satisfies the
following conditions.

  * (i)

The space $X$ is compact and the isolated points are dense in $X$.

  * (ii)

If $x\in X$ is not isolated, then the point $x$ is balanced for $R$.

We call a brick _isolated point cyclic_ , or _ip cyclic_ when it satisfies, in
addition,

  * (iii)

If $x$ is an isolated point, then there exists a $3-$cycle for $R$ which
contains $x$.

From the density of the isolated points, it follows that the $3-$cycle in
(iii) can be chosen to consist of isolated points.

If $X$ is finite then $(X,R)$ is a brick and if, in addition, $(X,R)$ is
regular, then it is ip cyclic by [9] Theorem 7.

######  Theorem 3.7.

If $(X_{1},R_{1})$ and the members of $\\{(Y_{x},S_{x}):x\in X\\}$ are all
bricks, with $(Y_{x},S_{x})$ trivial when $x$ is not isolated in $X_{1}$, then
the topological lexicographic product
$(X_{2},R_{2})=(X_{1},R_{1})\ltimes\\{(Y_{x},S_{x})\\}$ is a brick. If, in
addition, for each isolated point $x$, the brick $(Y_{x},S_{x})$ is ip cyclic,
then $(X_{2},R_{2})$ is ip cyclic.

###### Proof.

Compactness follows from Theorem 3.3.

If $x$ is isolated but $y\in Y_{x}$ is not, then since $y$ is assumed balanced
in $Y_{x}$, it follows that $(x,y)$ is balanced in $X_{2}$ by Theorem 3.3 (c).
Since $y$ is a limit of isolated points in $Y_{x}$, $(x,y)$ is a limit of
points isolated in $\\{x\\}\times Y_{x}$ and hence in $X_{2}$.

If $x$ is not isolated in $X_{1}$, then since $x$ is balanced in $X_{1}$, it
is balanced in $X_{2}$ by Addendum 3.4. If $x$ is the limit of a net
$\\{x_{k}\\}$ of isolated points in $X_{1}$ and $y_{k}$ is an isolated point
in $Y_{x_{k}}$ then $\\{(x_{k},y_{k})\\}$ is a net of isolated points in
$X_{2}$ which converges to $x$ in $X_{2}$ by Addendum 3.4 again.

Thus, $(Z,T)$ is a brick.

If $(x,y)$ is an isolated point and $(Y_{x},S_{x})$ is ip cyclic, then $y$ is
contained in a $3-$cycle $\\{y,y^{\prime},y^{\prime\prime}\\}$ in $Y_{x}$ and
so $(x,y)$ is contained in the $3-$cycle of points
$\\{(x,y),(x,y^{\prime}),(x,y^{\prime\prime})\\}$ in $X_{2}$.

∎

In general, we will call a topological tournament $(X,R)$ _ip cyclic_ when
every isolated point of $X$ is contained in a $3-$cycle.

##  4\.  Inverse Limits

An _inverse system_ $\\{(X_{i},f_{i}):i\in\mathbb{N}\\}$ is a sequence with
$f_{i}$ a function from $X_{i+1}$ to $X_{i}$ for all $i\in\mathbb{N}$. The
_inverse limit_ $X=\overleftarrow{Lim}\\{(X_{i},f_{i})\\}$ is given by

(4.1) $X\ =\ \\{x\in\prod_{i\in\mathbb{N}}\ X_{i}:f_{i}(x_{i+1})\ =\ x_{i}\ \
\text{for all}\ \ i\in\mathbb{N}\\},$

The functions $\pi_{i}:X\to X_{i}$ and $\pi_{i,i+1}:X\to X_{i}\times X_{i+1}$
are the projection mappings. Clearly, for all $i$:

(4.2) $f_{i}\circ\pi_{i+1}\ =\ \pi_{i}\quad\text{on}\ \ X.\hskip 28.45274pt$

We call $\\{(X_{i},f_{i}):i\in\mathbb{N}\\}$ a _surjective inverse system_
when each $f_{i}$ is a surjective map.

######  Proposition 4.1.

If $\\{(X_{i},f_{i}):i\in\mathbb{N}\\}$ is a surjective inverse system, then
for all $i\in\mathbb{N}$ $\pi_{i,i+1}$ maps $X$ onto $f_{i}^{-1}$ and
$\pi_{i}$ maps $X$ onto $X_{i}$.

###### Proof.

It is clear that $\pi_{i,i+1}$ maps into $f_{i}^{-1}$.

Let $(x_{i},x_{i+1})\in f_{i}^{-1}$. Inductively, for $j$ with $1\leq j<i$,
let $x_{i-j}$ be the point such that $f_{i-j}(x_{i-j+1})\ =\ x_{i-j}.$ Because
each $f_{k}$ is surjective, for $j$ with $1<j$ we can choose, inductively, a
point $x_{i+j}$ such that $f_{i+j-1}(x_{i+j})\ =\ x_{i+j-1}$. Thus,
$\pi_{i,i+1}$ maps onto $f_{i}^{-1}$. Since $f_{i}$ is surjective, it clearly
follows that $\pi_{i}:X\to X_{i}$ is onto as well.

∎

If each $f_{i}$ is a continuous map, then $X$ is a closed subset of
$\prod_{i\in\mathbb{N}}\ X_{i}$ with the latter given the product topology.
If, in addition, the spaces $X_{i}$ are compact, then the inverse limit space
$X$ is compact by the Tychonoff Product Theorem. In any case, the projection
maps are continuous.

######  Theorem 4.2.

Assume that $\\{(X_{i},f_{i})\\}$ is an inverse system with inverse limit $X$.
If, for each $i\in\mathbb{N}$, $R_{i}$ is a relation on $X_{i}$ such that
$f_{i}$ maps $R_{i+1}$ to $R_{i}$, then $\\{(R_{i},f_{i}\times f_{i})\\}$ is
an inverse system with inverse limit which we label $R$.

Identifying $\prod_{i\in\mathbb{N}}(X_{i}\times X_{i})$ with
$(\prod_{i\in\mathbb{N}}X_{i})\times(\prod_{i\in\mathbb{N}}X_{i})$ we can
regard $R$ as a relation on $X$ with

(4.3) $R\ =\ \bigcap_{i\in\mathbb{N}}\
(\pi_{i}\times\pi_{i})^{-1}(R_{i}).\hskip 56.9055pt$

If each $R_{i}$ is a tournament on $X_{i}$, then $R$ is a tournament on $X$
with $\pi_{i}$ mapping $R$ to $R_{i}$. For $x,x^{\prime}\in X$, we have
$(x,x^{\prime})\in R^{\circ}$ if and only if there exists $i\in\mathbb{N}$
such that $x_{j}=x^{\prime}_{j}$ for all $j<i$ and $(x_{i},x^{\prime}_{i})\in
R_{i}^{\circ}$.

If each $f_{i}:X_{i+1}\to X_{i}$ is a continuous map of topological spaces and
each $R_{i}$ is a topological tournament on $X_{i}$, then $R$ is a topological
tournament on $X$.

If each $R_{i}$ is a transitive tournament, then $R$ is transitive.

###### Proof.

It is clear that $\\{(R_{i},f_{i}\times f_{i})\\}$ is an inverse system and
with the above identification we can regard $R$ as a relation on $X$ such that
$\pi_{i}$ maps $R$ to $R_{i}$. Hence, $R\subset\bigcap_{i}\
(\pi_{i}\times\pi_{i})^{-1}(R_{i})$. On the other hand, if
$(x,x^{\prime})\in\bigcap_{i}\ (\pi_{i}\times\pi_{i})^{-1}(R_{i})$, then
$(x_{i},x^{\prime}_{i})\in R_{i}$ and $x,x^{\prime}\in X$ implies
$(x_{i},x^{\prime}_{i})=(f_{i}\times f_{i})(x_{i+1},x^{\prime}_{i+1})$. Hence,
$(x,x^{\prime})\in R$, proving (4.3).

Now assume that each $R_{i}$ is a tournament.

(4.4) $\displaystyle\begin{split}R\cap R^{-1}\ =\ &\bigcap_{i}\
(\pi_{i}\times\pi_{i})^{-1}(R_{i}\cap R_{i}^{-1})\\\ =\ \bigcap_{i}\
&(\pi_{i}\times\pi_{i})^{-1}(1_{X_{i}})\ =\ 1_{X}.\end{split}$

Therefore, $R$ is anti-symmetric.

Now assume that $(x,x^{\prime})\in(X\times X)\setminus R$. From (4.3) it
follows that for some $i_{0},\ (x_{i_{0}},x^{\prime}_{i_{0}})\in
R_{i_{0}}^{\circ-1}$.

If for some $i_{1},(x_{i_{1}},x^{\prime}_{i_{1}})\in R_{\circ i_{1}}$, then
applying (2.8) to the appropriate composition of the maps $f_{k}$ with would
obtain, with $i=\max(i_{0},i_{1})$ that $(x_{i},x^{\prime}_{i})\in
R_{i}^{\circ-1}\cap R_{i}^{\circ}$ which is impossible. Hence, for all
$i_{1}\in\mathbb{N},(x_{i_{1}},x^{\prime}_{i_{1}})\in R_{i_{1}}^{-1}$ and thus
$(x,x^{\prime})\in R^{-1}$. That is, $R$ is total and so is a tournament.

If $(x,x^{\prime})\in R^{\circ}$, then since $\pi_{j}$ maps $R$ to $R_{j}$, we
have $(x_{j},x^{\prime}_{j})\in R_{j}$ for all $i$. So if
$i=\min\\{j:x_{j}\not=x^{\prime}_{j}\\}$, then $(x_{i},x_{i}^{\prime})\in
R_{i}^{\circ}$. Conversely, if $(x_{i},x_{i}^{\prime})\in R_{i}^{\circ}$ then
since $\pi_{i}$ maps $R$ to $R_{i}$, we have we have $(x,x^{\prime})\in
R^{\circ}$.

Given the topological assumptions, it is clear that $R$ is closed and so is a
topological tournament.

Now assume that each $R_{i}$ is transitive and that
$(x,x^{\prime}),(x^{\prime},x^{\prime\prime})\in R$. We show that
$(x,x^{\prime\prime})\in R$. Clearly, we may assume that the three points are
distinct so that $(x,x^{\prime}),(x^{\prime},x^{\prime\prime})\in R^{\circ}$
There exists $i\in\mathbb{N}$ such that $x_{k}=x_{k}^{\prime}$ for all $k<i$
and $(x_{i},x_{i}^{\prime})\in R_{i}^{\circ}$. Similarly, for some
$j,x_{k}^{\prime}=x_{k}^{\prime\prime}$ for all $k<j$ and
$(x_{j}^{\prime},x_{j}^{\prime\prime})\in R_{j}^{\circ}$. If $i>j$, then
$x_{k}=x_{k}^{\prime}=x_{k}^{\prime\prime}$ for all $k<j$ and
$(x_{j},x_{j}^{\prime\prime})=(x_{j}^{\prime},x_{j}^{\prime\prime})\in
R_{j}^{\circ}$ so that $(x,x^{\prime\prime})\in R^{\circ}$. Similarly, if
$i<j$, $(x,x^{\prime\prime})\in R^{\circ}$. If $i=j$, then
$x_{k}=x_{k}^{\prime}=x_{k}^{\prime\prime}$ for all $k<i$ and
$(x_{i},x_{i}^{\prime}),(x_{i}^{\prime},x_{i}^{\prime\prime})\in
R_{i}^{\circ}$. By transitivity of $R_{i}$, we have
$(x_{i},x_{i}^{\prime\prime})\in R_{i}$. Antisymmetry and
$(x_{i},x_{i}^{\prime}),(x_{i}^{\prime},x_{i}^{\prime\prime})\in
R_{i}^{\circ}$ implies that $x_{i}\not=x_{i}^{\prime\prime}$. Hence, in this
case as well $(x,x^{\prime\prime})\in R^{\circ}$.

∎

We will call a sequence $\\{(X_{i},R_{i},f_{i}):i\in\mathbb{N}\\}$ an inverse
system of tournaments, when $\\{(X_{i},f_{i}):i\in\mathbb{N}\\}$ is an inverse
system, $R_{i}$ is a tournament on $X_{i}$ and $f_{i}$ maps $R_{i+1}$ to
$R_{i}$ for all $i$. We call the tournament $(X,R)$ the inverse limit of this
system when $X=\overleftarrow{Lim}\\{(X_{i},f_{i})\\}$ and
$R=\overleftarrow{Lim}\\{(R_{i},f_{i}\times f_{i})\\}$.

For every inverse system of tournaments,
$\\{(X_{i},R_{i},f_{i}):i\in\mathbb{N}\\}$ it will be convenient to assume
that there is a zero level with $(X_{0},R_{0},f_{0})$ with $(X_{0},R_{0})$ a
trivial tournament, i.e. $X_{0}$ is a singleton, and $f_{0}$ is the unique
function from $X_{1}$ to $X_{0}$.

######  Proposition 4.3.

If $\\{(X_{i},R_{i},f_{i}):i\in\mathbb{N}\\}$ is a surjective inverse system
of tournaments, then the limit tournament $(X,R)$ is arc cyclic, if and only
if $(X_{i},R_{i})$ is arc cyclic for every $i\in\mathbb{N}$.

###### Proof.

If $(X,R)$ is arc cyclic, then since $\pi_{i}$ maps $(X,R)$ onto
$(X_{i},R_{i})$, the latter is arc cyclic by Corollary 2.3.

Now assume that all the $(X_{i},R_{i})$’s are arc cyclic. If
$(x,x^{\prime})\in R^{\circ}$, then there exists $i\in\mathbb{N}$ such that
$x_{j}=x^{\prime}_{j}$ for all $j<i$ and $(x_{i},x^{\prime}_{i})\in
R_{i}^{\circ}$. Since $(X_{i},R_{i})$ is arc cyclic, there exists $z\in X_{i}$
such that $\\{x_{i},x^{\prime}_{i},z\\}$ is a $3-$cycle in $X_{i}$. Since
$\pi_{i}$ is surjective, there exists $x^{\prime\prime}$ such that
$\pi_{i}(x^{\prime\prime})=z$. Then $\\{x,x^{\prime},x^{\prime\prime}\\}$ is a
$3-$cycle in $X$.

∎

We call $\\{(X_{i},R_{i},f_{i}):i\in\mathbb{N}\\}$ an _inverse system of
topological tournaments_ when $\\{(X_{i},R_{i},f_{i})\\}$ is an inverse system
of tournaments with each $(X_{i},R_{i})$ a topological tournament and each
$f_{i}$ continuous. The inverse limit $(X,R)$ is then topological.

######  Addendum 4.4.

Assume that $\\{(X_{i},R_{i},f_{i})\\}$ and $\\{(Y_{i},S_{i},g_{i})\\}$ are
inverse systems of topological tournaments with limits $(X,R)$ and $(Y,S)$. If
for each $i$, the continuous function $h_{i}:X_{i}\to Y_{i}$ maps $R_{i}$ to
$S_{i}$ and $g_{i}\circ h_{i+1}=h_{i}\circ f_{i}$, then the product map
$\prod_{i}\ h_{i}:\prod_{i}X_{i}\to\prod_{i}Y_{i}$ defined by
$h(x)_{i}=h_{i}(x_{i})$ restricts to a continuous function $h:X\to Y$ which
maps $(X,R)$ to $(Y,S)$.

###### Proof.

Just as the family $\\{h_{i}\\}$ maps $\\{(X_{i},f_{i})\\}$ to
$\\{(Y_{i},g_{i})\\}$, the family $\\{h_{i}\times h_{i}\\}$ maps
$\\{(R_{i},f_{i}\times f_{i})\\}$ to $\\{(S_{i},g_{i}\times g_{i})\\}$ and
hence $(h\times h)(R)\subset S$.

∎

######  Examples 1.

(a) Let $\\{K_{i}\\}$ be a decreasing sequence of subsets of a set $X$ with
$k_{i}:K_{i+1}\to K_{i}$ the inclusion map. If $K\ =\
\bigcap_{i\in\mathbb{N}}\ K_{i}$, then map which associates to $x\in K$ the
constant sequence at $x$ is an identification of $K$ with the inverse limit of
$\\{(K_{i},k_{i})\\}$. The inverse map for this identification equals
$\pi_{i}$ for every $i$.

If $R$ is a tournament on $X$, then the identification is an isomorphism from
the restriction $R|K$ to the inverse limit of the system
$\\{(K_{i},R|K_{i},k_{i})\\}$.

(b) With $\\{Y_{i}\\}$ a sequence of spaces, let $X_{i}=\prod_{1\leq j\leq i}\
Y_{j}$, $f_{i}:X_{i+1}\to X_{i}$ be the projection on the first $i$
coordinates and $g_{i}:X_{i}\to Y_{i}$ be the $i^{th}$ coordinate projection.
Let $Y\ =\ \prod_{i\in\mathbb{N}}\ Y_{i}$. The map $q:X\to Y$ defined by
$q(x)_{i}=g_{i}(\pi_{i}(x))$ is an identification of $Y$ with the inverse
limit $X$ of $\\{(X_{i},f_{i})\\}$.

For $\\{(X_{i},R_{i},f_{i}):i\in\mathbb{N}\\}$ an inverse system of
topological tournaments with limit the topological tournament $(X,R)$ we let
$IS=\\{x\in X:x_{i}$ is an isolated point of $X_{i}$ for all
$i\in\mathbb{N}\\}$ .

Now assume that $(X_{1},R_{1})$ is a compact topological tournament.
Inductively we define $(X_{i+i},R_{i+1})$ to be the topological lexicographic
product $(X_{i},R_{i})\ltimes\\{(Y_{iz},S_{iz}):z\in X_{i}\\}$ with each
$(Y_{iz},S_{iz})$ a compact tournament and with $(Y_{iz},S_{iz})$ trivial when
$z$ is not isolated in $X_{i}$. Let $f_{i}:X_{i+1}\to X_{i}$ be the first
coordinate projection. By Theorem 3.3 each $(X_{i},R_{i})$ is a compact
topological tournament. Thus, $\\{(X_{i},R_{i},f_{i})\\}$ is a surjective
inverse system of topological tournaments which we will call a _lexicographic
inverse system_ . The limit system $(X,R)$ is a compact topological tournament
by Theorem 4.2.

If $(X_{1},R_{1})$ and each $(Y_{iz},S_{iz})$ is a brick, then,
inductively,$(X_{i},R_{i})$ is a brick and we will call
$\\{(X_{i},R_{i},f_{i})\\}$ a _lexicographic inverse system of bricks_ .

######  Theorem 4.5.

Assume that $\\{(X_{i},R_{i},f_{i})\\}$ is a lexicographic inverse system with
limit tournament $(X,R)$.

  * (a)

For each $i\in\mathbb{N}$ the projection map $\pi_{i}:X\to X_{i}$ is a
continuous, open surjection.

  * (b)

If $z\in X_{i}$ is not isolated, then $\\{\pi_{i}^{-1}(z)\\}$ is a singleton
subset $\\{x\\}$ of $X$, and a net $\\{x_{k}\\}$ in $X$ converges to $x$ in
$X$ if and only if $\\{\pi_{i}(x_{k})\\}$ converges to $z$ in $X_{i}$.

Now assume that $\\{(X_{i},R_{i},f_{i})\\}$ is a lexicographic inverse system
of bricks.

  * (c)

The set $IS$ is residual in $X$. That is, it is a dense $G_{\delta}$ subset of
$X$.

  * (d)

If for infinitely many $i\in\mathbb{N}$ and the tournament $(Y_{iz},S_{iz})$
has no terminal point for each isolated point $z\in X_{i}$, and for infinitely
many $i\in\mathbb{N}$ the tournament $(Y_{iz},S_{iz})$ has no initial point
each isolated point $z\in X_{i}$, then the limit tournament $(X,R)$ is
balanced. If, in addition, $X_{1}$ and each $Y_{iz}$ is countable, then $X$ is
a Cantor Set.

###### Proof.

(a): The $\pi_{j}$’s are surjective by Proposition 4.1. The basic open subsets
of $X$ can be written $\pi_{j}^{-1}(U)$ for $j$ arbitrarily large and $U$ open
in $X_{j}$. Because each $\pi_{j}$ is surjective by Proposition 4.1,
$\pi_{j}(\pi_{j}^{-1}(U))=U$. Choose $j>i$. Using (4.2) and induction we see
that $\pi_{i}(U)=f_{i}\circ\dots\circ f_{j-1}(U)$. This is open because each
$f_{j}$ is an open map by Theorem 3.3(a). Hence, $\pi_{i}$ is an open map. It
is clearly continuous.

(b): If $z\in X_{i}$ is not isolated, then $Y_{iz}$ is a singleton and so
$\\{f_{i}^{-1}(z)\\}$ is a singleton $\\{z^{\prime}\\}$ in $X_{i+1}$, and by
Theorem 3.3 $z^{\prime}$ is not isolated in $X_{i+1}$. Proceeding upwards by
induction we see that there is only one point $x$ with $x_{i}=z$. If
$\\{x_{k}\\}$ is a net in $X$ such that $\\{\pi_{i}(x_{k})\\}$ converges to
$z$ in $X_{i}$, then Addendum 3.4 implies that $\\{\pi_{i+1}(x_{k})\\}$
converges to $z^{\prime}$. Of course, by continuity
$\\{\pi_{i-1}(x_{k})=f_{i-1}(\pi_{i}(x_{k}))\\}$ converges to
$f_{i-1}(\pi_{i}(x))=\pi_{i-1}(x)$. Proceeding upwards and downwards by
induction we see that $\\{x_{k}\\}$ converges coordinatewise to $x$.

(c): If $Iso(X_{i})$ is the set of isolated points of $X_{i}$, then it is a
open subset of $X_{i}$ which is dense in $X_{i}$ because $X_{i}$ is a brick.
Because $\pi_{i}$ is continuous and open, the set $\pi_{i}^{-1}(Iso(X_{i}))$
is open and dense in $X$. By the Baire Category Theorem, the set $IS\ =\
\bigcap_{i}\ \pi_{i}^{-1}(Iso(X_{i}))$ is a dense $G_{\delta}$ subset of $X$.

(d): Let $x\in X$.

If $x_{i}\in X_{i}$ is not isolated, then since $(X_{i},R_{i})$ is a brick,
$x_{i}\in X_{i}$ is balanced and so we can choose a net $\\{z_{k}\\}$ in
$R_{i}^{\circ}(x_{i})$ which converges to $x_{i}$ in $X_{i}$. Choose $x_{k}$
so that $\pi_{i}(x_{k})=z_{k}$. Then $\\{x_{k}\\}$ is a net in $X$ which
converges to $x$. Because $\pi_{i}$ maps $R$ to $R_{i}$, we have $x_{k}\in
R^{\circ}(x)$. Hence, $x$ is right balanced and similarly it is left balanced.

Thus, if $x\not\in IS$, then it balanced in any case.

Now assume that $x\in IS$.

Fix $i$ arbitrarily large such that no $(Y_{iz},S_{iz})$ with $z$ isolated has
a terminal point and let $z=x_{i}$. The point $x_{i+1}=(z,y)$ with $y\in
Y_{iz}$. Since $(Y_{iz},S_{iz})$ has no terminal point, there exists
$y^{\prime}\in S_{iz}^{\circ}(y)$ and so $(z,y^{\prime})\in
R_{i+1}^{\circ}(z,y)$. There exists $x^{\prime}\in X$ with
$x^{\prime}_{i+1}=(z,y^{\prime})$ and so $x^{\prime}\in R^{\circ}(x)$.
Furthermore, $x^{\prime}_{i}=z=x_{i}$ and so $x_{j}=x^{\prime}_{j}$ for all
$j\leq i$. As $i$ was arbitrarily large, $x^{\prime}$ is arbitrarily close to
$x$ and so $x$ is right balanced. Similarly, $x$ is left balanced.

If $X_{1}$ and each $Y_{iz}$ is countable, then Theorem 3.3 and induction
imply each $X_{i}$ is a countable brick and certainly $X_{j}$ is not trivial
for $j>1$. Hence, the space $\prod_{i}\ X_{i}$ is a countable product of
compact, metrizable, totally disconnected spaces and so is a Cantor set. The
subset $X$ is therefore a compact, metrizable, totally disconnected space.
Since the tournament $R$ is balanced, $X$ has no isolated points and so is
itself a Cantor set.

∎

When $X_{1}$ and all the $Y_{iz}$’s are finite, then each $X_{i}$ is finite
and so consists of isolated points. We consider the case when for each $i$ the
$(Y_{iz},S_{iz})$’s are the same for all $z\in X_{i}$.

On a finite set of cardinality $n$ there are $2^{n(n-1)/2}$ tournaments. When
$n\geq 4$ the majority of these have no terminal nor initial point.

Let $S_{i}$ for $i\in\\\ Z_{+}=\\{0\\}\cup\mathbb{N}$ be a tournament on a
finite set $Y_{i}$ with infinitely many having no terminal point and with
infinitely many having no initial point. On the infinite product
$Y=\prod_{i\in\mathbb{Z}_{+}}Y_{i}$ define $S\ =\
\ltimes_{i\in\mathbb{Z}_{+}}\ S_{i}$ by

(4.5) $(y,z)\in S^{\circ}\quad\Longleftrightarrow\quad(y_{i},z_{i})\in S_{i}\
\ \text{for}\ \ i=\min\\{j:y_{j}\not=z_{j}\\},$

for $y,z$ distinct points of $Y$.

On the other hand, we can let $(X_{1},R_{1})=(Y_{0},S_{0})$ and inductively
for $i\in\mathbb{N}$ define
$(X_{i+1},R_{i+1})=(X_{i},R_{i})\ltimes(Y_{i},S_{i})$ with $f_{i}:X_{i+1}\to
X_{i}$ the first coordinate projection. Let $g_{1}:X_{1}\to Y_{0}$ be the
identity and for $i\in\mathbb{N}$ let $g_{i+1}:X_{i+1}\to Y_{i}$ be the second
coordinate projection.

It is clear that $\\{(X_{i},R_{i},f_{i})\\}$ an inverse system of topological
tournaments which is a lexicographic inverse system of bricks Let $(X,R)$ be
the limit. Thus, by Theorem 4.5 $(X,R)$ is a balanced tournament on a Cantor
set. Furthermore, the following is easy to check.

######  Theorem 4.6.

If we define $q:X\to Y$ by

(4.6) $q(x)_{i}\ =\ g_{i+1}(\pi_{i+1}(x)),\ \ \text{for}\ \
i\in\mathbb{Z}_{+}$

then $q$ is a homeomorphism from $X$ onto $Y$ which is a tournament
isomorphism from $R$ to $S$.

Thus, the relation $S$ is a balanced topological tournament on the Cantor set
$Y$.

##  5\.  Connectedness and Compactness

A _linear order_ is exactly a transitive tournament. When the space is
connected, a topological tournament is necessarily a linear order.

######  Theorem 5.1.

Let $R$ be a topological tournament on a space $X$.

  * (a)

If $A$ is a connected subset of $X$ and $x\in X\setminus A$, then either
$A\subset R^{\circ}(x)$ or $A\subset R^{\circ-1}(x)$.

  * (b)

If $X$ is connected, then for all $x\in X$ the sets $R(x)$ and $R^{-1}(x)$ are
connected. Furthermore, $R$ is transitive and so is a linear order on $X$.

###### Proof.

(a) Since $x\not\in A$, $A$ is the disjoint union of the relatively open
subsets $A\cap R^{\circ}(x)$ and $A\cap R^{\circ-1}(x)$. So if $A$ is
connected, one of these is empty.

(b) If $R(x)$ is not connected, then it contains a proper subset $A$ which is
clopen in the relative topology on $R(x)$. Replacing $A$ by its complement if
necessary, we may assume $x\not\in A$. Since $A$ is a closed subset of $R(x)$,
it is closed in $X$. Since $A$ is an open subset of $R^{\circ}(x)$, it is open
in $X$. Since $A$ is nonempty, $X$ is not connected. Applying the result to
$R^{-1}$ we see that $R^{-1}(x)$ is connected as well when $X$ is connected.

If $y\in R^{\circ}(x)$, then $R(y)$ is a connected set which meets
$R^{\circ}(x)$. By anti-symmetry, $x\not\in R(y)$. So (a) implies that
$R(y)\subset R^{\circ}(x)$. Hence, $R$ is transitive and so is a linear order.

∎

It follows that if $R$ is a topological tournament on $X$, then the
restriction of $R$ to any connected subset $A$ of $X$ is a linear order on
$A$.

######  Corollary 5.2.

For a compact space $X$, the set $E=\\{(x,y):x,y\in A$ with $A$ a connected
subset of $X\\}$ is a closed equivalence relation with equivalence classes the
components of $X$. The quotient space $X/E$ is totally disconnected. Let
$\pi:X\to X/E$ be the quotient map.

Assume that $X$ admits a topological tournament $R$. The relation
$R_{E}=(\pi\times\pi)(R)\subset X/E\times X/E$ is a topological tournament on
$X/E$ with $\pi$ a continuous, surjective tournament map from $R$ to $R_{E}$.
For every non-trivial component $A$ of $X$ there is an open subset $A^{\circ}$
of $X$ which is contained in $A$ and with the cardinality of $A\setminus
A^{\circ}$ at most two.

If, in addition, $X$ is metrizable, then for every $\epsilon>0$ the set of
components of $X$ with diameter at least $\epsilon$ is finite and so the set
of non-trivial components of $X$ is countable.

If $R$ is balanced, then $R_{E}$ is balanced and so $X/E$ has no isolated
points. If, in addition, $X$ is metrizable, then $X/E$ is a Cantor set.

###### Proof.

The equivalence classes of $E$ are clearly the components of $X$. For each
component $A$, the collection of clopen sets which contain $A$ form a base for
the neighborhood system of $A$. Any component which meets a clopen set is
contained in it. It follows that $E=\bigcap\\{B\times B\cup(X\setminus
B)\times(X\setminus B)\\}$ where $B$ varies over the clopen subsets of $X$.
Hence, $E$ is closed and $X/E$ is totally disconnected.

Obviously $R_{E}\cup R_{E}^{-1}=X/E$. By Theorem 5.1(a) if $A$ and $B$ are
distinct components of $X$ then either $A\times B\subset R^{\circ}$ or
$B\times A\subset R^{\circ}$. It follows that $R\cup
E=(\pi\times\pi)^{-1}(R_{E})$. Thus, $(\pi\times\pi)^{-1}(R_{E}\cap
R_{E}^{-1})=E$. It follows that $R_{E}$ is a topological tournament and that
$\pi$ is a tournament map.

We apply Theorem 2.4 to the surjective map $\pi$. Assume for $x\in
X/E,\pi^{-1}(x)$ is a non-trivial component $A$.

It then follows that $A^{\circ}=\pi^{*}(x)$ is an open subset of $X$ which
differs from $A$ by at most two points. Thus, the collection $\\{A^{\circ}\\}$
with $A$ varying over the non-trivial components of $X$ is a pairwise disjoint
collection of nonempty open subsets. If $X$ is metrizable, then it is totally
bounded and so for any $\epsilon>0$ for at most finitely many $A$ is it true
that $diamA^{\circ}\geq\epsilon$. Since $A$ is connected and so has no
isolated points, $diamA^{\circ}=diamA$. The metric result also follows
directly from Theorem 2.6.

If $M$ is the maximum for the compact linear order $R|A$, then $M$ is a
terminal point for $R|A$ and so $R^{\circ}(M)=\pi^{-1}(R_{E}^{\circ}(x))$. If
$x$ is not right balanced, then $R_{E}^{\circ}(x)$, and hence $R^{\circ}(M)$
as well, are clopen sets and so $M$ is not right balanced. With a similar
argument when $x$ is not left balanced, we see that if $R$ is balanced, then
$R_{E}$ is balanced. If $X$ is metrizable, then then $X_{E}$ is metrizable,
see [11] Theorem 5.20. Since $X/E$ is totally disconnected, it is a Cantor set
when it has no isolated points.

∎

For a closed relation $R$ on a compact metric space $X$, the map
$\overrightarrow{R}:X\to 2^{X}$ defined by $x\mapsto R(x)$ is upper
semicontinuous, where $2^{X}$ is the compact space of closed subsets of $X$
equipped with the Hausdorff metric, see, e.g. [1] Proposition 7.11.

######  Theorem 5.3.

If $R$ is a topological tournament on a compact metric space, then the map
$\overrightarrow{R}$ is an embedding, i.e. it is a homeomorphism onto its
image in $2^{X}$. In particular, it is lower semicontinuous as well as upper
semicontinuous.

###### Proof.

If $\\{x_{n}\\}$ is a sequence converging to $x\in X$ and $y\in R^{\circ}(x)$,
then since $R^{\circ}$ is open, eventually $(x_{n},y)\in R^{\circ}$. On the
other hand, if $y=x$, then $(x_{n},x_{n})\in R$. Each sequence converges to
$(x,y)$. It follows from [1] Exercise 7.4 and Proposition 7.11 that the map
$\overrightarrow{R}$ is lower semicontinuous and so is continuous.

If $y\in R^{\circ}(x)$, then $x\in R(x)\setminus R(y)$ by anti-symmetry and so
$R(x)\not=R(y)$. It follows that the map $\overrightarrow{R}$ is injective and
so is a homeomorphism onto its image by compactness.

∎

######  Theorem 5.4.

If $R$ is a topological tournament on a Cantor set $X$, then $R$ is regular if
and only if it is balanced.

###### Proof.

Clearly, if $R$ admits a terminal point or a initial point, then it is neither
regular nor balanced. So we may assume that for every $x\in X$, $R^{\circ}(x)$
and $R^{\circ-1}(x)$ are nonempty open subsets and since $x$ is not isolated,
it is either left or right balanced.

If $x$ is right balanced but not left balanced then $R(x)$ is a Cantor set
while $R^{-1}(x)$ consists of the Cantor set $R^{\circ-1}(x)$ together with an
isolated point $x$. Hence, $R$ is not regular. Similarly, if there exists a
point which is left balanced but not right balanced. It follows that if $R$ is
regular, then it is balanced.

Finally, if $x$ is balanced, then $R(x)$ and $R^{-1}(x)$ are Cantor sets and
so there is a homeomorphism $h_{x}:R(x)\to R^{-1}(x)$ with $h_{x}(x)=x$.
Define $h_{x}$ on $R^{-1}(x)$ to be $h_{x}^{-1}$.

∎

We will see that, in contrast with the finite case, an infinite regular
topological tournament need not be arc cyclic.

For a topological tournament, $R$ on $X$ and $x\in X$ the set $R(x)$ is clopen
if and only if $\overline{R^{\circ-1}(x)}\not=R^{-1}(x)$, i.e. $x$ is not left
balanced.

######  Theorem 5.5.

Let $R$ be a topological tournament on a compact metric space $X$. The set of
points $x$ which are not left balanced, i.e. for which $R(x)$ is clopen, is
countable. Similarly the set of points which are not right balanced is
countable. If $X$ has no isolated points, then the set of balanced points is
residual, i.e. it is a dense $G_{\delta}$ subset of $X$.

###### Proof.

As described in Background 1.10 (3) a compact metric space has only countably
many clopen subsets. By Theorem 5.3 the map $\overrightarrow{R}$ is injective
and so $\\{x:R(x)$ is clopen$\\}$ is countable. The union of this set and the
corresponding set for $R^{-1}$ is countable and so if $X$ has no isolated
points, the complement is a dense $G_{\delta}$ set by the Baire Category
Theorem.

∎

Without metrizability this result may fail.

On the real line $\mathbb{R}$, the linear order
$L_{\mathbb{R}}=\\{(t,t^{\prime}):t\leq t^{\prime}\\}$ is a transitive
topological tournament. Its restriction to $\\{\pm 1\\}(=\\{-1,+1\\})$ is an
arc. By Proposition 3.1 $L=L_{\mathbb{R}}\ltimes(L_{\mathbb{R}}|\\{\pm 1\\})$
is a linear order on $\mathbb{R}\times\\{\pm 1\\}$. When we use the associated
order topology, instead of the product topology, we obtain the _Sorgenfrey
Double Arrow_. For every $t\in\mathbb{R}$ let $t+=(t,+1),t-=(t,-1)$. Each
$L(t+)$ and each $L^{-1}(t-)$ is clopen. It follows that the space is not
metrizable, see Background 1.10 (3). The first coordinate projection to
$\mathbb{R}$ is a continuous surjective tournament mapping from $L$ to
$L_{\mathbb{R}}$. The subset $X=L(0+)\cap L^{-1}(1-)$ is compact with no
isolated points and the first coordinate projection is a continuous surjective
tournament mapping from the restriction of $L|X$ to the restriction of
$L_{\mathbb{R}}|I$ with $I$ the unit interval in $\mathbb{R}$.

The set of left balanced points and the set of right balanced points are
disjoint. Each is dense and the union is all of $X$. In particular, there are
no balanced points.

On the subset $\mathbb{R}\times\\{-1\\}$ the relative topology is not the
order topology. Instead the basis consists of half-open intervals
$(s_{-},t_{-}]$ with $s<t$. The space is non-metrizable and non-compact, but
it is separable and with no isolated points. When we restrict $L$ to this
subset, we obtain a topological tournament such that every $L^{-1}(t_{-})$ is
clopen. That is, there are no right balanced points. Every point is left
balanced.

On the other hand, we do have the following result in the general compact
case.

######  Theorem 5.6.

Let $R$ be a topological tournament on a compact space $X$. If the isolated
points are not dense, then the set of right balanced points is nonempty, in
fact, it is dense in the complement of the closure of the set of isolated
points. Similarly, the set of left balanced points is dense in the complement
of the closure of the set of isolated points.

###### Proof.

Let $U$ be a nonempty open subset of $X$ which contains no isolated points and
let $U_{1}$ be a nonempty open subset with $\overline{U_{1}}\subset U$. We
show that there exists $x\in U$ such that $R(x)=\overline{R^{\circ}(x)}$ or,
equivalently, $R^{-1}(x)$ is not clopen.

We may assume that $G=\\{x\in U_{1}:R^{-1}(x)$ is clopen $\\}$ is dense in
$U_{1}$. For if not, the required $x$ exists in $U_{1}$.

Choose $x_{1}\in G$. Assume we have constructed inductively
$x_{1},x_{2},\dots,x_{n}$ distinct points in $G$ such that for each $i$ with
$1<i\leq n$, $x_{i}\in\bigcap_{j<i}R^{-1}(x_{j})$. Hence,
$U_{1}\cap\bigcap_{j\leq n}R^{-1}(x_{j})$ is an open subset of $U_{1}$ which
contains $x_{n}$. Since $U_{1}$ contains no isolated points, there exists
$x_{n+1}\in G\cap[(U_{1}\cap\bigcap_{j\leq
n}R^{-1}(x_{j}))\setminus\\{x_{1},\dots,x_{n}\\}]$.

Let $x$ be a limit point of the sequence $\\{x_{n}\\}$ so that $x\in
U\supset\overline{U_{1}}$. By excluding one $x_{i}$ if necessary, we may
assume $x\not=x_{n}$ for any $n$. Since $x_{j}\in R^{-1}(x_{n})$ for all
$j>n$, it follows that $x\in R^{-1}(x_{n})$ and so $x_{n}\in R^{\circ}(x)$.
Thus, $x\in\overline{R^{\circ}(x)}$. Thus, $x$ is right balanced.

∎

We conclude this section with a useful tool.

######  Definition 5.7.

Let $(X,R)$ be a topological tournament and let
$F=\\{x_{1},x_{2},\dots,x_{n}\\}$ be a list of distinct points in $X$. A
_thickening_ of $F$ is a list $U_{F}=\\{U_{1},U_{2},\dots,U_{n}\\}$ of open
subsets of $X$ such that

  * •

For $i=1,\dots,n,\ \ x_{i}\in U_{i}$.

  * •

For $i,j=1,\dots,n,$ with $i\not=j$ and $z_{i}\in U_{i},z_{j}\in U_{j}$, we
have $z_{i}\rightharpoonup z_{j}$ if $x_{i}\rightharpoonup x_{j}$.

In particular, the open sets in $U_{F}$ are pairwise disjoint.

We call $U_{F}$ a _clopen thickening_ when every $U_{i}$ is clopen.

If $(X,R)$ is a compact topological tournament and
$F=\\{x_{1},x_{2},\dots,x_{n}\\}$ is any list of distinct points in $X$, then
there exists a thickening for $F$. In fact, the thickening can be chosen
uniformly.

######  Theorem 5.8.

If $(X,R)$ is a compact topological tournament, then for any neighborhood
$V_{1}$ of the diagonal $1_{X}$ there exists a neighborhood $V$ of the
diagonal such that whenever $F$ is a finite subset such that
$(x_{i},x_{j})\not\in V_{1}$ when $i\not=j$, then
$\\{V(x_{1}),\dots,V(x_{n})\\}$ is a thickening of $F$.

When $X$ is metrizable with metric $d$, then for every $\epsilon>0$, there
exists $\delta>0$ such that $d(x_{i},x_{j})\geq\epsilon$ when $i\not=j$
implies that $\\{V_{\delta}(x_{1}),\dots,V_{\delta}(x_{n})\\}$ is a thickening
of $F$.

If $X$ is totally disconnected, then we may choose $V$ to be a clopen
equivalence relation and so obtain a clopen thickening.

###### Proof.

We use induction on $n$. The result for $n=1$ is vacuous. We may use any
neighborhood of $1_{X}$.

Now assume that $V_{2}\subset V_{1}$ is a neighborhood of the diagonal such
that $\\{V_{2}(x_{1}),\dots,V_{2}(x_{n-1})\\}$ is a thickening of
$\\{x_{1},\dots,x_{n-1}\\}$ whenever $(x_{i},x_{j})\not\in V_{1}$ for
$i\not=j\leq n-1$. Note that the set of diagonal neighborhoods $V\subset
V_{2}$ is directed with intersection the diagonal. Suppose there existed
$F=\\{x_{1},\dots,x_{n}\\}$ such that no $V$ exists. Then for any such $V$
because $\\{V(x_{1}),\dots,V(x_{n-1})\\}$ is a thickening of
$\\{x_{1},\dots,x_{n-1}\\}$ there must exist $x_{1}(V),\dots,x_{n}(V)$ such
that $(x_{i},x_{i}(V))\in V$ for $i=1,\dots,n$, but for some $j_{V}<n$,
$x_{j_{V}}(V)\rightharpoonup x_{n}(V)$ while $x_{j(V)}\leftharpoonup x_{n}$ or
vice-versa. Assume the first. By restricting to a cofinal subset we may assume
that for some fixed $j<n$, $j_{V}=j$ for all $V$. Each net $\\{x_{i}(V)\\}$
has limit $x_{i}$ since $(x_{i},x_{i}(V))\in V$. By assumption,
$x_{j}\leftharpoonup x_{n}$. But $x_{j}(V)\rightharpoonup x_{n}(V)$ implies,
in the limit, $x_{j}\ \underline{\rightharpoonup}\ x$ violating anti-symmetry.
The argument for the reverse assumption is similar.

When $X$ is totally disconnected, the clopen equivalence relations form a
neighborhood base for the diagonal and so we may choose $V$ to be such.

∎

##  6\.  Group Tournaments

For a subset $A$ of a group $G$ we let $A^{-1}=\\{x^{-1}:x\in A\\}$. We let
$e$ denote the identity element.

######  Definition 6.1.

For a group $G$, a _game subset_ $A$ for $G$ is a subset such that

(6.1) $A\cap A^{-1}\ =\ \\{e\\},\quad\text{and}\quad A\cup A^{-1}\ =\ G.$

We let $A^{\circ}=A\setminus\\{e\\}$.

If $A$ is a game subset, then $A^{-1}$ is the reverse game subset. If $G$ is a
topological group and $A$ is closed, then $A^{\circ}=X\setminus A^{-1}$ is
open.

Clearly, a group admits a game subset if and only if it has no elements of
order two. In the finite case this says that $G$ has odd order.

If $h:G_{2}\to G_{1}$ is a group homomorphism and $A_{1},A_{2}$ are game
subsets for $G_{1}$ and $G_{2}$, respectively, then $h$ maps $A_{2}$ to
$A_{1}$ when $h(A_{2})\subset A_{1}$ or, equivalently, $A_{2}\subset
h^{-1}(A_{1})$. Since $h$ maps $A_{2}^{-1}$ to $A_{1}^{-1}$ it follows that

(6.2) $A_{2}^{\circ}\ \supset\ h^{-1}(A_{1}^{\circ}).\hskip 56.9055pt$

Thus, the preimage of $A_{1}^{\circ}$ is contained in $A_{2}^{\circ}$ and
$A_{2}$ is contained in the union of the preimage of $A_{1}^{\circ}$ and the
kernel $h^{-1}(e)$ of $h$.

If $H$ is a subgroup of $G$, then $H\cap A$ is a game subset for $H$ and the
inclusion maps $H\cap A$ to $A$.

If $A$ is a game subset for a group $G$ then the associated tournament
$\widehat{A}$ is defined by

(6.3) $\displaystyle\begin{split}\widehat{A}\ =\ \\{(x,y):x^{-1}y&\in
A\\}\quad\text{so that}\ \ \widehat{A}^{-1}\ =\ \widehat{A^{-1}},\\\ \text{and
so}\ \ \widehat{A}^{\circ}\ &=\ \\{(x,y):x^{-1}y\in A^{\circ}\\}.\end{split}$

Thus, $A=\widehat{A}(e)$ and $A^{\circ}=\widehat{A}^{\circ}(e)$. If
$h:G_{2}\to G_{1}$ is a group homomorphism, then $h$ maps the game subset
$A_{2}$ to the game subset $A_{1}$ if and only if it is a tournament map from
$\widehat{A_{2}}$ to $\widehat{A_{1}}$.

If $G$ is a topological group, then $A$ is a closed game subset if and only if
$\widehat{A}$ is a topological tournament.

For the results on the finite case of group games, see, e.g. [2].

The tournament $\widehat{A}$ is regular. Define for $x,y\in G$

(6.4) $\displaystyle\begin{split}h_{x}(y)\ =\ &xy^{-1}x\quad\text{so that}\ \
h_{x}(x)\ =\ x,\\\ \text{and}\quad&x^{-1}h_{x}(y)\ =\ y^{-1}x,\\\
\text{and}\quad&h_{x}\circ h_{x}=1_{G}.\end{split}$

Thus, $(x,y)\mapsto(x,xy^{-1}x)$ maps $\widehat{A}$ to $\widehat{A}^{-1}$ and
so $h_{x}(\widehat{A}(x))=\widehat{A}^{-1}(x)$.

The tournament $\widehat{A}$ on the group $G$ is _homogeneous_. For $x\in G$,
the left translation map $\ell_{x}$, defined by $\ell_{x}(y)=xy$, is an
automorphism of $\widehat{A}$. That is, $\ell_{x}$ is a bijection on $G$
mapping $\widehat{A}$ to itself.

For a topological group, the maps $h_{x}$ and $\ell_{x}$ are homeomorphisms.

######  Proposition 6.2.

Let $h:G_{2}\to G_{1}$ be a group homomorphism and $A_{1}$ be a game subset
for $G_{1}$. Then $A_{2}\subset G_{2}$ is a game subset for $G_{2}$ which is
mapped to $A_{1}$ by $h$ if and only if $A_{2}$ is the union of the disjoint
sets $h^{-1}(A_{1}^{\circ})$ and $B$ with $B$ a game subset for the kernel of
$h$, $H\ =\ h^{-1}(e)$.

If, in addition, $h$ is surjective, then there exists a retraction $p:G_{2}\to
H$ such that the product map $h\times p:G_{2}\to G_{1}\times H$ is a bijection
mapping $\widehat{A_{2}}$ isomorphically onto the lexicographic product
$\widehat{A_{1}}\ltimes\widehat{B}$.

If $h$ is a continuous group homomorphism between topological groups with non-
trivial kernel and $A_{1}$ is closed, then $A_{2}=B\cup h^{-1}(A_{1}^{\circ})$
is closed if and only if $B$ is closed and, in addition, the kernel $H$ is a
clopen subgroup.

###### Proof.

It is easy to check that if $B$ is a game subset for the kernel of $h$, then
$A_{2}=h^{-1}(A_{1}^{\circ})\cup B$ satisfies the conditions of (6.1) and is
mapped by $h$ to $A_{1}$.

Conversely, if $A_{2}$ is a game subset for $G_{2}$, then $B=A_{2}\cap
h^{-1}(e)$ is a game subset for the kernel and if $h$ maps $A_{2}$ to $A_{1}$,
then (6.2) implies that $A_{2}$ contains the game subset
$h^{-1}(A_{1}^{\circ})\cup B$. Clearly, if one game subset for $G_{2}$
includes another such, then the two are equal.

If $h$ is surjective, we can define a (not necessarily continuous) map
$j:G_{1}\to G_{2}$ such that $h\circ j=1_{G_{1}}$ with $j(e_{1})=e_{2}$.
Define $p(x)=j(h(x))^{-1}x$ so that $p$ maps $G_{2}$ into $H$ with $p=1_{H}$
on $H$. Since $j(h(x))p(x)=x$, the inverse map to $h\times p$ is given by
$(z,b)\mapsto j(z)b$. So $h\times p:G_{2}\to G_{1}\times B$ is a bijection.

If $h(x)\not=h(y)$, then $x^{-1}y\in A_{2}$ if and only if
$h(x^{-1}y)=h(x)^{-1}h(y)\in A_{1}$, i.e. $(x,y)\in\widehat{A_{2}}^{\circ}$ if
and only if $(h(x),h(y))\in\widehat{A_{1}}^{\circ}$.

If $h(x)=h(y)$, then $j(h(x))=j(h(y))$ and so $x^{-1}y=p(x)^{-1}p(y)$. Hence,
$(x,y)\in\widehat{A_{2}}$ if and only if $(p(x),p(y))\in\widehat{B}$.

It follows that $h\times p$ maps $\widehat{A_{2}}$ isomorphically onto
$\widehat{A_{1}}\ltimes\widehat{B}$.

In the topological case, the kernel $h^{-1}(e)$ is a closed subgroup because
of our standing assumption that all spaces are Hausdorff. If it is not open
then there exists a net $\\{a_{k}\\}$ in $G_{2}\setminus h^{-1}(e)$ which
converges to a point $x$ in the kernel. Replacing $a_{k}$ by $a_{k}^{-1}$ if
necessary and by going to a subnet we may assume that $a_{k}\in
h^{-1}(A_{1}^{\circ})$ for all $k$. If $x,y$ lie in the kernel with $x$ the
limit point of the net, then $\\{yx^{-1}a_{i}\\}$ is a net in
$h^{-1}(A_{1}^{\circ})$ which converges to $y$. Thus, all of $h^{-1}(e)$ is
contained in the closure of $h^{-1}(A_{1}^{\circ})$ which is contained in
$A_{2}$ when the latter is closed. If the kernel is non-trivial, then
$B^{\circ-1}$ is a nonempty subset of the kernel which is disjoint from
$A_{2}$. The contradiction shows that the kernel must be clopen.

∎

######  Proposition 6.3.

For $x\in G$ with $G$ a compact topological group, the set

(6.5) $\omega(x)\ =\ \bigcap_{n\in\mathbb{N}}\ \overline{\\{x^{i}:i\geq n\\}}$

is a nonempty closed subgroup of $G$.

###### Proof.

Since $\omega(x)$ is the intersection of a decreasing sequence of non-empty
compacta, it is nonempty and compact. It consists of the set of limit points
of the sequence $\\{x^{i}:i\in\mathbb{N}\\}$. So if $z\in\omega(x)$, then
$x^{-1}z$ is also a limit point of the sequence and so lies in $\omega(x)$.
Thus, $\\{y:y^{-1}z\in\omega(x)\\}$ is closed and contains $x^{i}$ for all
$i\in\mathbb{N}$. In particular, it contains $\omega(x)$. That is,
$\omega(x)^{-1}\omega(x)\subset\omega(x)$ and so $\omega(x)$ is a subgroup.

∎

######  Theorem 6.4.

If $A$ is a closed game subset on a compact topological group $G$, then
$(G,\widehat{A})$ is an arc cyclic tournament.

###### Proof.

By homogeneity it suffices to consider arcs with $x=e$ and so $y\in
A^{\circ}$. The arc $(e,y)$ is contained in a $3-$cycle if and only if $yA$
meets $A^{\circ-1}$. Assume now that $y\in A$ with $yA$ is disjoint from
$A^{\circ-1}$ and so $yA\subset A$. Inductively, for all $i\in\mathbb{N},\
y^{i}\in y^{i}A\subset y^{i-1}A$. In particular, the sequence $\\{y^{i}\\}$ is
contained in $yA$ and so $\omega(y)$ is contained in the closed set $yA$.
However, Proposition 6.3 implies that $\omega(y)$ is a subgroup and this
yields $e\in yA$ or, equivalently, $y\in A^{-1}$. Since $y\in A\cap A^{-1}$ we
have $y=e$. Thus, if $y\in A^{\circ}$ it must happen that $yA$ meets
$(A^{-1})^{\circ}$.

∎

The following is a topological version of the proof of [12] Theorem 3, which
in turn is an extension of [9] Theorem 7.

######  Corollary 6.5.

If $A$ is a closed game subset on an infinite compact topological group $G$,
then for every $n\geq 3$ each point of $G$ is contained in an $n-$cycle.

###### Proof.

The result for $n=3$ follows from Theorem 6.4. Now assume that
$C=\\{x_{1},\dots,x_{n}\\}$ is an $n-$cycle with $n\geq 3$. We may assume, by
multiplying by $x_{1}^{-1}$ if necessary, that $x_{1}=e$. We will construct an
$n+1-$cycle through $x_{1}$.

Case 1 Assume there exists $x\in G\setminus C$ such that $\widehat{A}(x)$ and
$\widehat{A}^{-1}(x)$ both meet $C$. By renumbering we may assume
$x_{1}\rightharpoonup x$. Let $k=\max\\{i:x_{j}\rightharpoonup x$ for all
$j\leq i\\}$. By assumption, $k<n$ and by definition $x\rightharpoonup
x_{k+1}$. Hence, $\\{x_{1},\dots,x_{k},x,x_{k+1},\dots,x_{n}\\}$ is an
$n+1-$cycle which contains all the points of $C$ and so includes the point
previously labelled $x_{1}$.

Case 2: Assume instead that with $Z_{+}=\\{x:C\subset\widehat{A}(x)\\}$ and
$Z_{-}=\\{x:C\subset\widehat{A}^{-1}(x)\\}$ we have $Z_{+}\cup
Z_{-}=G\setminus C$. Notice that in any case $Z_{+}\cup Z_{-}$ is disjoint
from $C$ since the points of $C$ lie on a cycle.

If $Z_{-}$ were empty, then for every point $x_{i}\in C$, we would have
$\widehat{A}(x_{i})=x_{i}A\subset C$. Thus would imply that $A$ is finite and
so $G=A\cup A^{-1}$ is finite. Similarly, $Z_{+}$ is nonempty.

Choose $z_{1}\in Z_{+},z_{2}\in Z_{-}$. We may assume that
$z_{2}\rightharpoonup z_{1}$. If instead $z_{1}\rightharpoonup z_{2}$, then
Corollary 6.4 there exits $z_{3}\in G$ such that $\\{z_{1},z_{2},z_{3}\\}$ is
a $3-$cycle. Because $z_{2}\rightharpoonup z_{3}$ it cannot happen that
$z_{3}\in C$. If $z_{3}\in Z_{+}$ then replace $z_{1}$ by $z_{3}$. If
$z_{3}\in Z_{-}$, then replace $z_{2}$ by $z_{3}$.

Assuming that $z_{2}\rightharpoonup z_{1}$ we obtain
$\\{x_{1},z_{2},z_{1},x_{3},\dots x_{n}\\}$ (omitting $x_{2}$) an $n+1-$cycle
containing $x_{1}$.

∎

######  Theorem 6.6.

If $A$ is a closed game subset on an infinite compact topological group $G$,
then $\widehat{A}$ is a balanced topological tournament.

###### Proof.

By homogeneity it suffices to show that $e$ is a balanced point. If it were
not then either $A^{\circ}$ or $A^{\circ-1}$ would be clopen and so both would
be clopen since the map $x\mapsto x^{-1}$ is a homeomorphism. In that case $e$
is an isolated point. By homogeneity all the points of $X$ would be isolated
and so, by compactness, $X$ would be finite.

∎

Now let $\\{G_{i}:i\in\mathbb{N}\\}$ be a sequence of finite groups of odd
order with $f_{i}:G_{i+1}\to G_{i}$ surjective group homomorphisms each with
non-trivial kernel $H_{i+1}$ so that the sequence of orders $\\{|G_{i}|\\}$ is
strictly increasing. Let $H_{1}=G_{1}$. Choose $A_{1}=B_{1}$ a game subset for
$G_{1}=H_{1}$ and $B_{i+1}$ a game subset for the kernel $H_{i+1}$.
Inductively, let $A_{i+1}\ =\ B_{i+1}\cup(f_{i})^{-1}(A_{i}^{\circ})$ which is
a game subset for $G_{i+1}$ mapped onto $A_{i}$ by $f_{i}$.

######  Theorem 6.7.

The sequence $\\{(G_{i},\widehat{A_{i}},f_{i}):i\in\mathbb{N}\\}$ is a
surjective inverse system of topological tournaments with limit $(G,R)$ a
compact, topological tournament and $G$ a Cantor set.

The space $G$ is a closed subgroup of the product topological group
$\prod_{i\in\mathbb{N}}\ G_{i}$ with closed game subset

(6.6) $A\ =\ \bigcap_{i\in\mathbb{N}}\ \pi_{i}^{-1}(A_{i})\ =\
\overleftarrow{Lim}\\{A_{i}\\}$

such that $R=\widehat{A}$.

###### Proof.

That $R$ is a topological tournament on the inverse limit $G$ follows from
Theorem 4.6.

It is clear that $G$ is a closed subgroup of the product group. It is easy to
check that the closed subset $A$ is a game subset for $G$ and that
$R=\widehat{A}$.

∎

######  Example 2.

The $3-$adic integers.

Consider the $3-$adic integers, with $\mathbb{Z}/3^{i}\mathbb{Z}$ and the
projection $f_{i}$ reduction mod $3^{i}$. The kernel of each $f_{i}$ is
isomorphic to $\mathbb{Z}/3\mathbb{Z}=\\{0,1,2\\}$. We can identify
$\mathbb{Z}/3^{i}\mathbb{Z}$ with the product $\\{0,1,2\\}^{\\{1,\dots,i\\}}$
with addition of two sequences pointwise (mod 3) but with carrying to the
right. The projection
$f_{i}:\mathbb{Z}/3^{i+1}\mathbb{Z}\to\mathbb{Z}/3^{i}\mathbb{Z}$ is a
surjective group homomorphism. So $\\{(\mathbb{Z}/3^{i}\mathbb{Z},f_{i})\\}$
is an inverse system of finite groups. As an additive topological group, the
inverse limit is identified with $\\{0.1,2\\}^{\mathbb{N}}$ with addition of
two sequences pointwise (mod 3) but with carrying to the right. We label this,
the group of $3-$adic integers by $\mathbb{Z}[3]$.

The identity element $e$ has $e_{i}=0$ for all $i$.

An example of a closed game subset $A$, let $A^{\circ}=\\{y\in G\setminus
e:y_{j}=1$ for $j=\min\\{k:y_{k}\not=0\\}\\}$. For each
$\mathbb{Z}/3^{i}\mathbb{Z}$ we let
$A_{i}^{\circ}=\\{y\in\mathbb{Z}/3^{i}\mathbb{Z}\setminus e:y_{j}=1$ for
$j=\min\\{k:y_{k}\not=0\\}\\}$.

Equipped with this game subset we will refer to the tournament
$(\mathbb{Z}[3],\widehat{A})$ as the _standard $3-$adic example_ . It is the
inverse limit of the system
$\\{(\mathbb{Z}/3^{i}\mathbb{Z},\widehat{A_{i}},f_{i})\\}$.

######  Lemma 6.8.

If $w$ is a homeomorphism on a Cantor set $X$ which induces a free
$\mathbb{Z}/2\mathbb{Z}$ action, i.e. $w\circ w=1_{X}$ and $w(x)\not=x$ for
all $x\in X$, then there exists a clopen subset $A$ of $X$ such that $X$ is
the disjoint union of $A$ and $w(A)$.

###### Proof.

We may choose a $w$ invariant ultra-metric $u$ on $X$, see Background 1.10
(9).

Because $u$ is an ultra-metric, the relation
$V_{\epsilon}=\\{(x,y):u(x,y)<\epsilon\\}$ is a clopen equivalence relation
for every $\epsilon>0$. Because $u$ is $w$ invariant, we have
$h(V_{\epsilon}(x))=V_{\epsilon}(w(x))$.

Choose $\epsilon$ so that $0<\epsilon<\min_{x\in X}\ u(x,w(x))$. The
equivalence classes $\\{V_{\epsilon}(x):x\in X\\}$ form a finite cover of $X$
by clopen sets. By choice of $\epsilon,\ V_{\epsilon}(w(x))$ is disjoint from
$V_{\epsilon}(x)$. So we can partition the cover by the collection of pairs
$\\{\\{V_{\epsilon}(x),V_{\epsilon}((w(x))\\}\\}$. Choose one member from each
pair and take the union to define $A$. Observe that there are $2^{n}$ choices
leading to distinct sets $A$ with $2n=|\\{V_{\epsilon}(x)\\}|$. By shrinking
$\epsilon$ we can increase the number of alternative sets $A$.

∎

######  Theorem 6.9.

Let $G$ be a topological group with the underlying space a Cantor set. There
exists a closed game subset $A$ for $G$ if and only if $G$ contains no
elements of order $2$.

###### Proof.

Clearly if $G$ contains an element of order $2$, then there is no game subset.
Now assume there are no such elements so that $w(x)=x^{-1}$ defines a
homeomorphism of $X$ which induces a free $\mathbb{Z}/2\mathbb{Z}$ action
except at the point $e$ where $w(e)=e$.

Choose $\\{U_{i}:i\in\mathbb{N}\\}$ a decreasing sequence of clopen
neighborhoods of $e$ with intersection $e$. For example, with $u$ the
ultrametric of the previous proof we may use $U_{i}=V_{1/i}(e)$. Replacing
$U_{i}$ by $U_{i}\cap U_{i}^{-1}$ for all $i$, we may assume that
$(U_{i})^{-1}=U_{i}$ for all $i$. Let $U_{0}=G$. By renumbering we may assume
that the sequence $\\{U_{0},U_{1},\dots\\}$ is strictly decreasing so that
$\\{X_{i}=U_{i-1}\setminus U_{i}:i\in\mathbb{N}\\}$ is a sequence of nonempty
clopen subsets which partition $G\setminus\\{e\\}$ and each of which is $w$
invariant.

For each $i$ use Lemma 6.8 to choose $A_{i}$ clopen in $X_{i}$ with
$\\{A_{i},w(A_{i})\\}$ a partition of $X_{i}$. Let $A^{\circ}\ =\ \bigcup_{i}\
A_{i}$. This is an open subset of $X$ with
$A=\overline{A^{\circ}}=A^{\circ}\cup\\{e\\}$. Thus, $A$ is a closed game
subset for $X$.

∎

######  Example 3.

The $2-$adic integers.

Consider the $2-$adic integers, with $\mathbb{Z}/2^{i}\mathbb{Z}$ and the
projection $f_{i}$ reduction mod $2^{i}$. The kernel of each $f_{i}$ is
isomorphic to $\mathbb{Z}/2\mathbb{Z}=\\{0,1\\}$. We can identify
$\mathbb{Z}/2^{i}\mathbb{Z}$ with the product $\\{0,1\\}^{\\{1,\dots,i\\}}$
with addition of two sequences pointwise (mod 2) but with carrying to the
right. The projection
$f_{i}:\mathbb{Z}/2^{i+1}\mathbb{Z}\to\mathbb{Z}/2^{i}\mathbb{Z}$ is a
surjective group homomorphism. So $\\{(\mathbb{Z}/2^{i}\mathbb{Z},f_{i})\\}$
is an inverse system of finite groups. As an additive topological group, the
inverse limit is identified with $\\{0.1\\}^{\mathbb{N}}$ with addition of two
sequences pointwise (mod 2) but with carrying to the right. We label this, the
group of $2-$adic integers by $\mathbb{Z}[2]$. Note that since
$\mathbb{Z}/2^{i}\mathbb{Z}$ has even order it does not admit a game subset.

With $\bar{0}=1,\bar{1}=0$ we define $\bar{y}$ for $y$ in $\mathbb{Z}[2]$ by
$(\bar{y})_{i}=\overline{y_{i}}$. With ${\mathbf{1}}=1000\dots$, it is clear
that $y+\bar{y}+{\mathbf{1}}={\mathbf{0}}$ where ${\mathbf{0}}=0000\dots$ is
the zero element of the additive group. So if $y=0^{i-1}1z$, then
$-y=0^{i-1}1\bar{z}$. Define $A_{i}=\\{0^{i-1}10z:z\in\mathbb{Z}[2]\\}$. This
is a clopen subset with $-A_{i}=\\{0^{i-1}11\bar{z}:z\in Y\\}$. So
$A=\\{e\\}\cup(\bigcup_{i}A_{i})$ is a game subset.

Equipped with this game subset we will refer to the tournament
$(\mathbb{Z}[2],\widehat{A})$ as the _standard $2-$adic example_.

The map $m_{k}$, multiplication by $k$ on $\mathbb{Z}[2]$, for any
$k\in\mathbb{N}$ odd, is an automorphism of the additive group $Y$.

It follows that if $h:G\to H$ is a surjective group homomorphism with $H$ a
finite group, then the order of $H$ is a power of $2$. For if not, since it is
necessarily abelian, it has a quotient group of odd order and so we may assume
that $H$ has odd order $k$. If $x\not=0$ in $H$, then there exists $y_{1}\in
G$ with $h(y_{1})=x$ and since $m_{k}$ is an automorphism of $G$ there exists
$y_{2}\in Y$ with $ky_{2}=y_{1}$. then $kh(y_{2})=h(y_{1})=x$. On the other
hand, $kz=0$ for all $z\in H$.

While every topological group on a Cantor set is an inverse limit of a
sequence of finite quotient groups, the $2-$adics provides an example where no
game subset can be obtained as a limit of game subsets from a sequence of
quotient groups.

##  7\.  Cycle Points

A tournament $R$ is transitive, and so is a linear order, if and only if
contains no $3-$cycle.

######  Definition 7.1.

Let $(X,R)$ be a topological tournament.

We say that $R$ is _nowhere locally transitive_ when there does not exist a
nonempty open subset $U$ of $X$ such that the restriction $R|U$ is transitive,
or, equivalently, when every nonempty open subset of $X$ contains a $3-$cycle.

We call $x\in X$ a _cycle point_ when every open set containing $x$ contains a
$3-$cycle which includes $x$.

Clearly a cycle point is balanced.

######  Lemma 7.2.

Let $(X,R)$ be a compact topological tournament.

If $V_{1}$ is a neighborhood of the diagonal $1_{X}$, then there exists a
neighborhood of the diagonal $V$ such that if $\\{x,y,z\\}$ is a $3-$cycle
with $(x,y)\in V$, then $(x,z),(y,z)\in V_{1}$. If $X$ is metrizable with
metric $d$ and $\epsilon>0$ there exists $\delta>0$ such that if $\\{x,y,z\\}$
is a $3-$cycle with $d(x,y)<\delta$, then $d(x,z)<\epsilon$ and
$d(y,z)<\epsilon$

Assume that $(x_{k},y_{k},z_{k})$ is a net in $X\times X\times X$ such that
for each $k$, $\\{x_{k},y_{k},z_{k}\\}$ is a $3-$cycle. If $\\{y_{k}\\}$ and
$\\{x_{k}\\}$ both converge to a point $x$, then $\\{z_{k}\\}$ converges to
$x$ as well.

###### Proof.

Suppose instead that for some $V_{1}>0$, we could construct for each $V$ a
$3-$cycles $\\{x_{V},y_{V},z_{V}\\}$ with $(x_{V},y_{V})\in V$ but with
$(x_{V},z_{V})\not\in V_{1}$. The collection of neighborhoods $V$ is directed
by inclusion with intersection the diagonal $1_{X}$. So we can regard
$\\{(x_{V},y_{V},z_{V})\\}$ as a net indexed by $V$. A limit point $(x,y,z)$
would satisfy $x=y$ but $x\not=z$. Since $(y,z),(z,x)\in R$, this would
violate anti-symmetry. In the metric case, the neighborhoods
$V_{\epsilon}=\\{d(x,y)<\epsilon\\}$ generate the neighborhoods of the
diagonal.

For the net $\\{(x_{k},y_{k},z_{k})\\}$, eventually $\\{(x_{k},y_{k})\\}$
enters $V$ and so eventually $\\{(x_{k},z_{k})\\}$ enters $V_{1}$. Since
$\\{x_{k}\\}$ converges to $x$, $\\{z_{k}\\}$ does as well.

∎

######  Theorem 7.3.

Let $(X,R)$ be a compact topological tournament. If $x$ is a non-isolated
point of $X$ and it has an arc cyclic neighborhood, then it is a cycle point
and so is balanced. So if $(X,R)$ is locally arc cyclic and $X$ has no
isolated points, then every point is a cycle point and $(X,R)$ is balanced.

###### Proof.

If $x$ is non-isolated, then it is either left or right balanced. If $x\in X$
is right balanced, there exists a net $\\{y_{k}\\}$ in $R^{\circ}(x)$ which
converges to $x$ and we may assume the net lies in an arc cyclic neighborhood
$U$. Because $U$ is an arc cyclic subset, we can choose for each $k$, a point
$z_{k}\in U$ such that $\\{x,y_{k},z_{k}\\}$ is a $3-$cycle. By Lemma 7.2,
$\\{z_{k}\\}$ converges to $x$. So if $U_{1}$ is any neighborhood of $x$,
eventually, the cycle $\\{x,y_{k},z_{k}\\}$ is contained in $U_{1}$. Thus, $x$
is a cycle point. Similarly, if $x$ is left balanced, it is a cycle point.
Since a cycle point is balanced, it follows that $(X,R)$ is balanced when it
is locally arc cyclic and there are no isolated points.

∎

######  Corollary 7.4.

If $A$ is a closed game subset for an infinite compact group $X$, then every
point of $X$ is a cycle point.

###### Proof.

Immediate from Theorem 6.4, Theorem 6.6 and Theorem 7.3.

∎

We have the following sharpening of Theorem 4.5.

######  Theorem 7.5.

Assume that $\\{(X_{i},R_{i},f_{i})\\}$ is a lexicographic inverse system of
topological tournaments with limit tournament $(X,R)$. If for infinitely many
$i\in\mathbb{N}$ the fiber $(Y_{iz},S_{iz})$ is ip cyclic for each $z$ an
isolated point of $X_{i}$, then every point $x$ of the subset $IS$ of $X$ is a
cycle point.

In particular, if $X_{1}$ is finite, and every $(Y_{iz},S_{iz})$ is finite and
point cyclic, then every point of $X=IS$ is a cycle point.

###### Proof.

Let $x\in IS$. Fix $i$ arbitrarily large so that the fibers $(Y_{iz},S_{iz})$
are ip cyclic for the isolated points $z$ of $X_{i}$, and let $z=\pi_{i}(x)$.
The point $\pi_{i+1}(x)=(z,y)$ with $y$ an isolated point in $Y_{iz}$. Since
$(Y_{iz},S_{iz})$ is ip cyclic, there exist $y^{\prime},y^{\prime\prime}\in
Y_{iz}$ so that $\\{y^{\prime},y,y^{\prime\prime}\\}$ is a $3-$cycle for
$(Y_{iz},S_{iz})$. There exist $x^{\prime},x^{\prime\prime}\in X$ with
$\pi_{i+1}(x^{\prime})=(z,y^{\prime}),\pi_{i+1}(x^{\prime\prime})=(z,y^{\prime\prime})$
and so $\\{x^{\prime},x,x^{\prime\prime}\\}$ is a $3-$cycle for $(X,R)$.
Furthermore, $\pi_{i}(x^{\prime\prime})=\pi_{i}(x^{\prime})=z=\pi_{i}(x)$ and
so $\pi_{j}(x)=\pi_{j}(x^{\prime})=\pi_{j}(x^{\prime\prime})$ for all $j\leq
i$. As $i$ was arbitrarily large, $x^{\prime}$ and $x^{\prime\prime}$ are
arbitrarily close to $x$ and so $x$ is a cycle point.

∎

Recall that if $\\{(X_{i},R_{i},f_{i})\\}$ is a lexicographic inverse system
of bricks, then $IS$ is a dense $G_{\delta}$ subset of $X$.

######  Theorem 7.6.

Let $h:(X_{2},R_{2})\to(X_{1},R_{1})$ be a continuous, surjective tournament
map of compact tournaments. If $y\in X_{1}$ is a cycle point, then $h^{-1}(y)$
is a singleton $\\{x\\}$ and $x\in X_{2}$ is a cycle point.

If every point of $X_{1}$ is a cycle point, then $h$ is a homeomorphism
mapping $R_{2}$ isomorphically onto $R_{1}$.

###### Proof.

Assume that $\\{(y^{\prime}_{k},y^{\prime\prime}_{k})\in X_{1}\times X_{1}\\}$
is a net converging to $(y,y)$ with
$\\{y^{\prime}_{k},y,y^{\prime\prime}_{k}\\}$ a $3-$cycle for all $k$. Since
$h$ is surjective, we can choose $(x^{\prime}_{k},x^{\prime\prime}_{k})\in
X_{2}\times X_{2}$ with
$h(x^{\prime}_{k})=y^{\prime}_{k},h(x^{\prime\prime}_{k})=y^{\prime\prime}_{k}$.
Since $h$ is a tournament map, $x^{\prime\prime}_{k}\rightharpoonup
x^{\prime}_{k}$. By Theorem 2.4 and Addendum 2.5 $h^{-1}(y)$ has a terminal
point $M$ and a initial point $m$ and every convergent subnet of
$\\{x^{\prime}_{k}\\}$ converges to $m$ and so, by compactness,
$\\{x^{\prime}_{k}\\}$ converges to $m$. Similarly,
$\\{x^{\prime\prime}_{k}\\}$ converges to $M$. Since
$x^{\prime\prime}_{k}\rightharpoonup x^{\prime}_{k}$ it follows that $M\
\underline{\rightharpoonup}\ m$. But $m$ is a initial point for $h^{-1}(x)$
and so $m\ \underline{\rightharpoonup}\ M$. It follows from anti-symmetry that
$m=M$ and so $h^{-1}(y)$ is a singleton.

If $h^{-1}(y)=\\{x\\}$ and $U$ is an open set containing $x$, then, by
compactness, there exists an open set $U_{1}$ containing $y$ with
$h^{-1}(U_{1})\subset U$. Any $3-$cyle containing $y$ in $U_{1}$ lifts to a
$3-$cycle in $U$ containing $x$. Hence, $x$ is a cycle point.

If every point of $X$ is a cycle point, then $h$ is a bijection and so is a
homeomorphism by compactness.

∎

######  Theorem 7.7.

Let $(X,R)$ be a compact topological tournament. If $x$ is a cycle point, then
the singleton $\\{x\\}$ is a $G_{\delta}$ set which is a component of $X$.

###### Proof.

Let $\pi:X\to X_{E}$ be the quotient map of Corollary 5.2. From Theorem 5.1 it
follows that a $3-$cycle in $X$ can meet a component in at most one point.
Hence, if $x$ is a cycle point in $X$, then $\pi(x)$ is a cycle point in
$X/E$. From Theorem 7.6 it follows that $\pi^{-1}(\\{\pi(x)\\})$ is the
singleton $\\{x\\}$ and so $\\{x\\}$ is a component.

Now assume that $x$ is a cycle point. There exists a $3-$cycle
$\\{a_{1},x,b_{1}\\}$ in $X$. $R^{\circ}(a_{1})\cap R^{\circ-1}(b_{1})$ is an
open set which contains $x$. Let $U_{1}$ be an open set with $x\in U_{1}$ and
with closure contained in $R^{\circ}(a_{1})\cap R^{\circ-1}(b_{1})$. Thus, for
every $z\in\overline{U_{1}},\ \\{a_{1},z,b_{1}\\}$ is a $3$-cycle,
Inductively, we define points
$\\{a_{1},\dots,a_{n}\\},\\{b_{1},\dots,b_{n}\\}$ and open sets
$\\{U_{1},\dots,U_{n}\\}$ such that for $i=2,\dots n,$

(7.1)
$\displaystyle\begin{split}a_{i},b_{i}\in&U_{i-1},\quad\overline{U_{i}}\subset
U_{i-1},\\\ \\{a_{i},z,b_{i}\\}\ \text{ is a }\ &3-\text{cycle, for all}\
z\in\overline{U_{i}}.\end{split}$

Then choose $\\{a_{n+1},x,b_{n+1}\\}$ a $3-$cycle in $U_{n}$ and thicken $x$
to an open set $U_{n+1}$ with closure contained in $U_{n}\cap
R^{\circ}(a_{n+1})\cap R^{\circ-1}(b_{n+1})$.

Let $(a,b)$ be a limit point of the sequence $\\{(a_{n},b_{n})\\}$ in $X\times
X$ and let $K=\bigcap_{n}U_{n}=\bigcap_{n}\overline{U_{n}}$. Since
$a_{i},b_{i}\in U_{n}$ for all $i>n$ it follows that $a,b\in K$. For all $z\in
K\subset U_{n},\ \\{a_{n},z,b_{n}\\}$ is a $3-$cycle. So in the limit
$(a,z),(z,b),(b,a)\in R$ for all $z\in K$. In particular, since $a,b\in K,\
(a,b),(b,a)\in R$ and so $a=b$ by anti-symmetry. Similarly, $(a,z),(z,b)\in R$
and $a=b$ implies $a=b=z$ for all $z\in K$. That is, $K$ is a singleton. Since
$x\in K,\ K=\\{x\\}$. Thus, $\\{x\\}$ is a $G_{\delta}$ set.

∎

For a compact space, a point is a $G_{\delta}$ point if and only if it has a
countable neighborhood base.

######  Theorem 7.8.

Assume $(X,R)$ is a compact topological tournament. If $R$ is nowhere locally
transitive, then $X$ is a totally disconnected space with no isolated points
and every nonempty open set contains a compact subset $K$ such that $R|K$ is
isomorphic to the standard $3-$adic example. Every point of $K$ is a cycle
point and so is $G_{\delta}$ point.

If, in addition, $x\in X$ is a cycle point, then for every open set $U$ with
$x\in U$, the compact set $K$ can be chosen with $x\in K$.

Conversely, if the cycle points for $R$ are dense in $X$, then $R$ is nowhere
locally transitive.

###### Proof.

Any non-trivial component of $X$ contains a nonempty open subset of $X$ by
Corollary 5.2 and by Theorem 5.1 the restriction of $R$ to this open set is
transitive. Hence, for a nowhere locally transitive tournament every component
is trivial. If $x$ were an isolated point, then $\\{x\\}$ would be an open
subset on which $R$ is trivially transitive. Hence, $X$ has no isolated
points.

Let $U$ be a nonempty open subset.

Because $R|U$ is not transitive and $R^{\circ}$ is open, we can choose a
$3-$cycle in $U$ and thicken it, using to get disjoint, nonempty, clopen sets
$K_{1}^{\epsilon}\subset U$ for $\epsilon=0,1,2$ so that $x\mapsto\epsilon$
for $x\in K_{1}^{\epsilon}$ defines a function $h_{1}$ from
$K_{1}=\bigcup_{\epsilon=0,1,2}K_{1}^{\epsilon}$ to
$\mathbb{Z}/3\mathbb{Z}=\\{0,1,2\\}$ which maps $R|K_{1}$ to $\widehat{A_{1}}$
on $\mathbb{Z}/3\mathbb{Z}$.

Assume that, inductively, we have defined $K_{i}$ a disjoint union of nonempty
clopen subsets $K_{i}^{y}$ for
$y\in\mathbb{Z}/3^{i}\mathbb{Z}=\\{0,1,2\\}^{\\{1,\dots,i\\}}$ so that
$x\mapsto y$ for $x\in K_{i}^{y}$ defines a function
$h_{i}:K_{i}\to\mathbb{Z}/3^{i}\mathbb{Z}$ mapping $R|K_{i}$ to
$\widehat{A_{i}}$ and for $y=z\epsilon$ with
$z\in\mathbb{Z}/3^{i-1}\mathbb{Z}$ and $\epsilon=0,1,2$ $K_{i}^{y}\subset
K_{i-1}^{z}$.

For the inductive step, for each $y\in Y_{i}$ choose a $3-$cycle in
$K_{i}^{y}$ and thicken it, using Theorem 5.8, to obtain disjoint nonempty
clopen subsets $K_{i+1}^{y\epsilon}\subset K_{i}^{y}$ for $\epsilon\in Y_{1}$
such that $x\mapsto\epsilon$ defines a function from
$\bigcup_{\epsilon=0,1,2}\ K_{i+1}^{y\epsilon}$ to $\mathbb{Z}/3\mathbb{Z}$
which maps $R|\bigcup_{\epsilon=0,1,2}\ K_{i+1}^{y\epsilon}$ to
$\widehat{A_{1}}$. With $K_{i+1}=\bigcup_{y\epsilon\in Y_{i+1}}\
K_{i+1}^{y\epsilon}$ $x\mapsto y\epsilon$ for $x\in K_{i+1}^{y\epsilon}$
defines the required function from $K_{i+1}$ to $\mathbb{Z}/3^{i+1}\mathbb{Z}$
taking $R|K_{i+1}$ to $\widehat{A_{i+1}}$.

Let $K$ be the intersection $\bigcap_{i\in\mathbb{N}}\ K_{i}\subset U$.

If $x\in U$ is a cyclic point, then we can make the choice so that $x\in
K_{i}^{0^{i}}$. In that case, $x\in K$.

With $K=\bigcap_{i}\ K_{i}$ we have that the restriction $R|K$ is identified
with the inverse limit of $(R|K_{i},k_{i}\times k_{i})$ with $k_{i}$ the
inclusion map from $K_{i+1}$ to $K_{i}$. Hence, the maps
$h_{i}:K_{i}\to\mathbb{Z}/3^{i}\mathbb{Z}$ defines the continuous limit map
$h:K\to\mathbb{Z}[3]$ which maps $R|K$ to $\widehat{A}$.

In the standard $3-$adic example every point $y\in\mathbb{Z}[3]$ is a cycle
point by Corollary 7.4. Hence, by Theorem 7.6, $h$ is a homeomorphism mapping
$R|K$ isomorphically onto the standard $3-$adic example. From it follows that
every point of $K$ is a cycle point for $R|K$. The $3-$cycles in $K$ through a
point $x\in K$ are $3-$cycles in $X$ and so each point of $K$ is a cycle point
for $R$.

The converse result is obvious.

∎

######  Theorem 7.9.

If $(X,R)$ is a balanced, compact topological tournament, then $R$ is nowhere
locally transitive if and only if the space $X$ is totally disconnected.

###### Proof.

If a compact tournament $(X,R)$ is nowhere locally transitive, then by Theorem
7.8 $X$ is totally disconnected.

Conversely, if $R$ is balanced and $U$ is a clopen subset of $X$, then the
restriction $R|U$ is balanced and so has no terminal or initial point. In
particular, since $U$ is compact, $R|U$ is not transitive. If the compact
space $X$ is totally disconnected, then every nonempty open subset contains a
nonempty clopen subset and so $R$ is nowhere locally transitive.

∎

This completes the proof of Theorem 1.4.

######  Corollary 7.10.

If an infinite compact group $X$ admits a closed game subset $A$, then $X$ is
a Cantor set.

###### Proof.

That $X$ is totally disconnected with no isolated points follows from
Corollary 7.4 together with Theorem 7.7, which also implies that the points of
$X$ are $G_{\delta}$ points. Hence, $e$ has a countable neighborhood base of
clopen subsets $U_{n}$. It follows that $V_{n}=\\{(x,y):x^{-1}y\in U_{n}\\}$
is a countable neighborhood base for $1_{X}$ by clopen subsets of $X\times X$.
For a compact space $X$ the set of neighborhoods of $1_{X}$ is a uniformity
which is metrizable if it has a countable base, see [11] Chapter 6 and in
particular, Theorem 6.13. Since $X$ is metrizable, it is a Cantor set.

∎

Together with Corollary 7.4 and Theorem 6.9 this completes the proof of
Theorem 1.3.

It follows that if $G$ is a nontrivial finite group of odd order and $K$ is an
uncountable set, then the product group $G^{K}$ is totally disconnected, with
no isolated points, and with no elements of order two, but since it is not
metrizable, it does not admit a closed game subset. Of course, since there is
no element of order two, there are many game subsets (none of which is
closed). In fact since such a product contains no $G_{\delta}$ points, it
follows from Theorem 7.9 and Theorem 7.8 that it admits no balanced
tournament. When the cardinality of $K$ at least ${\mathbf{c}}$, the
cardinality of the continuum, we will see below that the product group admits
no topological tournaments at all.

######  Question 7.11.

Let $(X,R)$ be a compact topological tournament. If every point of $X$ is a
cycle point, does it follow that $X$ is metrizable and so is a Cantor set?

I conjecture that the answer is affirmative.

##  8\.  LOTS Constructions

We have seen that a linear order on a set is exactly a tournament which is
transitive. If $L$ is a linear order on a set $X$ (usually written $\leq$),
then $L^{\circ}(x)$ is the set of points larger than $x$, and $L^{\circ-1}(x)$
is the set of points smaller than $x$. We omit the usual interval notation to
avoid confusion with ordered pairs.

A linearly ordered topological space, or _LOTS_ , $X$, is a space with a
linear order $L$, equipped with the _order topology_ which has subbase
$\\{L^{\circ}(x):x\in X\\}\cup\\{L^{\circ-1}(x):x\in X\\}$.

######  Theorem 8.1.

If $L$ linear order on $X$, then the order topology is Hausdorff and with
respect to the order topology $L$ is closed, and so is a topological
tournament on $X$.

If $X$ is compact and $L$ is closed, i.e. it is a topological tournament which
is transitive, then the topology on $X$ is the order topology obtained from
$L$. In particular, $X$ is a LOTS.

###### Proof.

Assume $b\in L^{\circ}(a)$. The pair $a,b$ is a _gap pair_ when there is no
point between them, i.e. $L^{\circ}(a)\ \cap\ L^{\circ-1}(b)\ =\ \emptyset$.
In that case, $L^{\circ-1}(b)=L^{-1}(a)$ and $L^{\circ}(a)=L(b)$ are disjoint
neighborhoods of $a$ and $b$, respectively. Furthermore, $L^{\circ-1}(b)\times
L^{\circ}(a)\subset L^{\circ}$.

If $c\ \in\ L^{\circ}(a)\ \cap\ L^{\circ-1}(b)$ then $L^{\circ-1}(c)$ and
$L^{\circ}(c)$ are disjoint neighborhoods of $a$ and $b$, respectively and
$L^{\circ-1}(c)\times L^{\circ}(c)\subset L^{\circ}$.

Thus, the LOTS $X$ is Hausdorff and $L^{\circ}$ is open. Hence, $L=(X\times
X)\setminus L^{\circ-1}$ is closed.

Conversely, if $L$ is a topological tournament, then each $L^{\circ}(x)$ and
$L^{\circ-1}(x)$ is an open subset of $X$. If $X_{ord}$ is the set $X$ with
the order topology, then the identity map $X\to X_{ord}$ is a continuous
bijection. If $X$ is compact, then since $X_{ord}$ is Hausdorff, the map is a
homeomorphism. That is, $X$ has the order topology.

∎

A LOTS is _complete_ when every bounded, nonempty subset $A$, has a supremum
$\sup A$ and an infimum $\inf A$. The LOTS $X$ is compact if and only if it is
complete and has a maximum point $M$ and a minimum point $m$. For details
about LOTS, see, e.g. [3] and its extension [4]. Regarding the order $L$ as a
topological tournament, a maximum is a terminal point and a minimum is an
initial point.

######  Lemma 8.2.

If $X$ is a complete LOTS, then every bounded sequence in $X$ has a convergent
subsequence.

###### Proof.

It suffices to recall the proof that a sequence $\\{x_{n}\\}$ in $X$ has a
monotone subsequence.

Call $n\in\mathbb{N}$ dominating in the sequence, if for all $m>n\ \
x_{n}\rightharpoonup x_{m}$.

If there are infinitely many dominating indices, then the restriction to those
indices is a monotone decreasing sequence. If there are only finitely many
dominating indices and $N$ is the largest such, then let $n_{1}=N+1$ and
inductively choose $n_{k+1}>n_{k}$ with $x_{n_{k+1}}\
\underline{\rightharpoonup}\ x_{n_{k}}$ which exists because $n_{k}$ is not
dominating. This is a monotone non-decreasing sequence.

A bounded monotone sequence converges to its supremum or infimum.

∎

######  Theorem 8.3.

If $Y$ is a non-trivial compact space and $I$ has cardinality at least
${\mathbf{c}}$, the cardinality of the continuum, then the compact product
space $Y^{I}$ does not admit any topological tournament.

###### Proof.

Let $P$ be the power set of $\mathbb{N}$. There is a surjection from $I$ onto
$P$ and an injection from the two point set $\\{0,1\\}$ into $Y$. This induces
a continuous embedding of $X=\\{0,1\\}^{P}$ into $Y^{I}$. It suffices to show
that $X$ does not admit a topological tournament.

Observe that $X$ contains no $G_{\delta}$ points. By Theorem 7.8 is will
suffice to show that any topological tournament on $X$ would have to be
nowhere locally transitive.

Suppose instead that on some non-empty clopen subset of $X$ there exists a
closed, transitive tournament. By restricting further to a basic open set
obtained by fixing finitely many coordinates we obtain a subset homeomorphic
to $X$ itself. It suffices to contradict the assumption that $X$ admits a
closed transitive tournament. By Theorem 8.1 the topology on $X$ is the
associated LOTS topology. From Lemma 8.2 it will suffice to produce a sequence
in $X$ with no convergent subsequence.

Each $A\in P$ is a subset of $\mathbb{N}$. Define $\\{x_{n}\in X\\}$ by

(8.1) $(x_{n})_{A}\ =\ \begin{cases}1\ \ \text{if}\ \ n\in A,\\\ 0\ \
\text{if}\ \ n\not\in A.\end{cases}$

Suppose that some subsequence $\\{x_{n_{k}}\\}$ converges. By going to a
further subsequence, we may assume $\\{n_{k}\\}$ is strictly increasing
varying over a subset $B$ of $\mathbb{N}$. The sequence
$\\{(x_{n_{k}})_{A}\\}$ converges to $1$ if and only if $A\cap B$ is a
cofinite subset of $B$ and $\\{(x_{n_{k}})_{A}\\}$ converges to $0$ if and
only if $(\mathbb{N}\setminus A)\cap B$ is a cofinite subset of $B$. Write $B$
as the disjoint union of two infinite subsets $B_{1}$ and $B_{2}$. Let
$A=B_{1}$. Since neither $A\cap B=B_{1}$ nor $(\mathbb{N}\setminus A)\cap
B=B_{2}$ is a cofinite subset of $B$, it follows that $\\{(x_{n_{k}})_{A}\\}$
does not converge.

∎

In a LOTS $X$ let $\\{x_{k}\\}$ be a net indexed by the directed set $D$ and
converging to $x$. The index set $D$ is partitioned by three subsets:
$\\{k:x_{k}\in L^{\circ-1}(x)\\},\ \\{k:x_{k}\in L^{\circ}(x)\\},\
\\{k:x_{k}=x\\}$. At least one of these sets is cofinal in $D$ and so by going
to a subnet we may assume that either

  * $-$

: $x_{k}\in L^{\circ-1}(x)$ for all $k$ in which case the net converges to $x$
from the left and $x$ is left balanced for $L$.

  * +

: $x_{k}\in L^{\circ}(x)$ for all $k$ in which case the net converges to $x$
from the right and $x$ is right balanced for $L$.

  * 0

: $x_{k}=x$ for all $k$ so that the net is constant at $x$.

If $(X_{1},L_{1})$ and $\\{(Y_{x},L_{x}):x\in X\\}$ are LOTS, then we will
denote by $(X_{2},L_{2})$ the lexicographic product
$(X_{1},L_{1})\ltimes\\{(Y_{x},L_{x})\\}$ as in (3.1). From Theorem 3.3 the
product tournament is transitive and so, when equipped with the order
topology, $X_{2}$ is a LOTS.

######  Proposition 8.4.

If $(X_{1},L_{1})$ is a complete LOTS and for each $x\in X_{1},(Y_{x},L_{x})$
is a compact LOTS with minimum $m_{x}$ and maximum $M_{x}$, then the LOTS
$(X_{2},L_{2})=(X_{1}\times\\{Y_{x}\\},L_{1}\ltimes\\{L_{x})\\})$ is complete
and the projection map $\pi:X_{2}\to X_{1}$ is a continuous, surjective
topological tournament map from $L_{2}$ on to $L_{1}$.

###### Proof.

For $A$ a bounded subset of $X_{2}$, the set $A_{1}=\pi(A)$ is a bounded
subset of $X_{1}$ and so it has a supremum, $a_{1}$.

If $a_{1}\in A_{1}$, then $\\{y\in Y_{a_{1}}:(a_{1},y)\in A\\}$ is nonempty
and so has a supremum $y_{1}\in Y_{a_{1}}$. In that case, $(a_{1},y_{1})$ is
the supremum of $A$.

If $a_{1}\not\in A_{1}$, then $(a_{1},m_{a_{1}})$ is the supremum of $A$ where
$m_{a_{1}}$ is the minimum of $Y_{a_{1}}$.

The $\pi$ preimage of $L_{1}^{\circ}(x)\subset X_{1}$ is
$L_{2}^{\circ}(x,M_{x})\subset X_{2}$ and the preimage of $L_{1}^{\circ-1}(x)$
is $L_{2}^{\circ-1}(x,m_{x})$. Hence, $\pi$ is continuous. It is clearly a
tournament map.

∎

Notice that , in contrast with the topological lexicographic products of
Theorem 3.3 the LOTS $(Y_{x},L_{x})$ can be non-trivial for any point $x$. If,
however, the LOTS is only non-trivial when $x$ is isolated in $X$, then the
order topology agrees with the topology in Theorem 3.3 when $X_{1}$ is
complete and each $Y_{x}$ is compact.

From the definition of the lexicographic product and the order topology, the
following is obvious.

######  Addendum 8.5.

If in $X_{2}$ a net $\\{(x_{k},y_{k})\\}$ converges to $(x,y)$, then
$\\{x_{k}\\}$ converges to $x$ in $X_{1}$.

  * $-$

: If $\\{x_{k}\\}$ converges to $x$ from the left, then $\\{(x_{k},y_{k})\\}$
converges to $(x,y)$ if and only if $y=m_{x}$, i.e. $y$ is the minimum for
$Y_{x}$.

  * +

: If $\\{x_{k}\\}$ converges to $x$ from the right, then $\\{(x_{k},y_{k})\\}$
converges to $(x,y)$ if and only if $y=M_{x}$, i.e. $y$ is the maximum for
$Y_{x}$.

  * 0

: If $\\{x_{k}\\}$ is constant at $x$, then $\\{(x_{k},y_{k})\\}$ converges to
$(x,y)$ if and only if $\\{y_{k}\\}$ converges to $y$ in $Y_{x}$.

When $(Y_{x},L_{x})=(Y,L)$ for all $x\in X$, then we write $X_{1}\ltimes Y$
for the product set equipped with linear order $L_{1}\ltimes L$. Notice that
the order topology is usually not the same as the product topology. For
example, if $b\in L^{\circ}(a)$ in $Y$, then for any $x\in X$, the set
$\\{x\\}\times(L^{\circ}(a)\cap L^{\circ-1}(b)$ is the interval
$L^{\circ}(x,a)\cap L^{\circ-1}(x,b))$ in $X\ltimes Y$ and so is open. The
Sorgenfrey Double Arrow described above is an example of such a lexicographic
product.

For our construction we begin with $S$ the topological tournament on a Cantor
set $C$ obtained from a closed game subset for a topological group structure
on $C$. From Corollary 7.4 it follows that every point of $C$ is a cycle point
for $S$.

Next, use the linear order $L_{\mathbb{N}}$ on the discrete set of natural
numbers $\mathbb{N}$. To the topological tournament $L_{\mathbb{N}}\ltimes S$
on $\mathbb{N}\times C$ we adjoin a terminal point $M$ to obtain the one-point
compactification of $\mathbb{N}\times C$. Every point of the resulting
topological tournament $S_{1}$ is a cycle point except the terminal point $M$
which is only left balanced. For its inverse $S_{0}$ every point is a cycle
point except for the initial point which is right balanced.

Because the one-point compactification is itself a Cantor set, we can use a
homeomorphism to move $S_{0}$ and $S_{1}$ and so obtain the tournaments on the
standard Cantor Set $C$ in the unit interval with the maximum $1$ the terminal
point for the tournament $S_{1}$ and with the minimum $0$ the initial point
for the tournament $S_{0}$.

Similarly, use the linear order $L_{\mathbb{Z}}$ on the discrete set of
integers $\mathbb{Z}$. To the topological tournament $L_{\mathbb{Z}}\ltimes S$
on $\mathbb{Z}\times C$ we adjoin a terminal point $M$ and initial point $m$
to obtain the two-point compactification of $\mathbb{Z}\times C$. We can use a
homeomorphism to obtain the topological tournament $S_{01}$ on the standard
Cantor Set $C$ with initial point equal to the minimum $0$ and terminal point
equal to the maximum $1$. Every point is a cycle point except for the left
balanced point $1$ and the right balanced point $0$.

Thus, on the Cantor Set $C$ we have four topological tournaments: $S,\ S_{0},\
S_{1},\ S_{01}$. Every point is a cycle point, and so is balanced, for each of
these tournaments except the the right balanced initial point $0$ for $S_{0}$
and $S_{01}$ and the left balanced terminal point $1$ for $S_{1}$ and
$S_{01}$.

######  Definition 8.6.

Let $(Y,L)$ be a non-trivial, compact LOTS with maximum $M$ and minimum $m$.
For each of the following types, $S$ is assumed to be a topological tournament
on $Y$ such that every point of $Y$ is balanced except for the terminal or
initial points for $S$ when such exist.

  * •

The tournament $S$ is type 0 when $S$ has no terminal point and the minimum
$m$ is an initial point for $S$ which is right balanced with respect to $S$.

  * •

The tournament $S$ is type 1 when $S$ has no initial point and the maximum $M$
is a terminal point for $S$ which is left balanced with respect to $S$.

  * •

The tournament $S$ is type 01 when $M$ is a terminal point for $S$ which is
left balanced with respect to $S$ and $m$ is a initial point for $S$ which is
right balanced with respect to $S$.

The existence of tournaments of each of these types for $Y$ requires that the
minimum $m$ be right balanced with respect to the order $L$ on $Y$ and that
the maximum $M$ be left balanced with respect to $L$, i.e. neither extremum is
isolated.

Let $X_{1}$ be a compact LOTS with order $L_{1}$ and with minimum $m$ and
maximum $M$. Of course, $m$ is not left balanced and $M$ is not right balanced
for $L_{1}$. For each point $x\in X$ we choose a compact LOTS $Y_{x}$ with
order $L_{x}$ and a topological tournament $S_{x}$ on $Y_{x}$ which satisfies
the following rules:

  * (i)

: If $x$ is balanced for $L_{1}$, then either $Y_{x}$ is a singleton with
$Y_{x}=\\{m_{x}\\}=\\{M_{x}\\}$ and so $L_{x}$ and $S_{x}$ are trivial, or
else $S_{x}$ is of type 01.

  * (ii)

If $x$ is left balanced for $L_{1}$, but not right balanced for $L_{1}$, then
$S_{x}$ is of type 0.

  * (iii)

If $x$ is right balanced, but not left balanced for $L_{1}$, then $S_{x}$ is
of type 1.

  * (iv)

If $x$ is an isolated point in $X$, then $S_{x}$ is balanced and so has no
terminal or initial point.

Thus, a terminal (or initial) point for the tournament $S_{x}$, when it
exists, coincides with the maximum (resp. the minimum) for the order $L_{x}$.

Now on the LOTS $(X_{2},L_{2})=(X_{1},R_{1})\ltimes\\{(Y_{x},L_{x})\\}$ we
define the tournament $R_{2}$ by:

(8.2) $((x,t),(y,s))\in
R_{2}^{\circ}\quad\Longleftrightarrow\quad\begin{cases}\qquad(x,y)\in
L_{1}^{\circ},\quad\text{or}\\\ x=y\ \ \text{and}\ \ (t,s)\in
S_{x}^{\circ}.\end{cases}$

It is clear the $R_{2}$ is just the tournament $L_{1}\ltimes\\{S_{x}\\}$ on
$X_{2}$ and the continuous surjection $\pi:X_{2}\to X_{1}$ is a tournament map
from $R_{2}$ to $L_{1}$. Notice that on $X_{2}$ we are using the order
topology obtained from $L_{2}$.

######  Theorem 8.7.

The tournament $R_{2}$ is a balanced topological tournament on the compact
LOTS $X_{2}$.

###### Proof.

Let $\\{((x_{k},t_{k}),(y_{k},s_{k}))\\}$ be a net in $R_{2}^{\circ}$ which
converges to
$((x,t),(y,s))$ in $X_{2}\times X_{2}$.

First observe that it cannot happen that $y\in L_{1}^{\circ-1}(x)$ since
$L_{1}^{\circ}$ is open and this would imply that eventually $y_{k}\in
L_{1}^{\circ-1}(x_{k})$ and so eventually $\\{((x_{k},t_{k}),(y_{k},s_{k}))\in
R_{2}^{\circ-1}$ violating anti-symmetry.

If $(x,y)\in L_{1}^{\circ}$, then $((x,t),(y,s))\in R_{2}^{\circ}$ as
required.

We are left with the cases when $x=y$. It cannot happen that at the same time
$\\{y_{k}\\}$ converges to $y=x$ from the left and $\\{x_{k}\\}$ converges to
$x$ from the right, because then $y=x$ and transitivity imply $y_{k}\in
L_{1}^{\circ-1}(x_{k})$ for all $k$ and so again
$\\{((x_{k},t_{k}),(y_{k},s_{k}))\in R_{2}^{\circ-1}$. Similarly, it cannot
happen that $\\{x_{k}\\}$ is constant and $\\{y_{k}\\}$ converges from the
left, and it cannot happen that $\\{x_{k}\\}$ converges from the right and
$\\{y_{k}\\}$ is constant.

If $\\{x_{k}\\}$ converges to $x$ from the left, then by Addendum 8.5
$t=m_{x}$ . Because $x$ is left balanced for $L_{1}$, $S_{x}$ is of type 0 or
type 01 and so $m_{x}$ is an initial point for $S_{x}$. Hence,
$((x,t),(y,s))=((x,m_{x}),(x,s))\in R_{2}$.

Similarly, if $\\{y_{k}\\}$ converges to $y$ from the right, then
$((x,t),(y,s))=((y,t),(y,M_{x}))\in R_{2}$.

The remaining possibility is that both $\\{x_{k}\\}$ and $\\{y_{k}\\}$ are
constant at $x$. In that case, $\\{(t_{k},s_{k})\in S_{x}^{\circ}\\}$