converges to $(t,s)$ in $Y_{x}\times Y_{x}$ by Addendum 8.5 and so $(t,s)\in
S_{x}$ which implies $(x,t),(y,s)\in R_{2}$.

We have proved that $R_{2}$ is closed.

Now let $(x,t)\in X_{2}$. If $t$ is neither a initial nor a terminal point for
$S_{x}$, then it is balanced for $S_{x}$ and so $(x,t)$ is balanced for
$R_{2}$. Note that on each $Y_{x}$ the relative topology induced from $X_{2}$
is that of $Y_{x}$.

If $t=m_{x}$ is a initial point for $S_{x}$, then $x$ is left balanced for
$L_{1}$ and so there exists a net $\\{x_{k}\\}$ which converges to $x$ from
the left. For any $y_{k}\in Y_{x_{k}}$, $(x_{k},x)\in L_{1}^{\circ}$ implies
$\\{((x_{k},y_{k}),(x,m_{x}))\in R_{2}^{\circ}\\}$ and $\\{(x_{k},y_{k})\\}$
converges to $(x,m_{x})$. Hence, $(x,m_{x})$ is left balanced.

Similarly, if $t=M_{x}$ is a terminal point for $S_{x}$, then $(x,M_{x})$ is
right balanced for $R_{2}$.

In particular, if $Y_{x}$ is trivial, then $(x,m_{x})=(x,M_{x})$ is balanced
for $R_{2}$.

Finally, assume that $Y_{x}$ is non-trivial.

In that case, the initial $m_{x}$ for $S_{x}$, when it exists, is right
balanced for $S_{x}$ and so there exists a net $t_{k}\in S_{x}^{\circ}(0)$
which converges to $m_{x}$ which implies that the net $\\{(x,t_{k})\in
R_{2}^{\circ}((x,m_{x}))\\}$ converges to $(x,m_{x})$. That is, $(x,m_{x})$ is
right balanced and so is balanced.

Similarly, if $t=M_{x}$ is a terminal point for $S_{x}$, then $(x,M_{x})$ is
balanced for $R_{2}$.

∎

We can make the following alterations in our choice for $S_{x}$

  * (v)

: If the minimum $m$ is right balanced for $L_{1}$, let $S_{m}$ be type 01
instead of type 1. If $m$ is isolated, let $S_{m}$ be type 0 instead of
balanced.

  * (vi)

: If the maximum $M$ is left balanced for $L_{1}$, let $S_{M}$ be type 01
instead of type 0. If $M$ is isolated, let $S_{M}$ be type 1 instead of
balanced.

It is easy to check the following.

######  Addendum 8.8.

If we alter our choices according to (v) we obtain a topological tournament
$R_{0}$ on $X_{2}$ which is of type 0.

If we alter our choices according to (vi) we obtain a topological tournament
$R_{1}$ on $X_{2}$ which is of type 1.

If we alter our choices according to both (v) and (vi) we obtain a topological
tournament $R_{01}$ on $X_{2}$ which is of type 01.

######  Examples 4.

Nonseparable Examples

(a) In [3] and [4] there is a rich supply of connected, complete, first
countable LOTS $X$ which are not separable. For example, let $I$ be the closed
interval in $\mathbb{R}$ with end-points $\pm 1$. The LOTS
$(X,L)=(\mathbb{R},L_{\mathbb{R}})\ltimes(I,L_{\mathbb{R}}|I)$ is the product
set $\mathbb{R}\times I$ equipped with the order topology from the
lexicographic product order $L$. Restrict to the compact subset $X_{1}$ which
is the closed interval in $\mathbb{R}\times I$ with minimum $m=(0,1)$ and
maximum $M=(1,0)$. For every $t\in\mathbb{R}$ with $0<t<1$, let $A_{t}$ be the
interval in $\\{t\\}\times I$ with end-points $(t,-\frac{1}{2})$ and
$(t,+\frac{1}{2})$. Thus, $\\{A_{t}\\}$ is an uncountable collection of pair-
wise disjoint, non-trivial intervals, illustrating that $X_{1}$ is not
separable. The LOTS is connected, equivalently every point of $X_{1}$ is
balanced with respect to the order, except for the right balanced minimum and
the left balanced maximum.

For all $x\in\ A\ =\ \bigcup_{t}\ A_{t}$ we let $Y_{x}$ be trivial set
$\\{0\\}$. For $x\in X_{1}\setminus A$ we let $Y_{x}$ be the standard Cantor
Set $C$ with order $L_{C}$ and tournaments $S_{0},S_{1}$ and $S_{01}$ chosen
as above. There are no isolated points in $X$ and so rule (iv) does not apply.
Let $(X_{2},L_{2})$ be the LOTS $(X_{1},L|X_{1})\ltimes\\{(Y_{x},L_{x}\\}$ and
let $R_{2}$ be the tournament on $X_{2}$ given by (8.2). For each $t$ between
$0$ and $1$, $A_{t}\times\\{0\\}$ is a closed interval in $X_{2}$ on which the
tournament $R_{2}$ is isomorphic to the order $L_{1}$ on $A_{t}$ and each of
these is a component of $X_{2}$. Contrast this with the countability result in
the metric case given in Corollary 5.2.

(b) Let $X_{1}$ be the unit interval in $\mathbb{R}$ with end-points $m=0$ and
$M=1$. For all $x\in X$ we let $Y_{x}$ be the standard Cantor Set $C$ which is
a LOTS with order $L_{C}$ inherited from $\mathbb{R}$. Let $S_{0},S_{1}$ and
$S_{01}$ be tournaments on $C$ chosen as above. Let $(X_{2},L_{2})$ be the
LOTS $(X_{1},L_{\mathbb{R}}|X_{1})\ltimes(C,L_{C})$ and let $R_{2}$ be the
tournament on $X_{2}$ given by (8.2). Every point which is not either equal to
$(t,0)$ for some $t$ with $0<t\leq 1$ or equal to $(t,1)$ for some $t$ with
$0\leq t<1$ is a cycle point for $R_{2}$ and each of these points has a Cantor
set neighborhood in $X_{2}$. Each of the remaining, exceptional, points is
balanced and with a countable neighborhood base, but with no separable
neighborhood.

We can alter $R_{2}$ to convert some of these exceptional points to cycle
points.

Fix $0<t<1$ and a strictly decreasing sequence $\\{t_{n}\\}$ in $X_{1}$
converging to $t$. Let $\\{A_{n}\\}$ be a sequence of pair-wise disjoint
clopen sets in $C$ arranged with $A_{n}<A_{n+1}$ and so that the sets converge
to the point $1$. Define the subset $Q$ of $R^{\circ}\subset X_{2}\times
X_{2}$ by

(8.3) $Q\ =\ \bigcup_{n=2}^{\infty}\ (\\{t\\}\times
A_{n})\times(\\{t_{n}\\}\times A_{1}).$

Observe that $Q$ is a clopen subset of $R^{\circ}$ with closure in
$X_{2}\times X_{2}$ given by $Q\cup\\{((t,1),(t,1))\\}$. So with

(8.4) $R_{Q}\ =\ (R\setminus Q)\cup Q^{-1},\hskip 56.9055pt$

we obtain a topological tournament on $X_{2}$. All of the cycle points for $R$
are still cycle points for $R_{Q}$, but in addition if $a_{n}\in A_{n}$ for
each $n$, then for $n\geq 2,\ \\{(t,a_{n}),(t,1),(t_{n},a_{1})\\}$ is a
$3-$cycle for $R_{Q}$ and so $(t,1)$ is a cycle point with no separable
neighborhood.

It is possible to use this procedure to convert a discrete countable
collection of exceptional points to become cycle points. However, this method
can’t be used to convert all of the exceptional points to cycle points. Notice
that if the question 7.11 has an affirmative answer, then no topological
tournament on the non-metric space $X_{2}$ could consist entirely of cycle
points.

A space $X$ has a LOTS topology if there exists a linear order on $X$ such
that the topology on $X$ is the order topology. A Cantor set and a finite set
have LOTS topologies and the results of our lexicographic product and inverse
limit constructions all have LOTS topologies. We saw in the proof of Theorem
8.3 that 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 have a LOTS topology and does not admit any
topological tournament.

######  Question 8.9.

If $(X,R)$ is a compact tournament, does the underlying space $X$ have to have
a LOTS topology?

##  9\.  Big Examples

In this section we perform the construction leading to Theorem 1.5. We use
ordinal numbers, see [10] or [8]. What we need is also in [4].

An ordinal number is a well-ordered set which is equal to the set of its
predecessors, beginning with $0=\emptyset$. That is,
$\alpha=\\{\beta<\alpha\\}$. The successor ordinal
$\alpha+1=\alpha\cup\\{\alpha\\}$. With its order topology a successor ordinal
$\alpha+1$ is a compact LOTS with minimum $0$ and maximum $\alpha$. The
successor ordinals in $\alpha+1$ form a dense set of isolated points. The
remaining, limit, ordinals are left balanced with respect to the order. By
well-ordering, no point is right balanced.

Let $\aleph$ be a limit ordinal so that the successor $\aleph+1$ is a compact
LOTS with minimum $0$ and maximum $\aleph$. We write $L_{\aleph}$ for the
order on $\aleph+1$. Let $A=\\{0\\}\cup\\{1/n:n\in\mathbb{N}\\}$ regarded as a
compact LOTS with the order $L_{A}$ from $\mathbb{R}$. Let
$A_{\alpha}=\\{0\\}$ for any non-limit ordinal, i.e. $0$ and all the successor
ordinals less than $\aleph$ and let $A_{\alpha}=A$ for all limit ordinals
contained in $\aleph+1$, including $\aleph$ itself. Let $(X_{0},L_{0})$ be the
LOTS $(\aleph+1,L_{\aleph})\ltimes\\{(A_{\alpha},L_{A})\\}$. The projection
map to $\aleph+1$ is continuous, and in this case, the injective map
$\alpha\to(\alpha,0)$ is continuous as well. So we will identify $\aleph+1$
with the subset $(\aleph+1)\times\\{0\\}\subset X_{0}$. Thus, $X_{0}$ consists
of $\aleph+1$ together with, for each limit ordinal $\alpha$, a decreasing
sequence of isolated points converging to $\alpha$. In the LOTS $X_{0}$ the
isolated points are dense and each non-isolated point is balanced with respect
to the order $L_{0}$. Thus, the LOTS $(X_{0},L_{0})$ is a brick in the sense
of Definition 3.6.

Now let $B$ be a finite set with odd cardinality and $L_{B}$ be a linear order
on $B$. Fix $S_{B}$ an arc cyclic tournament on $B$ so that $(B,S_{B})$ is an
ip cyclic brick. The LOTS $(B,L_{B})$ is also a brick, but not, of course, ip
cyclic.

For each isolated point $y\in X_{0}$ let $(B_{y},L_{y})=(B,L_{B})$ and
$(B_{y},S_{y})=(B,S_{B})$. for each non-isolated point $y$ we let
$(B_{y},S_{y})=(B_{y},L_{y})$ be trivial. Let the compact LOTS $(X_{1},L_{1})$
be the lexicographic product $(X_{0},L_{0})\ltimes\\{(B_{y},L_{y})\\}$. In
this case it is also topological lexicographic product. On $X_{1}$ we define
$R_{1}=L_{0}\ltimes\\{S_{y}\\}$ the topological lexicographic product using
the cycle tournaments. By Theorem 3.7 $(X_{1},R_{1})$ is a ip cyclic brick. We
call $(X_{1},R_{1})$ a Big Brick. The underlying space $X_{1}$ is a compact
LOTS with the ordering $L_{1}$.

If $p:X_{1}\to X_{0}$ is the first coordinate projection, then for each limit
ordinal $\alpha$ in $X_{0}$, $p^{-1}(\alpha)$ is a singleton set and we label
the point in this set by $\alpha$ as well. That is, we regard the limit
ordinals $\alpha\leq\aleph$ as points of $X_{1}$. Because the tournament on
$X_{0}$ is an order, it follows that no limit ordinal $\alpha$ in $X_{1}$ is
contained in a cycle in $X_{1}$.

We now perform a lexicographic inverse system of bricks using the Big Brick.

Begin with $(X_{1},R_{1})$ as above so that $X_{1}$ is a LOTS with order
$L_{1}$. Inductively, we are given the LOTS $(X_{i},L_{i})$ with the
tournament $R_{i}$ so that $(X_{i},R_{i})$ is a brick. We let
$(Y_{iz},S_{iz},L_{iz})=(X_{1},R_{1},L_{1})$ for all $z$ isolated in $X_{i}$
and $(Y_{iz},S_{iz},L_{iz})$ trivial for all $z$ non-isolated in $X_{i}$.

We let $(X_{i+1},L_{i+1})=(X_{i},L_{i})\ltimes\\{(Y_{iz},L_{iz})\\}$, i.e. the
LOTS which is the lexicographic product. Since $(Y_{iz},S_{iz},L_{iz})$
trivial for all $z$ non-isolated in $X_{i}$, this lexicographic product is at
the same time the topological lexicographic product.
$(X_{i+1},R_{i+1})=(X_{i},R_{i})\ltimes\\{(Y_{iz},R_{iz})\\}$. Since both of
these are topological lexicographic products, the space $X_{i+1}$ is the same
for both products and so the topology on $X_{i+1}$ is the LOTS topology
obtained from $L_{i+1}$. Let $f_{i}:X_{i+1}\to X_{i}$ be the first coordinate
projection which is an open, continuous surjection mapping $L_{i+1}$ to
$L_{i}$ and $R_{i+1}$ to $R_{i}$.

We let $(X,R,L)$ be the inverse limit of the inverse sequence
$\\{(X_{i},R_{i},L_{i},f_{i})\\}$, i.e. $(X,L)$ is the inverse limit of the
tournament inverse sequence $\\{(X_{i},L_{i},f_{i})\\}$ and $(X,R)$ is the
inverse limit of the tournament inverse sequence $\\{(X_{i},R_{i},f_{i})\\}$.

######  Theorem 9.1.

The ordered space $(X,L)$ is a compact, totally disconnected LOTS with no
isolated points and $R$ is a balanced topological tournament on $X$. The set
of cycle points for $R$ is the dense $G_{\delta}$ subset $IS$ of points $x\in
X$ such that each $x_{i}$ is isolated in $X_{i}$. In addition, every nonempty
open subset of $X$ has cardinality at least that of $\aleph$.

If the cardinality of $\aleph$ is countable, then $X$ is a Cantor set.

If $\aleph$ is uncountable, then there is a dense set of points which are not
$G_{\delta}$ points and no open subset is separable.

###### Proof.

By Theorem 4.5 $(X,R)$ is a balanced tournament with $X$ compact and totally
disconnected and similarly $L$ is a transitive topological tournament on $X$
so that the topology on $X$ is the order topology by Theorem 8.1.

By Theorem 4.5 again the subset $IS$ is a dense $G_{\delta}$ set and it
consists of cycle points by Theorem 7.5. Also by Theorem 4.5 each projection
map $\pi_{i}:X\to X_{i}$ is open as well as continuous.

A nonempty open subset contains some $\pi_{i}^{-1}(U)$ with $U$ an open
nonempty subset of $X_{i}$. There exists an $IS$ point $x$ in
$\pi_{i}^{-1}(U)$. So $x_{i}$ is isolated in $X_{i}$ and for every $y\in
Y_{ix_{i}}$, there exists a point $x^{\prime}\in X$ with
$x^{\prime}_{i}=x_{i}$ and $x^{\prime}_{i+1}=(x_{i},y)$. Thus,
$\pi_{i}^{-1}(x)\subset\pi_{i}^{-1}(U)$ contains a set of cardinality at least
that of $\aleph$.

Notice that if $x_{i}$ is an isolated point of $X_{i}$, then the map
$y\mapsto(x_{i},y)$ is a homeomorphism from $X_{1}$ onto a clopen subset of
$X_{i+1}$, inducing a tournament isomorphism from $(X_{1},R_{1})$ onto the
restriction $R_{i+1}|(\\{x_{i}\\}\times Y_{ix_{i}})$.

We saw above that if $\alpha$ is a limit ordinal in $Y$ then it is not
contained in a cycle in $Y$. It follows that the unique point $x^{\prime}\in
X$ with $x^{\prime}_{i+1}=(x_{i},\alpha)$ is not in any cycle contained in the
clopen set $\pi_{i}^{-1}(\\{x_{i}\\})$. Thus, $x^{\prime}$ is not a cycle
point.

It follows that the points of $IS$ are the only cycle points. By Theorem 7.7
every cycle point is a $G_{\delta}$ point.

Furthermore, because $\pi_{i+1}$ is a continuous, open map and because for the
above point $x^{\prime}$, $\pi_{i+1}^{-1}(x^{\prime})$ is a singleton, it
follows that $x^{\prime}$ is a $G_{\delta}$ point if an only if
$x^{\prime}_{i+1}$ is a $G_{\delta}$ point in the clopen set
$\pi_{i}^{-1}(\\{x_{i}\\})$ and so if and only if $\alpha$ is a $G_{\delta}$
point in $X_{1}$. This is true if and only if the limit ordinal $\alpha$ is
countable. Such a point is a $G_{\delta}$ point but is not a cycle point.

If $\aleph$ is countable, then the bricks are countable and so $X$ is a Cantor
set by Theorem 4.5.

If $\aleph$ is uncountable, the point $x^{\prime}$ with
$x^{\prime}_{i+1}=(x_{i},\aleph)$ is not a $G_{\delta}$ point. Thus, the
points which are not $G_{\delta}$ form a dense subset of $X$.

A nonempty clopen subset of $X$ is a compact LOTS with respect to the
restriction of the order $L$. A compact, separable LOTS can be embedded in
$\mathbb{R}$ and so every interior point would be a $G_{\delta}$ point. It
follows that no open subset is separable when $\aleph$ is uncountable. ∎

For example, if $\aleph$ is an uncountable ordinal but with cardinality less
than or equal to ${\mathbf{c}}$, the cardinality of the continuum, then $X$
has cardinality ${\mathbf{c}}$, that of the Cantor set, but contains no
separable nonempty open subset.

##  10\. WAC Tournaments and Prime Quotients

We begin with an extension of the concept of arc cyclicity.

######  Definition 10.1.

A topological tournament $(X,R)$ is called _weakly arc cyclic_ , or just a
_wac tournament_ , when $X$ is compact and every non-isolated point of $X$ is
a cycle point for $R$.

The topological tournament $(X,R)$ an _almost wac tournament_ when $X$ is
compact and every point of $X$ is either isolated, initial, terminal or a
cycle point.

We will write that the tournament $R$ is wac or almost wac when the underlying
space is understood.

######  Theorem 10.2.

(a) If $(X,R)$ is an almost wac tournament, then $X$ is totally disconnected.

(b) If $R$ is a compact locally arc cyclic tournament,e.g. a finite
tournament, then it is wac.

(c) If $R$ is a wac (or almost wac) tournament, then the reverse tournament
$R^{-1}$ is wac (resp. almost wac).

(d) If $R$ is a wac tournament (or an almost wac tournament) and $A$ is a
nonempty clopen subset of $X$, then the restriction $R|A$ is wac (resp. almost
wac). If $R$ is a wac tournament and $x\in X$, then the restriction $R|R(x)$
is almost wac with initial point $x$.

(e) If $R$ is a wac tournament, then any initial or terminal point for $R$ is
an isolated point.

(f) Let $h:(X_{2},R_{2})\to(X_{1},R_{1})$ be a surjective continuous map of
topological tournaments. If $(X_{2},R_{2})$ is wac (or almost wac), then
$(X_{1},R_{1})$ is wac (resp. almost wac) and for every non-isolated point $y$
of $X_{1}$, the set $h^{-1}(y)$ is a singleton subset $\\{x\\}$ with $x$ non-
isolated in $X_{2}$.

If $x$ is a terminal (or initial) point for $R_{2}$, then $h(x)$ is a terminal
(resp. initial) point for $R_{1}$. If $y$ is a terminal (or initial) point for
$R_{1}$ and $y$ is not isolated, then $h^{-1}(y)=\\{x\\}$ with $x$ terminal
(resp. initial) for $R_{2}$.

If $y$ is an isolated point of $X_{1}$, then the restriction
$R_{2}|h^{-1}(y))$ is wac (resp. almost wac).

There is an isomorphism to the topological lexicographic product

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

such that $\pi\circ\hat{h}=h$ where $\pi$ is the coordinate projection to
$(X_{1},R_{1})$. In particular, $h$ is an open map.

If $(X_{2},R_{2})$ is arc cyclic or locally arc cyclic, then $(X_{1},R_{1})$
satisfies the corresponding condition.

###### Proof.

(a): If $A$ were a non-trivial component of $X$, then it contains no isolated
points and by Corollary 5.2 it contains a non-empty open subset $U$ of $X$.
Furthermore, the restriction $R|A$ is an order and so the infinite open set
$U$ contains no cycle points. As at most two points are initial or terminal,
the tournament cannot be almost wac.

(b): A compact locally arc cyclic tournament is wac by Theorem 7.3.

(c): is obvious.

(d): If $x\in A$ and $A$ is an arbitrary subset, then $x$ non-isolated in $A$
or $x$ a cyclic point for $R|A$ implies the corresponding condition for $X$.
If $A$ is open, then the converse holds. If $A$ is clopen, then it is compact
as well.

So a point $x^{\prime}\in R^{\circ}(x)$ is isolated or a cycle point in $R(x)$
if and only if satisfies the corresponding property in $X$. Clearly, $x$ is
initial for the restriction to $R(x)$. Hence, $R|R(x)$ is almost wac when $R$
is wac.

(e): An initial or terminal point is not balanced and so is not a cycle point.
Such a point in a wac tournament is therefore isolated.

(f): Clearly, if $M$ is terminal for $X_{2}$, then $x\
\underline{\rightharpoonup}\ M$ for all $x\in X$ implies $h(x)\
\underline{\rightharpoonup}\ h(M)$. Since $h$ is surjective, $h(M)$ is
terminal for $X_{1}$. Similarly, $m$ initial for $X_{2}$ implies $h(m)$ is
initial for $X_{1}$.

If $y$ is an isolated point in $X_{1}$, then $h^{-1}(y)$ is clopen and so the
restriction $R_{2}|h^{-1}(y)$ is wac or almost wac by (c).

For $y$ a non-isolated point of $X_{1}$, let $U$ be an open set containing
$y$. We may assume $y$ is left balanced as the right balanced case is similar.
As it is left balanced, it is not an initial point for $X_{1}$. By Theorem 2.4
and Addendum 2.5, there is a unique point $x\in
h^{-1}(y)\cap\overline{h^{-1}(R_{1}^{\circ}(y))}$ and $x$ is an initial point
for the restriction $R_{2}|h^{-1}(y)$. So $x$ is a non-isolated point for the
wac tournament $(X_{2},R_{2})$. Since it is left balanced, it is not an
initial point for $X_{2}$.

Case 1 ($y$ is not terminal in $X_{1}$): In that case, $x$ is not terminal in
$X_{2}$. As it is not initial and not isolated, it is a cycle point Therefore,
there exists a $3-$cycle $\\{x,x^{\prime},x^{\prime\prime}\\}$ in $h^{-1}(U)$.
Because $x$ is an initial point for $R_{2}|h^{-1}(y)$, the cycle is not
contained in $h^{-1}(y)$. So by Proposition 2.2 it is mapped by $h$ to a
$3-$cycle $\\{y,y^{\prime},y^{\prime\prime}\\}$ in $U$. Thus, $y$ is a cycle
point and from Theorem 7.6 it follows that $h^{-1}(y)$ is the singleton
$\\{x\\}$.

Case 2 ($y=M_{1}$ is terminal in $X_{1}$): If $h^{-1}(M_{1})$ were to contain
more than one point, then $x$ is not terminal in $X_{2}$ since it is initial
in $h^{-1}(M_{1})$. Because $x$ is not isolated and neither initial nor
terminal, it would have to be a cycle point. As in Case 1, we would obtain a
$3-$cycle $\\{M_{1},y^{\prime},y^{\prime\prime}\\}$ in $U$. This is impossible
because $M_{1}$ is terminal. Hence, $h^{-1}(M_{1})$ is the singleton $\\{x\\}$
in this case as well. Finally, for all $x^{\prime}\in X_{2}\setminus
h^{-1}(M_{1})$, $h(x^{\prime})\rightharpoonup M_{1}$ implies
$x^{\prime}\rightharpoonup x^{\prime\prime}$ for all $x^{\prime\prime}\in
h^{-1}(M_{1})$. As the latter set is the singleton $\\{x\\}$ it follows that
$x$ is terminal in $X_{2}$.

Since $h^{-1}(y)$ is a singleton whenever $y$ is non-isolated, the topological
lexicographic product $(X_{1},R_{1})\ltimes\\{(h^{-1}(y),R_{2}|h^{-1}(y))\\}$
can be defined according to Definition 3.2. The map $\hat{h}$ defined by
$\hat{h}(x)=(h(x),x)$ with inverse $(y,x)\mapsto x$ is a bijection providing a
tournament isomorphism. From the definition of the basis in Definition 3.2 it
is clear that $\hat{h}$ is continuous. So it is a homeomorphism by
compactness. Form Theorem 3.3(a) it follows that $h$ is an open map.

If a neighborhood $U$ of the point in $h^{-1}(y)$ is an arc cyclic subset,
then since $h$ is an open map, $h(U)$ is a neighborhood of $y$ and Corollary
2.3 implies that it is an arc cyclic subset of $X_{1}$. In particular, if
$X_{2}$ is arc cyclic, then $X_{1}=h(X_{2})$ is arc cyclic.

∎

When $h:(X,R)\to(Y,S)$ is a surjective, continuous map of compact topological
tournaments, we will call it a _quotient map_.

Notice that if $(X,R)$ is a wac tournament and $(Y,S)$ is any arc cyclic,
compact topological tournament on a Cantor set, e.g. a tournament obtained
from a closed game subset on an infinite, compact topological group, then we
can perform a topological lexicographic product with base $(X,R)$ and with
$(Y_{x},S_{x})=(Y,S)$ for every isolated point $x\in X$. In the resulting
compact tournament every point is a cycle point and the tournament maps onto
$(X,R)$. If the answer to Question 7.11 is affirmative, then the lift is
metrizable and so $(X,R)$ is metrizable as well. Thus, it would then follow
that for any wac tournament the underlying space is metrizable and so has only
countably many isolated points.

We will later see that there exist wac tournaments which are not locally arc
cyclic. We pause to consider the stronger condition. Recall that for a
tournament $(X,R)$ a subset $A$ of $X$ is arc cyclic when every arc contained
in $A$ is contained in a $3-$cycle in $X$. Since the $3-$cycle need not be
contained in $A$, this condition is weaker than the assumption that the
restriction $R|A$ is an arc cyclic tournament on $A$.

######  Proposition 10.3.

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

(a) If $A$ is an arc cyclic subset, then its closure $\overline{A}$ is arc
cyclic. Any subset of $A$ is arc cyclic.

(b) If $\\{A_{i}\\}$ is a monotone family of arc cyclic subsets, then the
union $\bigcup_{i}A_{i}$ is arc cyclic.

(c) If $A$ is an arc cyclic subset and $x\in A$ has an arc cyclic
neighborhood, then there exists a neighborhood $U$ of $x$ such that $A\cup U$
is arc cyclic.

(d) Any arc cyclic subset, e.g. a singleton set, is contained in a maximal arc
cyclic subset which is a closed subset of $X$. If $A$ is a maximal arc cyclic
subset and $x\in A$ has an arc cyclic neighborhood, then $x$ is in the
interior of $A$.

###### Proof.

(a): If $x\rightharpoonup x^{\prime}$ with $x,x^{\prime}\in\overline{A}$, then
there is a net $\\{(x_{i},x_{i}^{\prime})\in R^{\circ}\cap(A\times A)$
converging to $(x,x^{\prime})$. By arc cyclicity, there exists
$x_{i}^{\prime\prime}\in X$ such that
$\\{x_{i},x^{\prime}_{i},x_{i}^{\prime\prime}\\}$ is a $3-$cycle. By
compactness we may assume $\\{x_{i}^{\prime\prime}\\}$ converges to
$x^{\prime\prime}\in X$. Since $R$ is closed,
$(x^{\prime},x^{\prime\prime}),(x^{\prime\prime},x)\in R$ and so by asymmetry,
$x^{\prime\prime}$ cannot equal either $x$ or $x^{\prime}$. Hence,
$\\{x,x^{\prime},x^{\prime\prime}\\}$ is a $3-$cycle. The subset result is
obvious.

(b): If $x\rightharpoonup x^{\prime}$ with $x,x^{\prime}\in\bigcup_{i}A_{i}$,
then for some $i$, $x,x^{\prime}\in A_{i}$ by monotonicity. Since $A_{i}$ is
arc cyclic there exists $x^{\prime\prime}$ such that
$\\{x,x^{\prime},x^{\prime\prime}\\}$ is a $3-$cycle.

(c): Suppose that $V_{1}$ is a neighborhood of the diagonal so that $V_{1}(x)$
is an arc cyclic subset. By Lemma 7.2 there exists a symmetric diagonal
neighborhood $V_{2}\subset V_{1}$ such that if $\\{x,y,z\\}$ is a $3-$cycle
with $(x,y)\in V_{2}$, then $(x,z),(y,z)\in V_{1}$ and so $(x,z)\not\in V_{1}$
or $(y,z)\not\in V_{1}$ implies $(x,y)\not\in V_{2}$. By Theorem 5.8 there
exists a symmetric neighborhood $V$ of the diagonal such that
$(x,y),(y,z),(x,z)\not\in V_{2}$ implies that $\\{V(x),V(y),V(z)\\}$ is a
thickening of $\\{x,y,z\\}$. I claim that $V(x)\cup A$ is an arc cyclic
subset.

If $x_{1},y_{1}\in V_{1}(x)$, then there exists $z_{1}$ such that
$\\{x_{1},y_{1},z_{1}\\}$ is a $3-$cycle, because $V_{1}(x)$ is an arc cyclic
set. Hence, $V(x)\cup(A\cap V_{1}(x))\subset V_{1}(x)$ is an arc cyclic
subset. Now suppose that $y\in A\setminus V_{1}(x)$. Because $A$ is an arc
cyclic subset, there exists $z$ such that $\\{x,y,z\\}$ is a $3-$cycle. It
cannot happen that $(x,z)\in V_{2}$ or $(y,z)\in V_{2}$ because either would
imply $(x,y)\in V_{1}$. Hence, $(x,y),(y,z),(x,z)\not\in V_{2}$. Hence,
$\\{V(x),V(y),V(z)\\}$ is a thickening of $\\{x,y,z\\}$. Thus, if $x_{1}\in
V(x)$, $\\{x_{1},y,z\\}$ is a $3-$cycle. That is, $V(x)\cup A$ is an arc
cyclic subset.

(d): Immediate from (b) and Zorn’s Lemma. By maximality, (a) implies that a
maximal arc cyclic set is closed. If $x\in A$ has a neighborhood which is an
arc cyclic subset, then, by maximality (c) implies that $A$ contains some
neighborhood of $x$.

∎

######  Corollary 10.4.

If $(X,R)$ be a locally arc cyclic topological tournament, then any maximal
arc cyclic subset is clopen.

###### Proof.

This is immediate from Proposition 10.3(d).

∎

######  Corollary 10.5.

If $h:(X_{2},R_{2})\to(X_{1},R_{1})$ is a quotient map such that $h^{-1}(y)$
is an arc cyclic subset of $X_{2}$ for all $y\in X_{1}$, then $(X_{2},R_{2})$
is locally arc cyclic if and only if $(X_{1},R_{1})$ is locally arc cyclic.

###### Proof.

Note first that if $h^{-1}(y)$ is an arc cyclic subset and
$\\{x,x^{\prime},x^{\prime\prime}\\}$ is a $3-$cycle with $x,x^{\prime}\in
h^{-1}(y)$, then by Proposition 2.2, $x^{\prime\prime}\in h^{-1}(y)$. Thus the
restriction $R_{2}|h^{-1}(y)$ is an arc cyclic tournament.

If $U$ is an open subset of $X_{1}$, then Corollary 2.3 and the restriction
assumptions imply that $U$ is an arc cyclic subset of $X_{1}$ if and only if
$h^{-1}(U)$ is an arc cyclic subset of $X_{2}$. Thus, if $(X_{1},R_{1})$ is
locally arc cyclic, then $(X_{2},R_{2})$ is because $h$ is surjective.

If $y\in X_{1}$, then the arc cyclic subset $h^{-1}(y)$ is contained in a
maximal arc cyclic subset $A$ of $X_{2}$. By Corollary 10.4 $A$ is clopen if
$(X_{2},R_{2})$ is locally arc cyclic. It then follows that there exists an
open subset $U$ of $X$ with $y\in U$ and such that $h^{-1}(U)\subset A$. Thus,
$h^{-1}(U)$ is arc cyclic. By Corollary 2.3 again it follows that $U$ is arc
cyclic. Hence, $(X_{1},R_{1})$ is locally arc cyclic.

∎

For a compact topological tournament $(X,R)$ we define the subsets
$Q,Q^{\circ}$ of $X\times X\times X$;

(10.1) $\displaystyle\begin{split}Q\ =\ \\{(x,y,z):&(x,z),(z,y)\in R\\}\
\cup\\\ &\\{(x,y,z):(y,z),(z,x)\in R\\},\\\ Q^{\circ}\ =\
\\{(x,y,z):&(x,z),(z,y)\in R^{\circ}\\}\ \cup\\\ &\\{(x,y,z):(y,z),(z,x)\in
R^{\circ}\\}\end{split}$

We regard $Q$ and $Q^{\circ}$ as relations from $X\times X$ to $X$. Clearly,
$Q$ is closed and $Q^{\circ}$ is open.

The following properties are easy to check.

(10.2) $\displaystyle\begin{split}\\{x\\}\ =\ \ &Q(x,x),\qquad\emptyset\ =\
Q^{\circ}(x,x).\\\ \\{x,y\\}\ \subset\ \ &Q(x,y)\ =\
Q(\\{x,y\\}\times\\{x,y\\}).\\\ Q(x,y)\setminus\\{x,y\\}\ =\ \
&Q^{\circ}(x,y)\ =\ Q^{\circ}(\\{x,y\\}\times\\{x,y\\}).\end{split}$

For any subset $A$ of $X,\ \ Q^{\circ}(A\times A)$ is open. $Q(A\times A)\ =\
Q^{\circ}(A\times A)\cup A$. So, if $A$ is closed, then $Q(A\times A)$ is
closed and $Q(A\times A)\setminus A=Q^{\circ}(A\times A)\setminus A$ is open.

######  Lemma 10.6.

$Q^{\circ}(\overline{A}\times\overline{A})=Q^{\circ}(A\times A)$.

###### Proof.

If $R^{\circ}(z)$ and $R^{\circ-1}(z)$ meet $\overline{A}$, then they meet
$A$.

∎

######  Lemma 10.7.

For a compact tournament $(X,R)$ assume $x\rightharpoonup y$ in $X$.

If $y$ is left balanced or $x$ is right balanced, then the open set
$R^{\circ}(x)\cap R^{\circ-1}(y)$ is nonempty and is contained in
$Q^{\circ}(x,y)$.

###### Proof.

The open set $R^{\circ}(x)$ contains $y$. If $y$ is left balanced, then
$R^{\circ-1}(y)$ meets $R^{\circ}(x)$ and the intersection is clearly
contained in $Q^{\circ}(x,y)$. Similarly, if $x$ is right balanced.

∎

######  Definition 10.8.

For a compact topological tournament $(X,R)$, a subset $A$ of $X$ is called
_$Q$ invariant_ when $Q(A\times A)\subset A$, or, equivalently, when
$Q^{\circ}(A\times A)\subset A$.

Clearly, every singleton subset is $Q$ invariant.

######  Proposition 10.9.

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

  * (a)

A subset $A$ is $Q$ invariant if and only if for all $z\in X\setminus A$,
either $A\subset R(z)$ or $A\subset R^{-1}(z)$.

  * (b)

If $A$ is $Q$ invariant, then the closure $\overline{A}$ is $Q$ invariant.

  * (c)

If $\\{A_{i}\\}$ is a family of $Q$ invariant sets, then the intersection
$\bigcap\\{A_{i}\\}$ is $Q$ invariant.

  * (d)

If $\\{A_{i}\\}$ is a monotone family of $Q$ invariant sets, then the union
$\bigcup\\{A_{i}\\}$ is $Q$ invariant.

  * (e)

If $(X,R)$ is arc cyclic and $A$ is a $Q$ invariant set, then the restriction
$(A,R|A)$ is arc cyclic.

###### Proof.

(a) is obvious and (b) follows from Lemma 10.6.

(c) is obvious, and for (d) if $x,y\in\bigcup\\{A_{i}\\}$, then monotonicity
implies that for some $i$, $x,y\in A_{i}$. Hence, $Q(x,y)\subset A_{i}$.

(e): If $(x,y)$ is an arc in $A$, then there is a $3-$cycle $\\{x,y,z\\}$ in
$X$. By $Q$ invariance, $z\in A$.

∎

######  Theorem 10.10.

If $h:(X_{2},R_{2})\to(X_{1},R_{1})$ is continuous map of compact topological
tournaments, then for every $Q$ invariant subset $B$ of $X_{1}$, the pre-image
$h^{-1}(B)$ is a $Q$ invariant subset of $X_{2}$. In particular for every
$y\in X_{1}$, the set $h^{-1}(y)$ is a closed, $Q$ invariant subset of
$X_{2}$.

If $A\subset X_{2}$ is $Q$ invariant and $h$ is surjective, then the image
$h(A)$ is a $Q$ invariant subset of $X_{1}$. If $A$ is a proper subset of
$X_{2}$ and $X_{1}$ has no initial nor terminal point, then $h(A)$ is a proper
subset of $X_{1}$.

Conversely, if $A$ is a nonempty, closed subset of a compact space $X$, then
$1_{X}\cup(A\times A)$ is a closed equivalence relation, with $\pi_{A}:X\to
X/A$ the quotient map to the space with $A$ smashed to a point. If $A$ is
clopen, then the point $\pi_{A}(A)$ is isolated in $X/A$. If $A$ is $Q$
invariant for the topological tournament $R$ on $X$, then
$R_{A}=(\pi_{A}\times\pi_{A})(R)$ is a topological tournament on $X/A$ and
$\pi_{A}$ maps $R$ to $R_{A}$.

###### Proof.

If $x,x^{\prime}\in h^{-1}(B)$ and $z\in X_{2}$ with
$x^{\prime}\rightharpoonup z\rightharpoonup x$, then
$h(x^{\prime})\ \underline{\rightharpoonup}\ h(z)\
\underline{\rightharpoonup}\ h(x)$ and so $h(z)\in B$ because $B$ is $Q$
invariant. Hence, $z\in h^{-1}(B)$.

Now assume $h$ is surjective. If $x\in X_{2}\setminus A$ and $z=h(x)$, then
since $A$ is $Q$ invariant either $A\subset R_{2}(x)$ or $A\subset
R_{2}^{-1}(x)$. Hence, $h(A)\subset R_{1}(z)$ or $h(A)\subset R_{1}^{-1}(z)$.
If $h(A)=X_{1}$, then $z=h(x)$ is either an initial or terminal point for
$X_{1}$. If $z\in X_{1}\setminus h(A)$, then there exists $x$ such that
$h(x)=z$ and, necessarily, $x\in X_{2}\setminus A$. So $h(A)\subset R_{1}(z)$
or $h(A)\subset R_{1}^{-1}(z)$ implies that $h(A)$ is $Q$ invariant.

If $A$ is a closed, $Q$ invariant subset, then (a) of Proposition 10.9 implies
that $R_{A}$ is a tournament. It is closed by compactness. If
$A=\pi_{A}^{-1}(\pi_{A}(A))$ is clopen, it follows that the point $\pi_{A}(A)$
is clopen by definition of the quotient topology.

∎

Now let $(X,R)$ be a compact tournament and let $A_{0}$ be a non-trivial
subset of $X$. Inductively, define $A_{n+1}=Q(A_{n}\times A_{n})$. If $A_{0}$
is closed, then, inductively, we see that $A_{n}$ is closed for all $n$.
Clearly, if $A_{0}$ is $Q$ invariant, then $A_{n}=A_{0}$ for all $n$.

######  Proposition 10.11.

For $(X,R)$ a compact tournament and $A_{0}$ be a non-trivial subset of $X$,
$\\{A_{n}\\}$ is a non-decreasing sequence. The union $\bigcup_{n}\\{A_{n}\\}$
is the smallest $Q$ invariant subset of $X$ which contains $A_{0}$, its
closure $\overline{\bigcup_{n}\\{A_{n}\\}}$ is the smallest closed, $Q$
invariant subset of $X$ which contains $A_{0}$.

If $A_{0}$ is closed, then all of the $A_{n}$’s are closed and each
$A_{n}\setminus A_{0}$ is open.

###### Proof.

Since $A$ is contained in $Q(A\times A)$, it follows that $\\{A_{n}\\}$ is an
increasing sequence.

If $z\in Q(x,y)$ with $x,y\in\bigcup_{n}\\{A_{n}\\}$, then for some $n$
$x,y\in A_{n}$ and so $z\in A_{n+1}$. It follows that $\bigcup_{n}\\{A_{n}\\}$
is $Q$ invariant and is clearly the smallest $Q$ invariant subset of $X$ which
contains $A_{0}$. Hence,$\overline{\bigcup_{n}\\{A_{n}\\}}$ is the smallest
closed, $Q$ invariant subset of $X$ which contains $A_{0}$.

If $A$ is closed, then $A$ is contained in the closed set $Q(A\times A)$ and
$Q(A\times A)\setminus A=Q^{\circ}(A\times A)\setminus A$ so that $Q(A\times
A)\setminus A$ is open.

Thus, if $A_{0}$ is closed, then $\\{A_{n}\\}$ is an increasing sequence of
closed sets and each $A_{n}\setminus A_{n-1}$ is open. Hence, $A_{n}\setminus
A_{0}=\bigcup_{i=1}^{n}A_{i}\setminus A_{i-1}$ is open.

∎

######  Lemma 10.12.

Let $(X,R)$ be a compact tournament and $x$ be an isolated point or a cycle
point of $X$. For all $n\geq 3$ if $x$ is contained in the closure of
$A_{n-3}$, then $x$ is contained in the interior of $A_{n}$.

In particular, if $(X,R)$ is wac, then the closure of of $A_{n-3}$ is
contained in the interior of $A_{n}$ for all $n\geq 3$.

###### Proof.

If $n>3$ and we define $A_{0}^{\prime}=A_{n-3}$ then in the associated
sequence $A_{n}^{\prime}$ we have $A_{3}^{\prime}=A_{n}$ and so it suffices to
prove the result for $n=3$.

Let $x\in\overline{A_{0}}$. If $x$ is isolated, then is in the interior of
$A_{0}$ and so of that of $A_{3}$.

Assume $x$ is a cycle point and choose $y\in A_{0}$ with $y\not=x$. Assume
that $y\rightharpoonup x$. There exists a $3-$cycle
$\\{x,x^{\prime},x^{\prime\prime}\\}$ contained in $R^{\circ}(y)$. By Theorem
5.8, there exists $\\{U_{x},U_{x^{\prime}},U_{x^{\prime\prime}}\\}$ a
thickening contained in $R^{\circ}(y)$. Let $z\in A_{0}\cap U_{x}$. Because
$U_{x^{\prime\prime}}\subset R^{-1}(z)\cap R(y)$ it follows that
$U_{x^{\prime\prime}}\subset A_{1}$. Since
$\\{z,z^{\prime},x^{\prime\prime}\\}$ is a $3-$cycle for any $z^{\prime}\in
U_{x^{\prime}}$ it follows that $U_{x^{\prime}}\subset A_{2}$. Similarly,
$U_{x}\subset A_{3}$. Hence, $x$ is in the interior of $A_{3}$.

∎

######  Theorem 10.13.

If $(X,R)$ is a wac tournament, then any non-trivial $Q$ invariant subset $A$
of $X$ is clopen.

If $A_{0}$ is a non-trivial closed subset of $X$, then for the increasing
sequence $\\{A_{n}\\}$ of closed sets with $A_{n}=Q(A_{n-1}\times A_{n-1})$
for $n\geq 3,\ \ A_{n}$ is clopen. Furthermore, for sufficiently large $n$,
$A_{n}$ is $Q$ invariant and so equals $\overline{\bigcup_{n}\\{A_{n}\\}}$.

###### Proof.

If $A$ is $Q$ invariant, then with $A_{0}=A$, $A_{3}=A$. So Lemma 10.12
implies that the closure of $A$ is contained in the interior of $A$, i.e. $A$
is clopen.

If $A_{0}$ is an arbitrary non-trivial closed set, then for the sequence of
closed sets, $A_{n}$, each $A_{n}\setminus A_{0}$ is open by Proposition
10.11. But from Lemma 10.12, for $n\geq 3$ $A_{n}=A_{n}\setminus A_{0}\cup
IntA_{3}$ and so it is open and therefore clopen.

The union $\bigcup_{n}\\{A_{n}\\}$ is $Q$ invariant and so it is clopen and
equals its closure. The sequence $\\{A_{n}\\}$ is an open cover of the closed
set $\bigcup_{n}\\{A_{n}\\}$ and so it has a finite subcover. Hence, for large
enough $n$, $A_{n}=\bigcup_{n}\\{A_{n}\\}=\overline{\bigcup_{n}\\{A_{n}\\}}$.

∎

Recall that $(X,R)$ is almost wac when every point of $X$ is either isolated,
initial, terminal or a cycle point.

######  Addendum 10.14.

Let $(X,R)$ be an almost wac tournament. Assume that $A$ is a non-trivial $Q$
invariant subset $A$ of $X$. If for $x$ terminal or initial, either $x\in A$
or $x\not\in\overline{A}$, then $A$ is clopen. In particular, if $A$ is
closed, then it is clopen.

###### Proof.

By Lemma 10.12 if $x$ is an isolated point or cycle point with
$x\in\overline{A}$, then $x$ is in the interior of $A$. By assumption on
$(X,R)$ this applies to every point which is not initial or terminal. If $M$
is a terminal point in $\overline{A}$ then by assumption $M\in A$. If $x\in
A\setminus\\{M\\}$, then for every point $x^{\prime}\in
R^{\circ}(x)\setminus\\{M\\}$ we have $x\rightharpoonup
x^{\prime}\rightharpoonup M$ and so $x^{\prime}\in A$ because $A$ is $Q$
invariant. Hence, $M\in R^{\circ}(x)\subset A$ and so $M$ is in the interior
of $A$. Similarly, for an initial point $m$.

∎

Remark: Note that if $M$ is a terminal point, then $X\setminus\\{M\\}$ is a
proper $Q$ invariant subset which is not closed unless $M$ is isolated.

######  Definition 10.15.

For $(X,R)$ a non-trivial compact tournament, a subset $A$ is a _maximal $Q$
invariant subset_, when it is a proper, closed, $Q$ invariant subset of $X$
such that $X$ is the only closed, $Q$ invariant subset which properly contains
$A$, i.e. $A\subset A^{\prime}$ with $A^{\prime}$ a closed, $Q$ invariant
subset, then either $A^{\prime}=A$ or $A^{\prime}=X$.

Since $\emptyset$ is contained in every singleton and $X$ is non-trivial,
$\emptyset$ is never maximal.

######  Theorem 10.16.

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

If $A\subset X$ is maximal $Q$ invariant of $X_{2}$ , then either the image
$h(A)$ equals $X_{1}$ or else $h(A)$ is a maximal $Q$ invariant of $X_{1}$. In
the latter case, $A=h^{-1}(h(A))$. If $X_{1}$ has no initial or terminal
point, then $h(A)$ is a maximal $Q$ invariant of $X_{1}$.

If $B$ is a maximal $Q$ invariant subset $B$ of $X_{1}$, then any closed $Q$
invariant subset $A$ of $X_{2}$ which properly contains the pre-image
$h^{-1}(B)$ maps onto $X_{1}$, i.e. $h(A)=X_{1}$. In particular, if $X_{1}$
has no initial or terminal point, then $h^{-1}(B)$ is a maximal $Q$ invariant
subset of $X_{2}$.

###### Proof.

Assume $A$ is maximal. If $B$ is a proper, closed, $Q$ invariant subset of
$X_{1}$ which contains $h(A)$, then $h^{-1}(B)$ is a proper, closed $Q$
invariant subset of $X_{2}$ which contains $A$ and so equals $A$. Applied to
$B=h(A)$ when it is a proper subset of $X_{1}$, we obtain $A=h^{-1}(h(A))$. If
$X_{1}$ has no initial or terminal point, then by Theorem 10.10, $h(A)$ is a
proper subset of $X_{1}$.

If $A$ is a closed, $Q$ invariant set which properly contains $h^{-1}(B)$,
then $h(A)$ is a closed, $Q$ invariant set of $X_{1}$ which properly contains
$B$. Hence, by maximality $h(A)=X_{1}$. If $X_{1}$ has no initial or terminal
point, then by Theorem 10.10 again, $A$ cannot be a proper subset of $X_{2}$.
Hence, $h^{-1}(B)$ is maximal.

∎

######  Theorem 10.17.

For $(X,R)$ a non-trivial, wac tournament, every proper $Q$ invariant subset
is contained in a maximal $Q$ invariant subset.

###### Proof.

By Theorem 10.13 every non-trivial $Q$ invariant subset is clopen. If
$\\{A_{i}\\}$ is a monotone family of proper non-trivial $Q$ invariant
subsets, then by Proposition 10.9(d) the union $\bigcup\\{A_{i}\\}$ is $Q$
invariant and so is clopen. The cover $\\{A_{i}\\}$ has a finite subcover and
so, by monotonicity the union equals $A_{i}$ for some $i$. This implies that
the union is proper. It follows that if $A$ is a proper non-trivial $Q$
invariant subset, we can apply Zorn’s Lemma to the family of proper $Q$
invariant subsets which contain $A$ and so obtain a maximal element.

For a singleton $\\{x\\}$ either it is contained in a non-trivial $Q$
invariant subset which is then contained in a maximal subset, or else the
singleton itself is maximal.

∎

######  Definition 10.18.

A tournament $(Y,P)$ is called a _prime topological tournament_ when it is
compact and non-trivial and every singleton subset is a maximal $Q$ invariant
subset. That is, $Y$ itself is the only closed, non-trivial $Q$ invariant
subset.

If $(X,R)$ is a compact topological tournament, then a surjective map
$\pi:(X,R)\to(Y,P)$ of topological tournaments with $(Y,P)$ prime is called a
_prime quotient map_ and $(Y,P)$ is called a _prime quotient_ for $(X,R)$.

An arc $(Y_{0},P_{0})$, i.e. a tournament on a two point set, is prime. When
$(X,R)$ admits a quotient map onto an arc then we say it has an _arc
quotient_.

######  Proposition 10.19.

If $(X,R)$ has an initial or terminal point which is either isolated or
contained in a non-trivial, proper, closed $Q$ invariant subset, then $(X,R)$
has an arc quotient. In particular, if $(X,R)$ is wac and has an initial or
terminal point, then it has an arc quotient.

###### Proof.

If $M$ is an isolated terminal point, then $M\mapsto 1$ and $x\mapsto 0$ for
all $x\not=M$ defines a quotient map to the arc on $\\{0,1\\}$ with
$0\rightharpoonup 1$. If $A$ is a non-trivial, proper, closed $Q$ invariant
subset which contains $M$, then by Addendum 10.14, $A$ is clopen. $x\mapsto 1$
for $x\in A$ and $x\mapsto 0$ otherwise defines a quotient map to the arc.
Similarly for an initial point $m$.

An initial or terminal point for a wac tournament is isolated by Theorem
10.2(d).

∎

######  Theorem 10.20.

Assume $(X,R)$ is a non-trivial almost wac tournament with no arc quotient.

(a) Every non-trivial, proper, closed $Q$ invariant subset is contained in a
maximal $Q$ invariant subset which is clopen and does not contain an initial
or terminal point. In particular, if $x$ is an initial or terminal point, then
$\\{x\\}$ is a maximal $Q$ invariant set.

(b) $h:(X,R)\to(X_{1},R_{1})$ is a quotient map with $(X_{1},R_{1})$ non-
trivial, then $(X_{1},R_{1})$ is almost wac. If $x$ is a terminal (or initial)
point of $X$, then $h(x)$ is terminal (resp. initial) point of $X_{2}$. If $y$
is a terminal (or initial) point of $X_{1}$, then $h^{-1}(y)$ is a singleton
$\\{x\\}$ with $x$ terminal (resp. initial) point of $X$.

If $A$ is a proper, closed $Q$ invariant subset of $X$ , then $h(A)$ is a
proper, closed $Q$ invariant subset of $X_{1}$.

###### Proof.

(a): If $X$ has no initial or terminal point then it is wac and we apply
Theorem 10.17 directly. We may assume that $X$ has a terminal point $M$.

Proposition 10.19 implies that an initial or terminal point is not contained
in a non-trivial proper, closed $Q$ invariant set since $(X,R)$ does not admit
an arc quotient. In particular, an initial or terminal point is not isolated.
Hence, for such a point $x,\ \ \\{x\\}$ is a maximal $Q$ invariant subset.

As in the proof of Theorem 10.17 we consider $\mathcal{A}=\\{A_{i}\\}$ a
monotone family of proper non-trivial, closed $Q$ invariant subsets. By
Addendum 10.14 each $A_{i}$ is clopen. By Proposition 10.9(b) and (d) the
closure of the union $\overline{\bigcup A_{i}}$ is $Q$ invariant. When we show
it is proper, we can apply Zorn’s Lemma as in Theorem 10.17. Assume instead
that $\overline{\bigcup_{i}A_{i}}=X$. Fix $y\in\bigcap_{i}A_{i}$.

Let $x$ be a point of $\overline{\bigcup_{i}A_{i}}$.

If $x\in X$ is isolated, then $x\in\overline{\bigcup_{i}A_{i}}$ implies
$x\in\bigcup_{i}A_{i}$. If $x$ is a cycle point, we may assume
$y\rightharpoonup x$. We follow the proof of Lemma 10.12. There exists a
$3-$cycle $\\{x,x^{\prime},x^{\prime\prime}\\}$ contained in $R^{\circ}(y)$.
By Theorem 5.8, there exists $\\{U_{x},U_{x^{\prime}},U_{x^{\prime\prime}}\\}$
a thickening contained in $R^{\circ}(y)$. For some $i$ there exists $z\in
A_{i}\cap U_{x}$. It then follows as in the Lemma that $U_{x}\subset A_{i}$
and, in particular, $x\in A_{i}$. Thus, the union contains every point of $X$
except for terminal and initial points.

Observe first that the terminal point $M$ is not in $A_{i}$ for any $i$ since
these are proper non-trivial, closed $Q$ invariant subsets.

Since $M$ is terminal, $M\in R^{\circ}(y)$. Define the clopen sets

$\begin{split}B_{i}=R^{\circ}(y)\setminus A_{i}=R(y)\setminus A_{i},\\\
B^{\prime}_{i}=R^{\circ-1}(y)\setminus A_{i}=R^{-1}(y)\setminus
A_{i}.\end{split}$

,

Because the family $\mathcal{A}=\\{A_{i}\\}$ is monotone, it is directed by
inclusion.

For all $i,$ $M\in B_{i}$. If $m$ is an initial point, then $m\in
B_{i}^{\prime}$ for all $i$. If there is no initial point, then eventually
$B^{\prime}_{i}$ is empty because then $\\{A_{i}\\}$ is a covering of
$R^{-1}(y)$ and so has a finite subcover. By monotonicity $R^{-1}(y)$ will
then be contained in $A_{i}$ for some $i$.

Because $A_{i}$ is $Q$ invariant we have for every $z\in B_{i}$ either
$A_{i}\subset R^{\circ}(z)$ or $A_{i}\subset R^{\circ-1}(z)$. I claim that for
some $i_{0},\ \ A_{i}\subset R^{\circ-1}(z)$ for all $z\in B_{i}$ and for all
$A_{i}\supset A_{i_{0}}$.

If not, then for a cofinal collection of $A_{i}$’s there exists $z_{i}\in
B_{i}$ such that $z_{i}\rightharpoonup x$. The only possible limit points of
this net are $M$ or an initial point. However, an initial point does not lie
in the closed set $R(y)$. Hence, the net $\\{z_{i}\\}$ (indexed by the cofinal
subset of the monotone family $\mathcal{A}$) converges to $M$. Since
$z_{i}\rightharpoonup x$, this would yield $M\rightharpoonup x$, violating the
condition that $M$ be terminal. It then follows that for $A_{i}\supset
A_{i_{0}},\ \ x^{\prime}\rightharpoonup z$ for all $x^{\prime}\in A_{i}$ and
$z\in B_{i}$.

Now I claim that for some $i_{1}$ with $A_{i_{1}}\supset A_{i_{0}}$,
$z^{\prime}\rightharpoonup z$ for all $z^{\prime}\in B^{\prime}_{i}$ and $z\in
B_{i}$. This is vacuously true if there is no initial point and so
$B^{\prime}_{i}$ is eventually empty. If not, then we can choose
$(z_{i},z^{\prime}_{i})\in R^{\circ}\cap(B_{i}\times B^{\prime}_{i})$, indexed
again by a cofinal subset of $\mathcal{A}$. The limit is $(M,m)$ with $m$ an
initial point. However, this pair is in $R^{\circ-1}$ rather than in $R$.

Thus, for $A_{i}\supset A_{i_{1}}$ we have $x\rightharpoonup z$ for all
$(x,z)\in(X\setminus B_{i})\times B_{i}$. This implies that such $B_{i}$’s are
proper clopen $Q$ invariant subsets which contain $M$. This contradicts the
assumption that $(X,R)$ has no arc quotient.

This contradiction implies that $\overline{\bigcup_{i}A_{i}}$ is a proper
subset of $X$. So, at long last, we may apply Zorn’s Lemma and show that every
non-trivial, proper, closed $Q$ invariant subset is contained in a maximal $Q$
invariant subset. Since the maximal $Q$ invariant subsets are closed, they are
clopen by Addendum 10.14.

(b): By Theorem 10.2(f), $(X_{1}.R_{1})$ is almost wac when $(X,R)$ is. Since
$(X,R)$ does not have an arc quotient, $(X_{1},R_{1})$ does not. Hence an
initial point or terminal point in $X_{1}$ is not isolated. It follows from
Theorem 10.2(f) again that the pre-image of a terminal point (or initial
point) is a singleton terminal point (resp. a singleton initial point).

If $A$ is a proper, closed $Q$ invariant subset of $X$, then by Theorem 10.10
$h(A)$ is a $Q$ invariant subset of $X_{1}$. If $A$ is a singleton, then since
$X_{1}$ is non-trivial, $h(A)$ is a proper subset. So we may assume $A$ is
non-trivial and so it is clopen by Addendum 10.14. If $h(A)=X_{1}$, then the
proof of Theorem 10.10 shows that for $x\not\in A,\ \ h(x)$ is either an
initial or terminal point of $X_{1}$. We have seen that the pre-image of an
initial or terminal point is a singleton. This would imply that $X\setminus A$
consists of at least one and at most two points and these are isolated. Thus,
$X$ would have an isolated initial or terminal point and so $(X,R)$ would have
an arc quotient.

∎

######  Theorem 10.21.

(a) A compact, non-trivial tournament $(Y,P)$ is prime if and only if whenever
$h:(Y,P)\to(Z,T)$ is a quotient map with $(Z,T)$ non-trivial, $h$ is a
homeomorphism and so is an isomorphism from $(Y,P)$ to $(Z,T)$.

(b) Assume that $h:(X_{1},R_{1})\to(X_{2},R_{2})$ is a quotient map. Let
$(Y,P)$ is a prime tournament such that either $Y$ had no initial or terminal
point, or else $(Y,P)$ is almost wac but not an arc. If
$\pi_{1}:(X_{1},R_{1})\to(Y,P)$ is a quotient map, then $\pi_{1}$ factors
through $h$ to uniquely define the continuous surjection $\pi_{2}:X_{2}\to Y$
such that $\pi_{2}\circ h=\pi_{1}$. Furthermore,
$\pi_{2}:(X_{2},R_{2})\to(Y,P)$ is a quotient map.

(c) If a compact tournament $(X,R)$ admits a prime quotient map
$\pi:(X,R)\to(Y,P)$ such that either $(Y,P)$ has no initial nor terminal point
or else $(Y,P)$ is almost wac but not an arc, then $\pi$ is unique up to
isomorphism. That is, if $\pi_{1}:(X,R)\to(Y_{1},P_{1})$ is a prime quotient
map, then there exists a homeomorphism $h:Y_{1}\to Y$ such that
$\pi=h\circ\pi_{1}$ and $h:(Y_{1},P_{1})\to(Y,P)$ is a tournament isomorphism.

###### Proof.

(a): Assume $(Y,P)$ is prime. For $y\in Y$, $\\{y\\}$ is a maximal $Q$
invariant subset and so by Theorem 10.16, $h(y)\not=Z$ (because $Z$ is not
trivial) implies $y=h^{-1}(h(y))$ and so $h$ is bijective. By compactness it
is a homeomorphism and so is an isomorphism from $(Y,P)$ to $(Z,T)$.

If $(Y,P)$ is not prime, then it contains a proper, closed $Q$ invariant
subset $A$. The projection $\pi_{A}$ obtained by smashing $A$ to a point as in
Theorem 10.10 provides a quotient map on $(Y,P)$ which is not an isomorphism.

(b): For $y\in X_{2},\ h^{-1}(y)$ is closed and $Q$ invariant in $X_{1}$ and
so $\pi_{1}(h^{-1}(y))$ is closed and $Q$ invariant in $Y$, see Theorem 10.10.
Because $Y$ is prime either $\pi_{1}(h^{-1}(y))$ is a singleton or else
$\pi_{1}(h^{-1}(y))=Y$. By Theorem 10.10 $\pi_{1}(h^{-1}(x))$ is a proper
subset of $Y$ if it has no initial nor terminal point. By Theorem 10.20
$\pi_{1}(h^{-1}(x))$ is a proper subset of $Y$ if $(Y,P)$ is wac and is not an
arc (and so does not have an arc quotient by (a)). By assumption on $Y$ it
follows that $\pi_{1}(h^{-1}(y))$ is a singleton for every $y\in X_{2}$.
Hence, there is a, necessarily unique, map $\pi_{2}:X_{2}\to Y$ such that
$\pi_{2}\circ h=\pi_{1}$. By compactness $\pi_{2}$ is continuous. It is
clearly surjective and maps $R_{2}$ to $P$.

(c): The existence of the continuous surjection $h$ follows from (b). Since
$(Y_{1},P_{1})$ is prime, (a) implies that $h$ is an isomorphism.

∎

######  Proposition 10.22.

If $(Y,P)$ is a prime topological tournament, then $Y$ is totally
disconnected.

###### Proof.

From Theorem 5.1 it follows that any component $A$ of $Y$ is a $Q$ invariant
subset on which $P$ restricts to an order. If $A$ were not trivial, then it
would contain a non-trivial proper subinterval, $B$, which is itself a closed,
$Q$ invariant subset. This would imply that $(Y,P)$ is not prime.

∎

######  Theorem 10.23.

Assume that $\pi:(X,R)\to(Y,P)$ is a prime quotient map with $(X,R)$ a wac
tournament with no arc quotient.

  * (a)

The tournament $(Y,P)$ is wac with no initial or terminal point.

  * (b)

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

  * (c)

If $y$ is an isolated point of $Y$, then $\pi^{-1}(y)$ is a maximal $Q$
invariant subset of $X$.

  * (d)

If $y$ is a non-isolated point of $Y$, then $\pi^{-1}(y)$ is a singleton
subset $\\{x\\}$ which is a maximal $Q$ invariant subset of $X$. The point $x$
is non-isolated in $X$ and so the points $x$ and $y$ are cycle points.

###### Proof.

(a), (b): By Theorem 10.2(f) , $(Y,P)$ is wac. If it had an initial or
terminal point, then it would have an arc quotient by Proposition 10.19.
Composing with $\pi$ we would obtain an arc quotient map for $(X,R)$, contra
assumption.

Corollary 2.3 implies that $(Y,P)$ is arc cyclic when $(X,R)$ is.

(c), (d): The set $\pi^{-1}(y)$ is a proper $Q$ invariant subset of $X$ by
Theorem 10.10. Let $A$ be a proper $Q$ invariant subset of $X$ which contains
$\pi^{-1}(y)$. By Theorem 10.10, again, $\pi(A)$ is a proper $Q$ invariant
subset of $Y$ which contains $y$. Because $Y$ is prime, it equals $\\{y\\}$
and so $A=\pi^{-1}(y)$. Thus, $\pi^{-1}(y)$ is maximal. The remaining results
follow from Theorem 10.2(f) again.

∎

######  Theorem 10.24.

Assume $(X,R)$ is a almost wac tournament which does not have an arc quotient.
The tournament $(X,R)$ admits a prime quotient map, unique up to isomorphism.

If $(X,R)$ is arc cyclic, then it does not have an arc quotient and its prime
quotient is arc cyclic.

###### Proof.

If $(X,R)$ contains no proper, non-trivial $Q$ invariant subset, then it is
already prime and the identity is a prime quotient map. Otherwise Theorem
10.20(a) implies there exists a maximal, non-trivial $Q$ invariant subset $A$.
We can use Zorn’s Lemma to obtain a maximal collection
$\mathcal{A}=\\{A_{i}\\}$ of pair-wise disjoint, maximal non-trivial $Q$
invariant subsets which contains $A$. By Theorem 10.20(a) the sets $A_{i}$ are
clopen. Since $(X,R)$ does not have an arc quotient, no $A_{i}$ contains a
terminal or initial point by Proposition 10.19.

Claim: Let $d$ be any continuous pseudo-metric on $X$. For any $\epsilon>0$,
there are only finitely many $A_{i}$’s with $d-$diameter greater than
$\epsilon$.

###### Proof.

If not we can choose two sequences $\\{x_{n}\\},\\{y_{n}\\}$ such that

  * •

$x_{n},y_{n}\in A_{i_{n}}$ with $A_{i_{n}}$ disjoint from $A_{i_{m}}$ when
$n\not=m$.

  * •

$d(x_{n},y_{n})\geq\epsilon$

  * •

The sequence of pairs $\\{(x_{n},y_{n})\\}$ has a limit point $(x,y)$ with
$d(x,y)\geq\epsilon$ and so we may assume $x\in R^{\circ}(y)$.

First we eliminate the possibility that $y$ is initial and $x$ is terminal.
Were this so, then for any $z\in X\setminus\\{x,y\\}$, e.g. $z\in A$, we would
have $(x,y)\in R^{\circ}(z)\times R^{\circ-1}(z)$. Then for infinitely many
$n,\ \ (x_{n},y_{n})\in R^{\circ}(z)\times R^{\circ-1}(z)$ which implies $z\in
A_{i_{n}}$ since $A_{i_{n}}$ is $Q$ invariant. This is impossible since the
$A_{n}$’s are pairwise disjoint.

Now assume that $x$ is not terminal. We follow the proof of Lemma 10.12. As
$x$ is not initial because $y\rightharpoonup x$ and it is not isolated as it
is a limit point, it is a cycle point and we may choose a $3-$cycle
$\\{x,x^{\prime},x^{\prime\prime}\\}$ contained in $R^{\circ}(y)$. We thicken
$\\{y,x,x^{\prime},x^{\prime\prime}\\}$ to
$\\{U_{y},U_{x},U_{x^{\prime}},U_{x^{\prime\prime}}\\}$. Then for infinitely
many $n,\ \ (x_{n},y_{n})\in U_{x}\times U_{y}$ . For any $z\in
U_{x^{\prime\prime}}$, $z\in R^{\circ}(y_{n})\cap R^{\circ-1}(x_{n})$. Because
these $A_{n}$’s are $Q$ invariant, we have $z\in A_{n}$ for infinitely many
$n$. Again, this is impossible.

We use a similar argument if $y$ is not initial and so is a cycle point.

∎

Now define the equivalence relation
$E_{\mathcal{A}}=1_{X}\cup\bigcup_{i}\\{A_{i}\times A_{i}\\}$. If
$\\{(x_{k},y_{k})\\}$ is a net in $E_{A}$ converging to $(x,y)$, then either
for some cofinal set of indices $k$ the pairs $(x_{k},y_{k})$ lie in a single
$A_{i}\times A_{i}$, in which case $(x,y)\in E_{\mathcal{A}}$ since the
$A_{i}$’s are closed, or else eventually the sequence leaves any finite
collection of $A_{i}$’s. The Claim implies that for any continuous pseudo-
metric $d\ \ \lim_{n}\ d(x_{n},y_{n})=0$ and so $d(x,y)=0$. Since the
continuous pseudo-metrics generate the unique uniformity on $X$, see [11]
Chapter 6, it follows that $x=y$ and so $(x,y)\in E_{\mathcal{A}}$.

Now as in Theorem 10.10 define the quotient map $\pi$ from $X$ to
$X_{\mathcal{A}}$, the space of $E_{\mathcal{A}}$ equivalence classes with the
quotient topology. Let $R_{\mathcal{A}}=(\pi\times\pi)(R)$ and as before it is
a topological tournament with $\pi:(X,R)\to(X_{\mathcal{A}},R_{\mathcal{A}})$
a continuous surjective tournament map. Since each $A_{i}$ is proper,
$X_{\mathcal{A}}$ is non-trivial and by Theorem 10.20(b) it is almost wac.
Since $(X,R)$ has no arc quotient, neither does
$(X_{\mathcal{A}},R_{\mathcal{A}})$.

By definition of the quotient topology, $A_{i}=\pi^{-1}(\pi(A_{i}))$ clopen
implies that the point $\pi(A_{i})$ is an isolated point. Because $A_{i}$ is
maximal, Theorem 10.16 and Theorem 10.20(b) implies that $\\{\pi(A_{i})\\}$ is
maximal.

Now assume that $B$ is a proper, closed $Q$ invariant subset of
$X_{\mathcal{A}}$. If $B$ contains some $\pi(A_{i})$, then it equals
$\\{\pi(A_{i})\\}$ by maximality.

Now suppose that $B$ is disjoint from all of the $\pi(A_{i})$’s. Assume $B$ is
non-trivial, so that $\pi^{-1}(B)$ is a non-trivial proper $Q$ invariant
subset of $X$. Hence, it is contained in a maximal $Q$ invariant subset
$A^{\prime}$. It follows from Theorem 10.16 and Theorem 10.20(b) again that
$\pi(A^{\prime})$ is a maximal $Q$ invariant subset of $X_{\mathcal{A}}$. As
it contains $B$, it is not equal to any $\pi(A_{i})$ and so is disjoint from
all of them. It follows that $A^{\prime}$ is a non-trivial maximal $Q$
invariant subset of $X$ which is disjoint from all the $A_{i}$’s. This
contradicts the maximality of the family $\\{A_{i}\\}$. So we see that $B$ had
to be trivial.

Thus, every proper $Q$ invariant subset of $X_{\mathcal{A}}$ is a singleton
and so $(X_{\mathcal{A}},R_{\mathcal{A}})$ is prime.

Uniqueness up to isomorphism follows from Theorem 10.21.

If $(X,R)$ is arc cyclic, then every quotient is arc cyclic by Corollary 2.3.
Since the arc is not arc cyclic, $(X,R)$ does not have an arc quotient and its
prime quotient is arc cyclic.

∎

From the proof we obtain the following.

######  Addendum 10.25.

If $(X,R)$ is an almost wac tournament which does not admit an arc quotient,
then the maximal $Q$ invariant subsets of $X$ are the elements of
$\mathcal{A}=\\{A_{i}\\}$ together with the singletons $\\{x\\}$ for $x\in
X\setminus\bigcup_{i}\\{A_{i}\\}$. In particular, any two distinct maximal $Q$
invariant subsets are disjoint.

###### Proof.

Let $A^{\prime}$ be an arbitrary non-trivial maximal $Q$ invariant subset and
let $\mathcal{A}^{\prime}$ be a maximal collection of pair-wise disjoint non-
trivial maximal $Q$ invariant subsets which includes $A^{\prime}$. We obtain
the prime quotient map $\pi^{\prime}:X\to X_{\mathcal{A}^{\prime}}$ as before.
By Theorem 10.21 we obtain the homeomorphism $h:X_{\mathcal{A}^{\prime}}\to
X_{\mathcal{A}}$ such that $\pi=h\circ\pi^{\prime}$. It follows that
$(\pi^{\prime})^{-1}(h^{-1}(z))=\pi^{-1}(z)$ for all $z\in X_{\mathcal{A}}$.
Hence, $\mathcal{A}=\mathcal{A}^{\prime}$ and so $A^{\prime}\in\mathcal{A}$.
We saw in the proof above that for each $x\not\in\bigcup_{i}A_{i}$ the
singleton $\\{x\\}$ is maximal.

∎

Now we consider what happens when a wac tournament has an arc quotient.

######  Theorem 10.26.

If a wac tournament $(X,R)$ has an arc quotient, then any prime quotient of
$(X,R)$ is an arc.

###### Proof.

Suppose that $\pi_{0}:(X,R)\to(Y_{0},P_{0})$ and
$\pi_{1}(X,R)\to(Y_{1},P_{1})$ are prime quotient maps with $(Y_{0},P_{0})$ an
arc.

We first observe that $(Y_{1},P_{1})$ has an initial or a terminal point. Were
this not true then by Theorem 10.21(b) there would be a surjective map from
$(Y_{0},P_{0})$ onto $(Y_{1},P_{1})$. Since $|Y_{0}|=2$ and $Y_{1}$ is non-
trivial, the map would have to be an isomorphism and so $(Y_{1},P_{1})$ would
have both an initial and a terminal point.

Since $(Y_{1},P_{1})$ is wac, the existence of an initial or terminal point
implies that $(Y_{1},P_{1})$ has a quotient map onto an arc $(Y_{3},P_{3})$.
Since $(Y_{1},P_{1})$ is prime, this quotient map is an isomorphism and so
$(Y_{1},P_{1})$ is an arc.

∎

Thus, in any case, a wac tournament has a prime quotient, unique up to
isomorphism. However, when the tournament has an arc quotient, the quotient
map need not be unique up to isomorphism.

######  Definition 10.27.

Suppose that $L$ is a linear order on a non-trivial finite set $I$ so that
$(I,L)$ is a non-trivial finite, transitive tournament. A quotient map
$\pi:(X,R)\to(I,L)$ is called an _order quotient map_. It is called a _maximum
order quotient map_ when for each $i\in I$, the restriction $R|\pi^{-1}(i)$
does not have an arc quotient.

######  Theorem 10.28.

Assume that for a compact tournament $(X,R),\ \ \pi:(X,R)\to(I,L)$ and
$\pi_{1}:(X,R)\to(I_{1},L_{1})$ are order quotient maps with $\pi$ maximum.
There exists a surjective tournament map (i.e. an order-preserving surjection)
$h:(I,L)\to(I_{1},L_{1})$ such that $\pi_{1}=h\circ\pi$. In particular,
$|L|\geq|L_{1}|$.

If $\pi_{1}$ is also maximum, then $h$ is an isomorphism and so $|I|=|I_{1}|$.
Conversely, $h$ is an isomorphism if $|I|=|I_{1}|$.

Thus, the maximum quotient map, if it exists, is unique up to isomorphism.

###### Proof.

If $\pi_{1}(\pi^{-1}(i))$ contains more than one point for any $i\in I$, then
clearly, the restriction $R|\pi^{-1}(i)$ admits an arc quotient. Because $\pi$
is maximum, it follows that each $\pi_{1}(\pi^{-1}(i))$ is a singleton and so
the map $h$ is defined as usual.

Since $h$ is a surjection between finite sets, it follows that
$|I|\geq|I_{1}|$. If $\pi_{1}$ is also maximum, then $|I_{1}|\geq|I|$ and so
$|I|=|I_{1}|$. Since $h$ is a surjection, it is a bijection (and so an
isomorphism) if and only if $|I|=|I_{1}|$.

∎

######  Theorem 10.29.

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

###### Proof.

Begin with $\pi_{0}:(X,R)\to(I_{0},L_{0})$, with $(I_{0},L_{0})$ an arc. If it
is not maximum, then we construct, inductively, a finite or infinite sequence
of finite orders $(I_{k},L_{k})$ surjective, but not bijective tournament maps
$f_{k}:(I_{k+1},L_{k+1}))\to(I_{k},L_{k})$ and continuous tournament
surjections $\pi_{k}:(X,R)\to(I_{k},L_{k})$ such that
$f_{k}\circ\pi_{k+1}=\pi_{k}$.

If at stage $k$, the map $\pi_{k}$ is not maximum, then for some $i\in L_{k}$,
the restriction $(\pi_{k}^{-1}(i),R|\pi_{k}^{-1}(i))$ admits an arc quotient.
It is easy to see that we can split the point $i$, to obtain
$f_{k}:(I_{k+1},L_{k+1})\to(I_{k},L_{k})$ with $|f_{k}^{-1}(i)|=2$ and a lift
$\pi_{k+1}:(X,R)\to(I_{k+1},L_{k+1})$.

This process terminates when $\pi_{k}$ is maximum.

In fact it must terminate. If it did not, then $\\{(I_{k},L_{k},f_{k})\\}$
would be an inverse system of finite, transitive tournaments with the inverse
limit $(I,L)$ an infinite compact LOTS. Furthermore, the maps $\pi_{k}$ would
induce a quotient map $\pi:(X,R)\to(I,L)$. A LOTS has no cycle points. An
infinite compact LOTS has some non-isolated points. Since the continuous
surjective image of a wac tournament is wac, it follows that a wac tournament
cannot map onto an infinite compact LOTS.

∎

Notice that $X\ =\ \\{0\\}\cup\\{1/n:n\in\mathbb{N}\\}$ with the order $L$
inherited from $\mathbb{R}$ is an almost wac tournament which does not admit a
maximum order quotient.

##  11\.  Classification of WAC Tournaments

We first separate the class of wac tournaments into three types .

  * •

Type 1: $(X,R)$ is Type 1 when it is non-trivial and does not have an arc
quotient.

  * •

Type 2: $(X,R)$ is Type 2 when it has an arc quotient (and so is non-trivial).

  * •

Type 3: $(X,R)$ is Type 3 when it is trivial.

Each wac tournament has a so-called _base quotient map_ a continuous,
surjective surjective tournament map $\pi:(X,R)\to(Y,P)$ which is unique up to
isomorphism. If $(X,R)$ is Type 1, then its base quotient map is its prime
quotient map, as in Theorem 10.24. If $(X,R)$ is Type 2, then the base
quotient map is its maximum order quotient map, as in Theorem 10.29. If
$(X,R)$ is Type 3, then the base quotient map is the isomorphism onto any
trivial tournament.

For a quotient map $h:(X,R)\to(Y,P)$ of topological tournaments, if $(X,R)$ is
wac, then $(Y,P)$ is wac by Theorem 10.2(f) and for every $y\in Y$, the
restriction $R|h^{-1}(y)$ is wac because if the $Q$ invariant set $h^{-1}(y)$
is not trivial, then it is clopen.

######  Definition 11.1.

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)

$h_{i}=f_{i}\circ h_{i+1}$.

  * (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 11.2.

A wac tournament $(X,R)$ admits a classifier system
$\\{(X_{i},R_{i},f_{i})\\}$ with maps $\\{h_{i}\\}$.

If $\\{(X_{i},R_{i},f_{i})\\}$ is a classifier system with maps $\\{h_{i}\\}$,
then the map $h:(X,R)\to\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ given by
$h(x)_{i}=h_{i}(x)$ is an isomorphism.

If $\\{(X_{i}^{\prime},R_{i}^{\prime},f_{i}^{\prime})\\}$ with maps
$\\{h_{i}^{\prime}\\}$ is another classifier system for $(X,R)$, then there
exist isomorphisms $q_{i}:(X_{i},R_{i})\to(X_{i}^{\prime},R_{i}^{\prime})$
such that for all $i$:

(11.1) $f^{\prime}_{i}\circ q_{i+1}\ =\ q_{i}\circ f_{i},\quad\text{and}\quad
q_{i}\circ h_{i}\ =\ h_{i}^{\prime}.$

###### Proof.

Begin with $h_{1}:(X,R)\to(X_{1},R_{1})$ a base quotient map. Assume that
$(X_{i},R_{i})$ with maps $h_{i}$ have been constructed for $i\leq n$ and with
projection $f_{i}$ for $i\leq n-1$ so that conditions (i) and (iii) of
Definition 11.1 hold for $i\leq n-1$.

If $x_{n}\in X_{n}$ is a non-isolated point, then by Theorem 10.2(f) ,
$h_{n}^{-1}(x_{n})$ is a singleton set and so
$(h_{n}^{-1}(x_{n}),R|h^{-1}_{n}(x_{n}))$ is a type 3, trivial, wac. If
$x_{n}$ is isolated, then $h_{n}^{-1}(x_{n})$ is clopen and
$(h_{n}^{-1}(x_{n}),R|h^{-1}_{n}(x_{n}))$ is a wac. For any $x_{n}$, let
$\pi_{nx_{n}}:(h_{n}^{-1}(x_{n}),R|h^{-1}_{n}(x_{n}))\to(Y_{nx_{n}},P_{nx_{n}})$
be a base quotient map. Let
$(X_{n+1},R_{n+1})=(X_{n},R_{n})\ltimes\\{(Y_{nx_{n}},P_{nx_{n}})\\}$ be the
lexicographic product. It is a topological lexicographic product because
$(Y_{nx_{n}},P_{nx_{n}})$ is trivial whenever $x_{n}$ is non-isolated in
$X_{n}$. The map $f_{n}$ is the first coordinate projection for the
lexicographic product. The map $h_{n+1}$ is defined by

(11.2) $h_{n+1}(x)\ =\ (h_{n}(x),\pi_{nh_{n}(x)}(x))\quad\text{for all}\ \
x\in X.$

Because $h_{n}$ and each $\pi_{xn}$ is a quotient map, it easily follows that
$h_{n+1}$ is a quotient map. It follows from Theorem 10.2(f) again that
$(X_{n+1},R_{n+1})$ is a wac tournament. Finally, conditions (i) and (iii) of
Definition 11.1 hold for $i=n$.

Thus, by inductive construction, we obtain $\\{(X_{i},R_{i},f_{i})\\}$ a
classifier system with maps $\\{h_{i}\\}$.

Clearly, the map $h:(X,R)\to\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ is a
quotient map and so the inverse limit space is wac. It suffices to prove that
$h$ is injective, i.e. the pre-image of every point is a singleton. If
$x\in\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ is non-isolated, this
follows from Theorem 10.2(f) again. In particular, if $x_{n}\in X_{n}$ is non-
isolated for any $n$, then with
$g_{i}:\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}\to(X_{i},R_{i})$ is the
projection given by $x\mapsto x_{i}$, the sets $g_{n}^{-1}(x_{n})$ and
$h_{n}^{-1}(x_{n})=h^{-1}(g_{n}^{-1}(x_{n}))$ are singletons consisting of
non-isolated points by Theorem 10.2(f) yet again.

Now assume that $x\in\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ is isolated
so that $x_{i}$ is isolated in $X_{i}$ for all $i$. For any
$x\in\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\},\ \ \\{g_{i}^{-1}(x_{i})\\}$
is a decreasing sequence of closed sets with intersection $\\{x\\}$. When $x$
is isolated, $\\{x\\}$ is clopen and so the sequence stabilizes and so for
some $n$ , $g_{i}^{-1}(x_{i})=\\{x\\}$ for all $i\geq n$. In particular,
$(Y_{nx_{n}},P_{nx_{n}})$ is trivial because $g_{n+1}$ maps the singleton
$\\{x\\}$ onto $f_{n}^{-1}(x_{n})=\\{x_{n}\\}\times Y_{nx_{n}}$. Now by
construction
$\pi_{nx_{n}}:(h_{n}^{-1}(x_{n}),R|h_{n}^{-1}(x_{n}))\to(Y_{nx_{n}},P_{nx_{n}})$
is a base quotient map. Since $(Y_{nx_{n}},P_{nx_{n}})$ is trivial, the
tournament $(h_{n}^{-1}(x_{n}),R|h_{n}^{-1}(x_{n}))$ is type 3 and so
$h_{n}^{-1}(x_{n})$, which contains (and so equals) $h^{-1}(x)$ is a
singleton.

Thus, we can use $h$ to identify $(X,R)$ with the inverse limit
$\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ so that $h_{i}$ is identified
with the projection map $g_{i}$. That is, for $x\in X$, $h_{i}(x)=x_{i}$ in
$X_{i}$.

Given two different classifiers, the construction of the maps $\\{q_{i}\\}$ is
an obvious induction using the uniqueness up to isomorphism of the base
quotient maps. Observe that by Theorem 10.2(f), the quotient maps $f_{i}$ and
$f^{\prime}_{i}$ are isomorphic to projections from topological lexicographic
products and so they are open maps. Hence, continuity of $q_{i+1}$ follows
from that of $q_{i}$.

∎

Remark: Thus, the classifier system for a wac tournament is unique up to
isomorphism.

If $(X,R)$ is an almost wac tournament which does not have an arc quotient, so
that an initial or terminal point is non-isolated, then $(X,R)$ also has a
classifier, unique up to isomorphism.

######  Addendum 11.3.

For a wac tournament $(X,R)$ the topological inverse system of topological
lexicographic products used to construct the classifier satisfies the
following properties for every point $x$ in $X$ (which we identify with the
inverse limit space) and for every $n\in\mathbb{N}$, .

  * (i)

If $(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$ is a non-trivial, finite linear
order, then $(Y_{nx_{n}},P_{nx_{n}})$ does not have an arc quotient.

  * (ii)

If $(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$ is a trivial tournament, then
$(Y_{nx_{n}},P_{nx_{n}})$ is a trivial tournament.

###### Proof.

(i): If $(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$ is a non-trivial, finite linear
order, then

$\pi_{(n-1)x_{n-1}}:(h_{n-1}^{-1}(x_{n-1}),R|h^{-1}_{n-1}(x_{n-1}))\to(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$

is a maximum order quotient and with $x_{n}=(x_{n-1},y)$ we have
$h_{n}^{-1}(x_{n})=(\pi_{(n-1)x_{n-1}})^{-1}(y)$. The restriction of $R$ to
this set does not have an arc quotient by definition of the maximum order
quotient. Since
$\pi_{nx_{n}}:(h_{n}^{-1}(x_{n}),R|h_{n}^{-1}(x_{n}))\to(Y_{nx_{n}},P_{nx_{n}})$
is a base quotient map, it follows that $(Y_{nx_{n}},P_{nx_{n}})$ does not
have an arc quotient.

(ii): If $(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$ is trivial, i.e. type 3, then
since

$\pi_{(n-1)x_{n-1}}:(h_{n-1}^{-1}(x_{n-1}),R|h^{-1}_{n-1}(x_{n-1}))\to(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$

is a base quotient map, it follows that
$(h_{n-1}^{-1}(x_{n-1}),R|h^{-1}_{n-1}(x_{n-1}))$ is trivial and so also is
$(Y_{nx_{n}},P_{nx_{n}})$. If $(X_{1},R_{1})$ is trivial, then $(X,R)$ is
trivial and so every $(Y_{nx_{n}},P_{nx_{n}})$ is trivial.

∎

######  Corollary 11.4.

Let $(X,R)$ be a wac tournament with classifier system
$\\{(X_{i},R_{i},f_{i})\\}$ with maps $\\{h_{i}\\}$. The following conditions
are equivalent.

  * (i)

The tournament $(X,R)$ is arc cyclic.

  * (ii)

For every $i\in\mathbb{N}$ the tournament $(X_{i},R_{i})$ is arc cyclic.

  * (iii)

The base tournament $(X_{1},R_{1})$ is arc cyclic and for every $x\in X$ and
$i\in\mathbb{N}$, the restriction $(h_{i}^{-1}(x_{i}),R|h_{i}^{-1}(x_{i}))$
has an arc cyclic base quotient.

  * (iv)

For every non-trivial $Q$ invariant subset $A$ of $X$, the restriction
$(A,R|A)$ has an arc cyclic prime quotient.

The following conditions are equivalent.

  * (i’)

The tournament $(X,R)$ is locally arc cyclic.

  * (ii’)

There exists $i\in\mathbb{N}$ such that $(X_{i},R_{i})$ is locally arc cyclic
and for every $x\in X$ and $j\geq i$, the restriction
$(h_{j}^{-1}(x_{j}),R|h_{j}^{-1}(x_{j}))$ has an arc cyclic base quotient.

  * (iii’)

There exists $i\in\mathbb{N}$ such that such that $(X_{i},R_{i})$ is locally
arc cyclic and for every $x\in X$ the restriction
$(h_{i}^{-1}(x_{i}),R|h_{i}^{-1}(x_{i}))$ is arc cyclic.

###### Proof.

(i) $\Rightarrow$ (iv): By Proposition 10.9(e) the restriction $R|A$ is arc
cyclic when $R$ is arc cyclic and $A$ is $Q$ invariant. Hence, when $A$ is
non-trivial it has a unique prime quotient which is arc cyclic by Corollary
2.3.

(iv) $\Rightarrow$ (iii): Since $X$ itself is $Q$ invariant, the base quotient
$(X_{1},R_{1})$ is arc cyclic. Since a trivial tournament is vacuously arc
cyclic and the restriction to any of $Q$ invariant subsets $h_{i}^{-1}(x_{i})$
has an arc cyclic base quotient.

(iii) $\Rightarrow$ (ii): By uniqueness we may assume that
$\\{(X_{i},R_{i},f_{i})\\}$ is given by the inductive construction in the
proof of Theorem 11.2. Proceed by induction. By assumption the base quotient
$(X_{1},R_{1})$ is arc cyclic.

Now assume that $(X_{n},R_{n})$ is arc cyclic. For every $x\in X$, the base
for $(h_{n}^{-1}(x_{n}),R|h_{n}^{-1}(x_{n})$ is $(Y_{nx},P_{nx})$ which is an
arc cyclic by assumption. From Corollary 2.3 applied to $f_{n}$ it follows
$(X_{n+1},R_{n+1})$ is arc cyclic.

(ii) $\Rightarrow$ (i): By Proposition 4.3 the inverse limit of an inverse
system of arc cyclic tournaments is arc cyclic.

(i’) $\Leftrightarrow$ (iii’): If $(X_{i},R_{i})$ is locally arc cyclic, and
each of the restrictions is arc cyclic, then $(X,R)$ is locally arc cyclic by
Corollary 10.5 applied to the quotient map $h_{i}$.

Now assume that $(X,R)$ is locally arc cyclic. Let $\\{U_{1},\dots,U_{m}\\}$
be an open cover of $X$ by arc cyclic subsets. Define $G_{j}=\\{x\in
X:h_{j}^{-1}(x_{j})\subset U_{t}$ for some $t=1,\dots,n\\}$. Since

$G_{j}\ =\ \bigcup_{t=1}^{n}\ X\setminus h_{j}(X\setminus U_{t})$

it follows that $G_{j}$ is open. Since $h_{j+1}^{-1}(x_{j+1})\subset
h_{j}^{-1}(x_{j})$ it follows that $G_{j}\subset G_{j+1}$. Because
$x=\bigcap_{j}h_{j}^{-1}(x_{j})$ it follows that each $x$ is contained in some
$G_{j}$.

From compactness, it follows that for some $i\ \ G_{i}=X$. That is, each
$h_{i}^{-1}(x)$ is a $Q$ invariant arc cyclic subset and so the restriction
$(h_{i}^{-1}(x_{i}),R|h_{i}^{-1}(x_{i}))$ is arc cyclic. By Corollary 10.5
again it follows that $(X_{i},R_{i})$ is locally arc cyclic.

(iii’) $\Leftrightarrow$ (ii’): It is clear that we can construct the
classifier for $(h_{i}^{-1}(x_{i}),R|h_{i}^{-1}(x_{i}))$ by starting with
$(f_{i}^{-1}(x_{i}),R_{i+1}|f_{i}^{-1}(x_{i}))$, which is isomorphic to the
base for $(h_{i}^{-1}(x_{i}),R|h_{i}^{-1}(x_{i}))$ and then by using the same
choices for the restricted lexicographic construction which uses the bases for
the restrictions $(h_{j}^{-1}(z_{j}),R|h_{j}^{-1}(z_{j}))$ with $z\in
h_{i}^{-1}(x_{i})$. The equivalence then follows from the equivalence of (i)
and (iii) applied to $(h_{i}^{-1}(x_{i}),R|h_{i}^{-1}(x_{i}))$.

∎

Above we began with a wac tournament. Now we would like to build the
classifier system directly, achieving the wac tournament as the limit.

######  Lemma 11.5.

Let $(X_{2},R_{2})$ be the topological lexicographic product
$(X_{1},R_{1})\ltimes\\{(Y_{x},S_{x}):x\in X\\}$. The tournament
$(X_{2},R_{2})$ is wac if and only if $(X_{1},R_{1})$ and each $(Y_{x},S_{x})$
is wac.

###### Proof.

If $(X_{2},R_{2})$ is wac, then the quotient $(X_{1},R_{1})$ and the
restriction to the clopen subsets $\\{x\\}\times Y_{x}$ are wac by Theorem
10.2.

Now assume that $R_{1}$ and each $S_{x}$ is wac. If $x$ is non-isolated in
$X_{1}$ then it is a cycle point and each $3-$cycle containing $x$ lifts to a
$3-$cycle containing the unique point in $\pi^{-1}(x)$. If $x$ is isolated,
then $(x,y)$ is non-isolated in $X_{2}$ if and only if $y$ is non-isolated in
$Y_{x}$ and if $\\{y,y^{\prime},y^{\prime\prime}\\}$ is a $3-$cycle in $Y_{x}$
then $\\{(x,y),(x,y^{\prime}),(x,y^{\prime\prime})\\}$ is a $3-$cycle in
$X_{2}$.

∎

Recall that when $X$ is the limit of an inverse system the set $IS$ consists
of those points $x\in X$ such that $x_{i}$ is isolated for all
$i\in\mathbb{N}$.

######  Theorem 11.6.

Let $\\{(X_{i},R_{i},f_{i})\\}$ be an inverse lexicographic system so that for
each $i\in\mathbb{N},\ \ (X_{i+1},R_{i+1})$ is the topological lexicographic
product $(X_{i},R_{i})\ltimes\\{(Y_{ix_{i}},P_{ix_{i}})\\}$ with $f_{i}$ the
first coordinate projection. Thus, if $x_{i}$ is non-isolated in $X_{i}$, then
$(Y_{ix_{i}},P_{ix_{i}})$ is trivial.

We assume that conditions (i) and (ii) of Addendum 11.3 hold and, in addition,

  * (iii)

The tournaments $(X_{1},R_{1})$ and each tournament $(Y_{ix_{i}},P_{ix_{i}})$
is either trivial, a non-trivial finite order, or a prime wac which not an
arc.

Let $(X,R)=\overleftarrow{Lim}\\{(X_{i},R_{i},f_{i})\\}$ with
$h_{i}:(X,R)\to(X_{i},R_{i})$ the coordinate projection map.

A point $x\in IS$ is a cycle point if and only if it satisfies the following:

  * (iv)

For infinitely many $i\in\mathbb{N}$, with $x_{i+1}=(x_{i},y_{i}),\ \ y_{i}$
is contained in a $3-$cycle in $Y_{ix_{i}}$.

A point $x\in IS$ is an isolated point in $X$ if and only if it satisfies the
following:

  * (iv’)

There exists $i\in\mathbb{N}$ such that $(Y_{ix_{i}},P_{ix_{i}})$ is trivial.

The limit tournament $(X,R)$ is wac if and only if (iv) or (iv’) holds for
every $x\in IS$. In particular, if $(Y_{ix_{i}},R_{ix_{i}})$ is either trivial
or point cyclic for every $x\in X$ and $i\in\mathbb{N}$, then $(X,R)$ is wac.

If there exists $k\in N$ such that for every $x\in X$ and $i\geq k,\ \
(Y_{ix_{i}},P_{ix_{i}})$ is arc cyclic, $(X,R)$ is locally arc cyclic and so
is wac.

If $(X_{1},R_{1})$ is arc cyclic and for every $x\in X$ and $i\in\mathbb{N},\
\ (Y_{ix_{i}},P_{ix_{i}})$ is arc cyclic, $(X,R)$ is arc cyclic and so is wac.

When $(X,R)$ is wac, the inverse system $\\{(X_{i},R_{i},f_{i})\\}$ is a
classifier for $(X,R)$.

###### Proof.

First assume that $x\not\in IS$ and let $n$ be the smallest value such that
$x_{n}$ is not isolated in $X_{n}$. If $n=1$, then since $(X_{1},R_{1})$ is
wac, it follows that $x_{1}$ is a cycle point in $X_{1}$. If $n>1$, then
$x_{n-1}$ is isolated in $X_{n-1}$ and with $x_{n}=(x_{n-1},y)$ we have that
$y$ is non-isolated in the wac tournament
$(Y_{(n-1)x_{n-1}},P_{(n-1)x_{n-1}})$ and so $y$ is a cycle point in
$Y_{(n-1)x_{n-1}}$. Consequently, $x_{n}$ is a cycle point in $X_{n}$. By
Theorem 7.6, $\\{x\\}=h_{n}^{-1}(x_{n})$ and $x$ is a cycle point in $X$.

Now let $x$ be a point of $IS$.

If for infinitely many $i\in\mathbb{N}$, with $x_{i+1}=(x_{i},y_{i}),\ \
y_{i}$ is contained in a $3-$cycle in $Y_{ix_{i}}$, then it follows, as in the
proof of Theorem 7.5 that $x$ is a cycle point.

Conversely, assume that $x$ is a cycle point. For $j$ arbitrarily large,
$x_{j}$ isolated implies that $h_{j}^{-1}(x_{j})$ is a neighborhood of $x$ and
so contains a $3-$cycle $\\{x,x^{\prime},x^{\prime\prime}\\}$. Let $k+1$ be
the minimum index $i$ such that $x_{i}\not=x^{\prime}_{i}$ so that $k\geq j$.
Because $h_{k+1}^{-1}(x_{k+1})$ and $h_{k+1}^{-1}(x^{\prime}_{k+1})$ are
disjoint and $Q$ invariant, it cannot happen that $x^{\prime\prime}$ lies in
either them. Hence, $\\{x_{k+1},x^{\prime}_{k+1},x^{\prime\prime}_{k+1}\\}$ is
a $3-$cycle in $X_{k+1}$. If
$x_{k+1}=(x_{k},y),x^{\prime}_{k+1}=(x_{k},y^{\prime})$ and
$x^{\prime\prime}_{k+1}=(x_{k},y^{\prime\prime})$, then
$\\{y,y^{\prime},y^{\prime\prime}\\}$ is a $3-$cycle in $Y_{kx_{k}}$ as
required.

If, instead, $(Y_{ix_{i}},P_{ix_{i}})$ is trivial for some $i$, then Condition
(ii) implies, inductively, that $(Y_{jx_{j}},P_{jx_{j}})$ is trivial for all
$j\geq i$. It follows that $\\{x\\}=h_{i}^{-1}(x_{i})$ and so $x$ is isolated
in $X$.

Conversely, if $x\in IS$ is isolated, i.e. $\\{x\\}$ is clopen, then
$\\{x\\}=\bigcap_{i}h_{i}^{-1}(x_{i})$ implies that for some $i,\ \
\\{x\\}=h_{j}^{-1}(x_{j})$ for all $j\geq i$. So $\\{x_{j}\\}\times
Y_{jx_{j}}=f_{j}^{-1}(x_{j})=\\{x_{j+1}\\}$ for all $j\geq i$. Thus,
$(Y_{jx_{j}},P_{jx_{j}})$ is trivial for all $j\geq i$.

If $(Y_{ix_{i}},P_{ix_{i}})$ trivial, then $f_{i}^{-1}(x_{i})$ is a singleton.
If $(Y_{ix_{i}},P_{ix_{i}})$ is point cyclic, then with
$x_{i+1}=(x_{i},y_{i}),\ \ y_{i}$ is contained in a $3-$cycle in $Y_{ix_{i}}$.
Hence, the point cyclic assumption implies Condition (iv).

The arc cyclicity results follow from Corollary 11.4.

If $\\{(X^{\prime}_{i},R^{\prime}_{i},f^{\prime}_{i})\\}$ is a classifier for
$(X,R)$, we use Conditions (i) and (ii) to inductively construct the
isomorphisms $q_{i}:(X_{i},R_{i})\to(X_{i}^{\prime},R_{i}^{\prime})$ which
satisfy (11.1). Observe that Conditions (i)-(iii) imply that if
$(Y_{ix_{i}},P_{ix_{i}})$ is a non-trivial finite order, then
$(Y_{(i+1)x_{i+1}},P_{(i+1)x_{i+1}})$ is either a prime tournament and not an
arc, or else it is trivial. Either of these implies that

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

is a maximum order quotient map. ∎

######  Lemma 11.7.

Assume that $(X,R)$ is a prime tournament which is not an arc and that $x\in
X$. If $x$ is not initial, terminal or balanced, then it is contained in a
cycle. In particular, if $X$ is finite, then $R$ is point cyclic.

###### Proof.

Assume that $(X,R)$ is a prime tournament and $x\in X$ is not contained in a
cycle. Then for every $a,b\in X,\ \ a\rightharpoonup x\rightharpoonup b$
implies $a\rightharpoonup b$. If $x$ is neither terminal nor initial, then
both $R^{\circ-1}(x)$ and $R^{\circ}(x)$ are nonempty. If $x$ is not left
balanced, then $R^{\circ-1}(x)$ is closed and $R^{\circ-1}(x)\mapsto 0$ and
$R(x)\mapsto 1$ is a quotient map to an arc. Since $(X,R)$ is prime, it is an
arc. Similarly, if $x$ is not right balanced, then $(X,R)$ is an arc.

In a finite prime tournament, every point is isolated and so no point is
balanced. If it is not an arc, then it has no initial nor terminal point.

∎

######  Addendum 11.8.

In the construction of Theorem 11.6, assume that every
$(Y_{ix_{i}},P_{ix_{i}})$ is a finite tournament. A point $x\in X$ is a cycle
point if and only if it satisfies the condition:

  * (iv”)

For infinitely many $i\in\mathbb{N}$, with $(Y_{ix_{i}},P_{ix_{i}})$ is a
prime tournament which is not an arc.

So $(X,R)$ is wac if and only if (iv”) or (iv’) holds for every $x\in X$.

###### Proof.

Because the $(Y_{ix_{i}},P_{ix_{i}})$’s are finite, $X=IS$. Then Lemma 11.7
implies that conditions (iv) and (iv”) are equivalent.

∎

######  Examples 5.

Uncountably many distinct arc cyclic tournaments on the Cantor set.

Let $(Y_{0},P_{0})$ and $(Y_{1},P_{1})$ be the regular tournaments with
$|Y_{0}|=3$ and $|Y_{1}|=5$. So $Y_{0}$ consists of a single $3-$cycle. Each
of these is an arc cyclic, prime tournament. This is easy to check but we will
verify these statements in the next section.

(a) Let $\theta\in\\{0,1\\}^{\mathbb{N}}$. Let $\\{(X_{i},R_{i},f_{i})\\}$ be
the inverse system with $(X_{1},R_{1})=(Y_{\theta_{1}},P_{\theta_{1}})$ and
$(X_{i+1},R_{i+1})=(X_{i},R_{i})\ltimes(Y_{\theta_{i+1}},P_{\theta_{i+1}})$.
That is, we use the construction of Theorem 11.6 with
$(Y_{ix_{i}},P_{ix_{i}})=(Y_{\theta_{i+1}},P_{\theta_{i+1}})$ for all
$i\in\mathbb{N}$ and $x_{i}\in X_{i}$. It follows from the theorem that the
inverse limit $(X_{\theta},R_{\theta})$ is an arc cyclic tournament on a
Cantor set. Furthermore, if $\theta\not=\theta^{\prime}$, then uniqueness of
the classifiers implies that $(X_{\theta},R_{\theta})$ is not isomorphic to
$(X_{\theta^{\prime}},R_{\theta^{\prime}})$. If we exclude the countable set
of $\theta$’s which are eventually $0$ or eventually $1$, then remaining
uncountable family of tournaments are all group tournaments associated with
closed game subsets on the same group, namely a product of a countable number
of $\mathbb{Z}/3\mathbb{Z}$’s with a countable number of
$\mathbb{Z}/5\mathbb{Z}$’s. Alternatively, we can use the group structure
which is the product of the $3-$adics with the $5-$adics.

(b) The tournament $(Y_{1},P_{1})$ is a group tournament on the cyclic group
$\mathbb{Z}/5\mathbb{Z}$ and the only automorphisms of $(Y_{1},P_{1})$ are
translations by elements of the group, see, e.g. [2] Theorem 3.9. Choose
$x_{1},x_{2}\in\mathbb{Z}/5\mathbb{Z}$ with $x_{1}\not=e\not=x_{2}$ and
$x_{2}\not=x_{1}\not=x_{2}^{-1}$. For example, choose $x_{1},x_{2}$ the two
distinct members of the game subset $A$. There is then no automorphism of
$(Y_{1},P_{1})$ which maps the pair $\\{e,x_{1}\\}$ to $\\{e,x_{2}\\}$. Now
let $(X_{1},R_{1})=(Y_{1},P_{1})$ and let
$(X_{2},R_{2})=(X_{1},R_{1})\ltimes\\{(Y_{1x},P_{1x})\\}$ with
$(Y_{1x},P_{1x})=(Y_{1},P_{1})$ for $x=e,x_{1}$ and $=(Y_{0},P_{0})$
otherwise. Alternatively, let
$(X^{\prime}_{2},R^{\prime}_{2})=(X_{1},R_{1})\ltimes\\{(Y^{\prime}_{1x},P^{\prime}_{1x})\\}$
with $(Y^{\prime}_{1x},P^{\prime}_{1x})=(Y_{1},P_{1})$ for $x=e,x_{2}$ and
$=(Y_{0},P_{0})$ otherwise. Now fix $\theta\in\\{0,1\\}^{\mathbb{N}}$. Let
$(X,R)=(X_{2},R_{2})\ltimes(X_{\theta},R_{\theta})$ and
$(X^{\prime},R^{\prime})=(X^{\prime}_{2},R^{\prime}_{2})\ltimes(X_{\theta},R_{\theta})$.
Then $(X,R)$ and $(X^{\prime},R^{\prime})$ are arc cyclic tournaments on the
Cantor set which are not isomorphic despite the fact that the sets
$\\{(Y_{ix_{i}},P_{i,x_{i}}):x\in X\\}$ and $\\{(Y_{ix_{i}},P_{i,x_{i}}):x\in
X^{\prime}\\}$ are equal for every level $i$.

(c) Let $Y_{2}=\\{a_{1},a_{2},b_{1},b_{2},c\\}$ and on it define the
tournament $P_{2}$ to consist of

(11.3) $\displaystyle\begin{split}(a_{1},b_{1}),\ \
(b_{2},a_{2}),\quad(a_{1},b_{2}),&\ \ (a_{2},b_{1}),\quad(a_{1},a_{2}),\ \
(b_{1},b_{2}),\\\ (c,a_{1}),\ \ (c,a_{2}),&\quad(b_{1},c),\ \
(b_{2},c).\end{split}$

We have $3-$cycles $\\{a_{1},b_{i},c\\},\\{c,a_{i},b_{1}\\}$ for $i=1,2$ and
$\\{b_{1},b_{2},a_{2}\\}$. Thus, every arc is in a $3-$cycle except for
$(a_{1},a_{2})$. Also $b_{2}\in P_{2}(a_{1})\cap P_{2}^{-1}(a_{2})$.

It easily follows that $(Y_{2},P_{2})$ is prime and the maximal arc cyclic
subsets are $Y_{2}\setminus\\{a_{1}\\}$ and $Y_{2}\setminus\\{a_{2}\\}$.
Notice that the restriction of $P_{2}$ to neither of these subsets is arc
cyclic.

Let $\\{(X_{i},R_{i},f_{i})\\}$ be the inverse system with $(X_{1},R_{1})\ =\
(Y_{2},P_{2})$ and $(X_{i+1},R_{i+1})=(X_{i},R_{i})\ltimes(Y_{2},P_{2})$.

It follows from Addendum 11.8 that for the limit system $(X,R)$ every point of
$X$ is a cycle point and so $(X,R)$ is wac. On the other hand, if
$x,x^{\prime}\in X$ with $x_{i}=x^{\prime}_{i}$ and
$x_{i+1}=(x_{i},a_{1}),x^{\prime}_{i+1}=(x_{i},a_{2})$, then the arc
$(x,x^{\prime})$ is not contained in any $3-$cycle in $X$. It follows that $X$
contains no nonempty, open, arc cyclic subset.

##  12\.  Prime Tournament Constructions

Throughout our examples below, for a set $J$ when we consider the product
$J\times\\{-1,+1\\}$ we will write for $a\in J\ \ a-=(a,-1)$ and $a+=(a,+1)$
and similarly write $J\pm$ for $J\times\\{\pm 1\\}$.

###  12.1.  Doubles and Reduced Doubles

######  Examples 6.

(a) For $(J,P)$ a finite tournament, we follow [2] Section 6, to define the
_double_ $2(J.P)=(2J,2P)$ to be a tournament on $2J=\\{0\\}\cup
J\times\\{-1,+1\\}$.

The tournament $2P$ is defined as follows.

(12.1) $\displaystyle\begin{split}a\in J\quad&\Longrightarrow\quad
a-\rightharpoonup a+,\ a+\rightharpoonup 0,\ 0\rightharpoonup a-\ \ \text{in}\
2P.\\\ a\rightharpoonup b\ \ \text{in}\ P\quad&\Longrightarrow\quad
a+\rightharpoonup b+,\ a-\rightharpoonup b-,\ b+\rightharpoonup a-,\
b-\rightharpoonup a+\ \ \text{in}\ 2P.\end{split}$

The _reduced double_ $2^{\prime}(J,P)=(2^{\prime}J,2^{\prime}P)$ is the
restriction of the double to $J\times\\{-1,+1\\}$. That is, we remove the
point $0$. Thus, the double of a trivial tournament is a $3-$cycle and its
reduced double is an arc.

We will call the tournament $(J,P)$ _irreducible_ if for every pair $a\not=b$
in $J$ there exists $c\in J$ such that either $\\{a,b\\}\subset P^{\circ}(c)$
or $\\{a,b\\}\subset P^{\circ-1}(c)$. We will explain later the reason for the
label. Clearly, a tournament is irreducible when for every $a,b\in J,\ \
Q(a,b)\not=J$.

######  Theorem 12.1.

For a finite tournament $(J,P)$ the double $2(J,P)$ is regular, arc cyclic and
prime with $|2J|=2|J|+1$.

If $(J,P)$ is irreducible, then the reduced double $2^{\prime}(J,P)$ is arc
cyclic and prime with $|2^{\prime}J|=2|J|.$

###### Proof.

A double is always regular and so is arc cyclic. Directly, observe that if
$a\rightharpoonup b$ in $P$, then $\\{a-,b-,b+\\}$ and $\\{a+,b+,a-\\}$ are
$3-$cycles in $2^{\prime}J$. So if $U$ is a $Q$ invariant subset for
$2^{\prime}(J,P)$ and any pair among the four points $\\{a-,b-,a+,b+\\}$ other
than $\\{a+,b-\\}$ is contained in $U$ then all four points are contained in
$U$.

Furthermore, $\\{a+,0,b-\\}$ and $\\{a+,0,a-\\}$ are $3-$cycles in $2J$. It
easily follows directly that the double is arc cyclic and prime.

If $\\{a,b\\}\subset P^{\circ}(c)$, then $\\{a+,c-,b-\\}$ is a $3-$cycle in
$2^{\prime}J$. If $\\{a,b\\}\subset P^{\circ-1}(c)$, then $\\{a+,c+,b-\\}$ is
a $3-$cycle in $2^{\prime}J$. So if $\\{a+,b-\\}$ is contained in $U$ then
either $\\{c-,b-\\}$ or $\\{a+,c+\\}$ is contained in $U$ as well. In the
first case it follows as above that all four points of $\\{c-,b-,c+,b+\\}$ are
contained in $U$ and in the second, all four points of $\\{c-,a-,c+,a+\\}$ are
contained in $U$. It easily follows that the reduced double is arc cyclic and
prime.

The cardinality results are obvious.

∎

######  Corollary 12.2.

For every odd number $2n+1\geq 3$ there are arc cyclic, prime tournaments of
order $2n+1$. For every even number $2n\geq 8$ there are arc cyclic, prime
tournaments of order $2n$.

###### Proof.

Beginning with any tournament of order $n$, including the trivial tournament
with $n=1$, the double of a tournament of order $n$ is an arc cyclic, prime
tournaments of order $2n+1$.

Now begin with any tournament $(J_{0},P_{0})$. First attach two additional
points $m,M$ to get $(J_{1},P_{1})$ with $J_{1}=J_{0}\cup\\{m,M\\}$, and with
$P_{1}$ extending $P_{0}$ so that $m$ is initial and $M$ is terminal in
$J_{1}$. Now attach an additional point $p$ to get $(J_{2},P_{2})$ with
$J_{2}=J_{1}\cup\\{p\\}$ and with $P_{2}$ extending $P_{1}$ so that
$p\rightharpoonup m,\ p\rightharpoonup M$ and $x_{0}\rightharpoonup p$ in
$P_{2}$ for some $x_{0}\in J_{0}$. We check that $(J_{2},P_{2})$ is
irreducible.

The point $M$ is still terminal for $P_{2}$ and so any pair which does not
include $M$ is contained in $P_{2}^{-1}(M)$. Any pair $\\{x,M\\}$ with $x\in
J_{0}$ is contained in $P_{2}(m)$. This takes care of all pairs except for
$\\{m,M\\}\subset P_{2}(p)$ and $\\{p,M\\}\subset P_{2}(x_{0})$.

Thus, $(J_{2},P_{2})$ is irreducible and so its reduced double is arc cyclic
and prime. The smallest case of this is with $(J_{0},P_{0})$ trivial. In that
case $n=|J_{2}|=4$ and so $2n=8$. ∎

There also exists a prime tournament of order 6 which can be obtained from a
regular tournament of order 7 by removing a suitable point. However, any
tournament of order 4 has either a $3-$cycle or an arc as a quotient and so is
not prime.

If $J$ is the odd cyclic group $\mathbb{Z}/(2n+1)\mathbb{Z}$ and the
tournament $\widehat{A}$ is associated with the game subset
$A=\\{1,\dots,n\\}$, then the tournament $(J,\widehat{A})$ is isomorphic to
the double of the order $(I,L)$ of length $n$, see [2] Example 6.5. Hence, it
is arc cyclic and prime. In particular, with $n=2$ this applies to the unique
regular tournament of order $5$. Notice that if $2n+1$ is not a prime number,
then Proposition 6.2 implies that there is a game subset whose associated
tournament is isomorphic to a non-trivial lexicographic product and so is not
prime.

On the other hand, if the odd order group $J$ is a non-cyclic group with any
game subset or a cyclic group $\mathbb{Z}/(2n+1)\mathbb{Z}$ with game subset
$A^{\prime}$ such that $(J,\widehat{A^{\prime}})$ is not isomorphic to
$(J,\widehat{A})$ above, then by [2]Theorem 3.18, the tournament
$(J,\widehat{A^{\prime}})$ is irreducible, as well as regular. Hence, its
reduced double is prime.

(b) For any topological tournament $(J,P)$ a _generalized reduced double_
$2^{\prime}(J,P)=(2^{\prime}J,2^{\prime}P)$ is a topological tournament on
$J\times\\{-1,+1\\}$ such that the following conditions are satisfied.

  * (i)

The map $x+\mapsto x$ is an isomorphism from the restriction to
$J+=J\times\\{+1\\}$ to $(J,P)$ and $x-\mapsto x$ defines an isomorphism from
the restriction to $J-=J\times\\{-1\\}$ to $(J,P)$.

  * (ii)

The set $(2^{\prime}P)\cap(J-\times J+)$ is a surjective relation from $J-$ to
$J+$. That is, for every $a\in J$, there exist $a^{\prime},a^{\prime\prime}\in
J$ such that $a^{\prime}-\rightharpoonup a+$ and $a-\rightharpoonup
a^{\prime\prime}+$ in $2^{\prime}P$, i.e. $(2^{\prime}P)^{-1}(J+)\supset J-$
and $(2^{\prime}P)(J-)\supset J+$.

For the ordinary reduced double of a finite tournament, $a-\rightharpoonup a+$
for all $a\in J$ implies condition (ii).

The lexicographic product of the arc on $\\{-1,+1\\}$ with
$-1\rightharpoonup+1$ together with $(J,P)$ is a generalized reduced double.
However we will be primarily interested in the cases when $2^{\prime}(J,P)$ is
prime.

###  12.2.  Compact Countably Infinite Tournaments

######  Examples 7.

(a) Let $\mathbb{N}^{*}$ be the one point compactification of the set
$\mathbb{N}$ via the point $\infty$ at infinity.

We define the tournament $N_{0}=(\mathbb{N}^{*},L_{0})$ with $L_{0}$ the
linear order on $\mathbb{N}^{*}$, i.e.

(12.2) $i\rightharpoonup j\quad\Longleftrightarrow\quad
i<j\quad\text{including }\ \ j=\infty.$

We will write $\bar{N}_{0}$ for the reverse tournament
$(\mathbb{N}^{*},L_{0}^{-1})$.

The tournament $N_{0}$ has an arc quotient with infinitely many quotient maps
to the arc.

Notice that for any $i\in\mathbb{N}$, $|L_{0}^{-1}(i)|=i$. It follows that
$N_{0}$ is rigid, i.e. the only automorphism of $N_{0}$ is the identity.

We define the tournament $N_{1}=(\mathbb{N}^{*},L_{1})$ with $L_{1}$ the
linear order on $\mathbb{N}^{*}$ adjusted by reversing the arcs $(i,i+1)$ for
all $i\in\mathbb{N}$. Thus,

(12.3) $i+1\rightharpoonup i\rightharpoonup j\quad\Longleftrightarrow\quad
i+1<j\quad\text{including }\ \ j=\infty.$

The reverse tournament is $\bar{N}_{1}=(\mathbb{N}^{*},L_{1}^{-1})$.

######  Theorem 12.3.

The tournament $N_{1}$ is prime and rigid. Furthermore, the restriction to any
interval $\\{k:i\leq k\leq j\\}$ is prime provided $j-i\geq 2$.

###### Proof.

Let $U$ be a closed, non-trivial $Q$ invariant subset of $\mathbb{N}^{*}$.

Observe first that for any $i\in\mathbb{N},\ \ c_{i}=\\{i,i+2,i+1\\}$ is a
$3-$cycle.

Assume $i<j$ are in $U$.

If $1<i$, then $i\rightharpoonup i-1\rightharpoonup j$ implies that $i-1\in U$
and so, inductively, $i^{\prime}\in U$ for all $i^{\prime}<i$.

If $j<\infty$, then $i\rightharpoonup j+1\rightharpoonup j$ implies that
$j+1\in U$ and so $j^{\prime}\in U$ for all $j^{\prime}>j$ with
$j^{\prime}<\infty$. Hence, if $j=i+1,\ \ \mathbb{N}\subset U$.

If $j=i+2$, then $i+1\in U$ because of the $3-$cycle $c_{i}$. Again
$\mathbb{N}\subset U$.

If $j>i+2$, then $i\rightharpoonup k\rightharpoonup j$ for all $k$ with
$i+1<k<j-1$ implies that such $k$ are in $U$. Thus, $U$ contains every point
of $\mathbb{N}$ except possibly $i+1$ and, if $j$ is finite, $j-1$. When $j$
is finite, $j+1\in U$ and so $i\rightharpoonup j-1\rightharpoonup j+1$ implies
$j-1\in U$. The $3-$cycle $c_{i}$ then implies that $i+1\in U$.

It follows that $\mathbb{N}\subset U$. Since $U$ is closed, $\infty\in U$.

A similar argument shows that the restriction to an interval containing at
least three points is prime.

Assume that $h:N_{1}\to N_{1}$ is a continuous tournament map with image non-
trivial. Since $N_{1}$ is prime and not an arc, it follows that $h$ is a
tournament isomorphism onto its image. Hence, $h(\infty)=\infty$. Furthermore,
(12.3) implies that $h(i+1)=h(i)+1$. Thus, with $k=h(1)$ we have $h(i)=k+i-1$
for all $i$. In particular, if $h$ is surjective, it is the identity.

∎

(b) Let $2\mathbb{N}^{*}$ be the one-point compactification of
$\mathbb{N}\times\\{-1,+1\\}$ by the point $\infty$ at infinity. We will use
the label $2N=(2\mathbb{N}^{*},2L)$ for a tournament which satisfies

(12.4) $\displaystyle\begin{split}\mathbb{N}+\ =\
(2L)^{\circ-1}(\infty),&\quad\mathbb{N}-\ =\ (2L)^{\circ}(\infty),\\\ (i-,i+)\
\in\ 2L\quad&\text{for all}\ \ i\in\mathbb{N}.\end{split}$

There are two important examples

We define the tournament $2N_{0}=(2\mathbb{N}^{*},2L_{0})$ so that the
restriction of $2N_{0}$ to
$\mathbb{N}^{*}+=\mathbb{N}\times\\{+1\\}\cup\\{\infty\\}$ is isomorphic to
$N_{0}$ by $i+\mapsto i$ and the restriction of $2N_{0}$ to
$\mathbb{N}^{*}-=\mathbb{N}\times\\{-1\\}\cup\\{\infty\\}$ is isomorphic to
$\bar{N}_{0}$ by $i-\mapsto i$. In addition,

(12.5) $i-\rightharpoonup i+,(i+2)+\qquad i+\rightharpoonup j-\quad\text{for
all}\ \ j\not=i,i-2.$

######  Theorem 12.4.

The tournament $2N_{0}$ is arc cyclic, prime and rigid.

###### Proof.

If $i<j$ and $i+2\not=j$, then

$\\{i-,i+,j-\\},\\{i+,j+,i-\\},\\{i+,\infty,i-\\},\\{(i+2)+,\infty,i-\\}$

are $3-$cycles. Thus, $2N_{0}$ is arc cyclic.

Assume $U$ is a non-trivial, closed $Q$ invariant subset. If any pair in
$\\{i-,i+,(i+1)+,(i+1)-\\}$ except $\\{(i+1)+,(i+1)-\\}$ is contained in $U$,
then all four points are contained in $U$. Proceeding upward, we obtain
$j-,j+\in U$ for all $j\geq i$ as well as $\infty\in U$. If
$\\{(i+1)+,(i+1)-\\}\subset U$, then because $j-,j+\in U$ for all $j\geq i+1$
we have $\infty\in U$ because $U$ is closed. Because $\\{(i+2)+,\infty,i-\\}$
is a $3-$cycle, it follows that $i-\in U$. Since $i-,(i+1)-\in U$ it follows
that $\\{i-,i+,(i+1)+,(i+1)-\\}\subset U$. Thus, it follows that
$2\mathbb{N}^{*}\subset U$ and so $2N_{0}$ is prime.

Because any automorphism of $2N_{0}$ would have to fix $\infty$ and because
$N_{0}$ is rigid, it follows that $2N_{0}$ is rigid.

∎

If we fix $n\in\mathbb{N}$ with $n>3,\ \ 2L_{0}$ contains the countable set of
arcs $A_{n}\ =\ \\{((i+k)+,i-):i\in\mathbb{N},3<k\leq n\\}$. If we reverse the
arcs in any subset of $A_{n}$ we still have an arc cyclic, prime tournament.
An isomorphism between two such would have to be the identity on
$2\mathbb{N}^{*}$ by rigidity of $N_{0}$ and $\bar{N}_{0}$. Thus, for distinct
subsets of $A_{n}$ the resulting tournaments are not isomorphic. In this way
we obtain an uncountable number of distinct, countably infinite, compact, arc
cyclic, prime tournaments each with a single non-isolated point.

We define the tournament $2N_{1}=(2\mathbb{N}^{*},2L_{1})$ so that the
restriction of $2N_{1}$ to $\mathbb{N}^{*}+$ is isomorphic to $N_{1}$ by
$i+\mapsto i$ and the restriction to $\mathbb{N}^{*}-$ is isomorphic to
$\bar{N}_{1}$ by $i-\mapsto i$. In addition,

(12.6) $i-\rightharpoonup i+,\qquad i+\rightharpoonup j-\quad\text{for all}\ \
j\not=i.$

######  Theorem 12.5.

The tournament $2N_{1}$ is arc cyclic, prime and rigid. The restriction to
$\\{k-,k+:i\leq k\leq j\\}$ is arc cyclic and prime provided $j-i\geq 2$.

###### Proof.

If $i+1<j\leq\infty$, then

$\\{i-,i+,j-\\},\\{i-,i+,j+\\},\\{(i+1)-,(i+1)+,i+\\},\\{(i+1)+,i-,(i+1)-\\}$

are $3-$cycles with $j-=j+=\infty$ when $j=\infty$. It follows that $2N_{1}$
is arc cyclic.

Assume $U$ is a non-trivial, closed $Q$ invariant subset. If $U$ contains two
points of $N^{*}+$, then because $N_{1}$ is prime, it follows that
$N^{*}+\subset U$. Similarly, if $U$ contains two points of $N^{*}-$, then
$N^{*}-\subset U$. If either of these occurs then from the cycles it contains
all $i+,i-$ and so $2\mathbb{N}^{*}\subset U$.

Now we use the cycles listed above.

If $U$ contains $i+,i-$ for some $i$, then it contains $j+$ and $j-$ for all
$j>i+1$ and so again $2\mathbb{N}^{*}\subset U$.

Now assume $U$ contains $i+,j-$ with $i\not=j$. If $j=i+1$, then $(i+1)+\in
U$. If $j=i-1$, then $i-\in U$. If $j>i+1$, then $i-\in U$. If $j<i-1$, then
$j-\in U$. From the earlier computations it follows that $U=2\mathbb{N}^{*}$
in these casees as well.

Thus, $2N_{1}$ is prime.

A similar computation works for the restriction.

An automorphism must fix $\infty$. Again because $N_{1}$ is rigid, it follows
that $2N_{1}$ is rigid.

∎

###  12.3.  Adjusting Lexicographic Products

######  Examples 8.

(a) We assume that $(J,P)$ is a topological tournament with a generalized
reduced double $2^{\prime}(J,P)$ which is prime and arc cyclic. We also assume
that $J$ does not have both an initial point and a terminal point. For
example, in the finite case we may use $(J,P)$ any regular, irreducible
tournament as in that case Theorem 12.1 says that the reduced double
$2^{\prime}(J,P)$ is arc cyclic and prime.

We will start with a topological lexicographic product and then alter the arc
connections over certain pairs in the base of the product.

We begin with the topological lexicographic product of
$N_{0}\ltimes\\{(Y_{a},S_{a}):a\in\mathbb{N}^{*}\\}$ with
$(Y_{i},S_{i})=(J,P)$ for all $i\in\mathbb{N}$ and $(Y_{\infty},S_{\infty})$
trivial. So the total space $Y=(\mathbb{N}\times J)\cup\\{\infty\\}$.

Leaving the other arcs unchanged we define $(Y,S)$ so that for each
$i\in\mathbb{N}$, the restriction $S|[\\{i,i+1\\}\times J]$ is isomorphic to
$2^{\prime}(J,P)$ by the map $(i,x)\mapsto x-$ and $(i+1,x)\mapsto x+$ for
$x\in J$.

######  Theorem 12.6.

The tournament $(Y,S)$ is prime with non-isolated terminal point $\infty$.
Every point of $Y$ except for $\infty$ has an arc cyclic neighborhood.
Furthermore, if $i<j$, then the restriction of $S$ to $\\{k:i\leq k\leq
j\\}\times J$ is prime and locally arc cyclic.

###### Proof.

For each point $(i,x)$ $\\{i,i+1\\}\times J$ is an arc cyclic neighborhood by
assumption on $2^{\prime}(J,P)$. Hence, $(Y,S)$ is an almost wac tournament.
It follows from Addendum 10.14 that any closed, non-trivial $Q$ invariant
subset $U$ is clopen. If $U$ contains two points of $\\{i,i+1\\}\times J$,
then because $2^{\prime}(J,P)$ is prime, $U$ contains $\\{i,i+1\\}\times J$
and if $i>1$, then $U$ contains $\\{i-1,i\\}\times J$ as well. Proceeding
upwards and downwards, we obtain $\mathbb{N}\times J\subset U$ and since $U$
is closed, $\infty\in U$.

Now assume $(i,x),(j,y)\in U$ with $j>i+1$ so that
$(i,x)\rightharpoonup(j,y)$. If $J$ does not have a terminal point, then there
exists $x^{\prime}\in J$ so that $(x,x^{\prime})\in P^{\circ}$. Then
$(i,x)\rightharpoonup(i,x^{\prime})\rightharpoonup(j,y)$ and so
$(i,x^{\prime})\in U$ and as above $U=Y$. If $J$ does not have an initial
point, then there exists $y^{\prime}\in J$ so that $(y^{\prime},y)\in
P^{\circ}$. Then $(i,x)\rightharpoonup(j,y^{\prime})\rightharpoonup(j,y)$ and
so $(j,y^{\prime})\in U$ and as above $U=Y$.

If $(i,x),\infty\in U$ then for any $j>i+1$,
$(i,x)\rightharpoonup(j,x)\rightharpoonup\infty$ and so $(j,x)\in U$. As
before this implies $U=Y$.

The same arguments work for the restriction to $\\{k:i\leq k\leq j\\}\times
J$.

∎

The maximal arc cyclic subsets of $L$ are all of the form $\\{i,i+1\\}\times
J$. In particular, the isomorphism class of the restriction $\\{k:i\leq k\leq
j\\}\times J$ is determined by the length $j-i$ since the restriction has
exactly $j-i$ maximal arc cyclic subsets.

If we had used $N_{1}$ instead of $N_{0}$ in the above construction we would
have obtained the same tournament $(Y,S)$ since $N_{1}$ was obtained from
$N_{0}$ by reversing the $(i,i+1)$ arcs.

Notice that if $(Y,S)$ is a compact tournament with a terminal point $M$ which
is not isolated, and so is left balanced, it cannot happen that every arc not
connected to $M$ is contained in a $3-$cycle. For suppose that $\\{y_{n}\\}$
is a sequence in $Y\setminus\\{M\\}$ which converges to $M$ and that $x\in
Y\setminus\\{M\\}$. Since $M\in S^{\circ}(x)$, eventually $x\rightharpoonup
y_{n}$. Suppose $z_{n}\in Y$ with $z_{n}\rightharpoonup x\rightharpoonup
y_{n}$. We may assume $\\{z_{n}\\}$ converges to a point $z$ so that $z\
\underline{\rightharpoonup}\ x$ and, in particular, $z\not=M$. Hence,
$z\rightharpoonup M$. Since $(z_{n},y_{n})$ converges $(z,M)$, eventually
$(z_{n},y_{n})\in S^{\circ}$ and so eventually $\\{z_{n},x,y_{n}\\}$ is not a
$3-$cycle. That is, eventually the pair $\\{x,y_{n}\\}$ is not contained in
any $3-$cycle.

For the arc $(\\{0,1\\},L)$ with $\\{1\\}=L^{\circ}(0)$ we consider the
lexicographic product with $(Y,S)$, defining
$(\tilde{Y},\tilde{S})=(\\{0,1\\},L)\ltimes(Y,S)$. The first coordinate
projection map, $\pi$, is an arc quotient map. However, we have another prime
quotient map $\tilde{\pi}:(\tilde{Y},\tilde{S})\to(Y,S)$ given by, with $x\in
L$:

(12.7) $\tilde{\pi}(0,x)\ =\ x,\quad\text{and}\quad\tilde{\pi}(1,x)\ =\
\infty.$

Thus, the prime quotients of $(\tilde{Y},\tilde{S})$ exist, but are not
unique.

(b) Now assume that $(Z,P)$ is a topological tournament with
$\\{Z-,Z+\\}$ a partition of $Z$ by two disjoint clopen subsets such that the
relation $P\cap(Z-\times Z+)$ is surjective. We will write $P\pm$ for the
restriction $P|Z\pm$.

For example, we may use $(Z,P)$ equal to the generalized double
$2^{\prime}(J,P)$ as in part (a).

Now let $2N=(2\mathbb{N}^{*},2L)$ be a tournament satisfying (12.4).

To define $(K,T)$ we begin with the topological lexicographic product of
$2N\ltimes\\{(Y_{a},S_{a}):a\in 2\mathbb{N}^{*}\ \\}$ such that for all
$i\in\mathbb{N}$

(12.8) $\displaystyle\begin{split}(Y_{i+},S_{i+})\ =\ &(Z+,P+),\\\
(Y_{i-},S_{i-})\ =\ &(Z-,P-)\end{split}$

and with $(Y_{\infty},S_{\infty})$ trivial. Thus, the total space
$K=(\mathbb{N}+\times Z+)\cup(\mathbb{N}-\times Z-)\cup\\{\infty\\}$.

Leaving the other arcs unchanged we define $(K,T)$ so that for each
$i\in\mathbb{N}$, the restriction to $[(\\{i-\\}\times Z-)\cup(\\{i+\\}\times
Z+)]$ is isomorphic to $(Z,P)$ by the map $(i-,z-)\mapsto z-$ for $z-\in Z-$
and $(i+,z+)\mapsto z+$ for $z+\in Z+$.

Recall that we defined a section for a topological lexicographic product. In
this case, given any function $\tilde{\xi}:\mathbb{N}+\cup\ \mathbb{N}-\to Z$
with $\tilde{\xi}(\mathbb{N}+)\subset Z+$ and $\tilde{\xi}(\mathbb{N}-)\subset
Z-$, the associated section $\xi:2\mathbb{N}^{*}\to K$ is defined by
$\xi(i\pm)=(i\pm,\tilde{\xi}(i\pm))$ and $\xi(\infty)=\infty$. Lemma 3.5 says
that any section $\xi$ is continuous and induces an isomorphism from $2N$ to
the restriction of the corresponding lexicographic product to the image of
$\xi$. We will restrict ourselves to sections which satisfy the condition

(12.9) $\tilde{\xi}(i-)\rightharpoonup\tilde{\xi}(i+)\ \ \text{in}\ \ P\ \
\text{for all}\ \ i\in\mathbb{N}.$

This will imply that $\xi$ is a tournament isomorphism from $2N$ to the
restriction $T|j(2\mathbb{N}^{*})$.

######  Theorem 12.7.

If $2N$ and $(Z,P)$ are both arc cyclic (or both prime) tournaments then
$(K,T)$ is an arc cyclic (resp. prime) tournament.

###### Proof.

Let $U$ be a closed, non-trivial $Q$ invariant subset of $K$.

First consider a pair of points in $[(\\{i-\\}\times Z-)\cup(\\{,i+\\}\times
Z+)]$ for some $i$. If $(Z,P)$ is arc cyclic then such a pair is contained in
an arc in $[(\\{i-\\}\times Z-)\cup(\\{,i+\\}\times Z+)]$ Furthermore, if
$(Z,P)$ is prime and $U$ contains such a pair, then it contains all of
$[(\\{i-\\}\times Z-)\cup(\\{,i+\\}\times Z+)]$.

Given any other sort of pair, the assumption that $P\cap(Z-\times Z+)$ is
surjective implies that there exists a section $\xi$ which contains the pair
and, in particular, so that $\xi$ satisfies condition (12.9). If $2N$ is arc
cyclic, then any such pair is contained in a $3-$cycle in
$j(2\mathbb{N}^{*})$. It follows that $(K,T)$ is arc cyclic when $2N$ and
$(Z,P)$ are arc cyclic.

If $2N$ is prime, it follows that if $U$ contains a pair of points in
$j(2\mathbb{N}^{*})$ then it contains all of $j(2\mathbb{N}^{*})$.

Hence, $U$ contains $j(2\mathbb{N}^{*})$ for some section $\xi$. If
$\xi^{\prime}$ is another section which agrees with $\xi$ at some pair of
points, then $U$ contains $\xi^{\prime}(2\mathbb{N}^{*})$. By thus varying the
sections, we see that $U$ contains $(\mathbb{N}+\times
Z+)\cup(\mathbb{N}-\times Z-)$. Since $U$ is closed, it contains all of $K$.
Thus, $(K,T)$ is prime when $2N$ and $(Z,P)$ are prime.

∎

For two special cases we can use $2N=2N_{0}$ and $2N=2N_{1}$ which are arc
cyclic, prime tournaments by Theorems 12.4 and 12.5. We use the labels
$(K,T_{0})$ and $(K,T_{1})$ for these special cases. Thus we have

######  Corollary 12.8.

If $(Z,P)$ is an arc cyclic, prime tournament, then the tournaments
$(K,T_{0})$ and $(K,T_{1})$ are arc cyclic and prime.

If we use $(Z,P)$ equal to the reduced double of part (a), then
$K=[(\mathbb{N}-\ \cup\ \mathbb{N}+)\times J]\cup\\{\infty\\}$.

###  12.4.  The Attachment Construction

######  Definition 12.9.

For a tournament $(X,R)$ a subset $E$ of $X$ is called a _spanning set_ when
it satisfies the following equivalent conditions.

  * (i)

The set $E$ meets every input and output set, i.e. for all $x\in X$,
$R^{\circ}(x)\cap E$ and $R^{\circ-1}(x)\cap E$ are nonempty.

  * (ii)

There does not exists $x\in X$ such that either $E\subset R(x)$ or $E\subset
R^{-1}(x)$.

  * (iii)

The images $R^{\circ}(E)$ and $(R^{\circ})^{-1}(E)$ each equal all of $X$.

  * (iv)

For every $x\in X$ there exist $x^{\prime},x^{\prime\prime}\in E$ such that
$x^{\prime}\rightharpoonup x\rightharpoonup x^{\prime\prime}$.

If $E$ is a spanning set, then $Q(E\times E)=X$. Conversely, if $(X,R)$ is
balanced and $E$ is open, then $Q(E\times E)=X$ implies that $E$ is a spanning
set, because in that case for any $x\in E$, both $R^{\circ}(x)$ and
$R^{\circ-1}(x)$ meet $E$.

If $E$ is a spanning set, then any subset which contains $E$ is a spanning
set. We will be primarily interested in sets $E$ such that both $E$ and its
complement are spanning sets.

Spanning sets need not be large. Let $(J,P)$ be a finite tournament with three
points $a,b,c\in J$ with $b\rightharpoonup c$. It is easy to check that in the
double $2(J,P)$ each $3-$cycle $\\{0,a-,a+\\}$ $\\{b-,b+,c+\\}$ is a spanning
set and so each has a spanning set complement as well. If $(J,P)$ has no
initial point, then $J-$ and $J+$ are complementary spanning sets for the
reduced double $2^{\prime}(J,P)$.

######  Proposition 12.10.

Let $(J,P)$ be a finite, regular tournament of size $2n+1$. There are at least
${2n+1\choose n}-(2n+1)(2n+2)$ separate spanning sets $A$ with size $|A|=n$.
If $n\geq 6$, then there are at least $n(2n+1)(n+1)$ such sets. For each such
spanning set, the complement is a spanning set as well.

###### Proof.

Each $P(x)$ and $P^{-1}(x)$ has size $n+1$ and so contains $n+1$ subsets of
size $n$. Hence, there are at most $2(2n+1)(n+1)$ sets of size $n$ which are
contained in some $P(x)$ or $P^{-1}(x)$. Hence, $J$ contains at least
${2n+1\choose n}-(2n+1)(2n+2)$ subsets of size $n$ which are not contained in
any $P(x)$ or $P^{-1}(x)$. We can write the difference as

(12.10)
$(2n+1)(n+1)[(\frac{2n}{n+1}\frac{2n-1}{n}\cdots\frac{n+5}{6}\frac{n+4}{10}\frac{n+3}{12})(n+2)-2]$

Cancelling the initial $2$ into the $12$ and observing that $2n-1\geq n+1$
when $n\geq 2$, we see that when $n\geq 6$, then the parenthesized expression
is greater than $1$.

If $B$ is the complement of one of these sets $A$, then $|B|=n+1$. So if $B$
is contained in some $P(x)$, then it equals $P(x)$ and so $A$ is disjoint from
$P^{\circ}(x)$, contra the assumption that $A$ is a spanning set. Similarly,
$B$ cannot be contained in any $P^{-1}(x)$. Thus, $B$ is a spanning set.

∎

Clearly, if $(X,R)$ has an initial or terminal point, then it does not admit a
spanning set.

######  Theorem 12.11.

If the topological tournament $(X,R)$ has no initial or terminal point, then
the entire space $X$ is a spanning set.

Assume that the compact topological tournament $(X,R)$ has no initial or
terminal point and $X$ has no isolated points. If $U$ is an open spanning set,
then $U$ contains a pairwise disjoint sequence of finite subsets
$\\{H_{i}:i\in\mathbb{N}\\}$ all with the same cardinality and such that each
is a spanning set.

###### Proof.

Assume that the topological tournament $(X,R)$ has no initial or terminal
point. It is clear that $X$ is a spanning set.

If $U$ is an open spanning set, then for every $z\in X$, there exist $z-,z+\in
U$ such that $z\in R^{\circ}(z-)\cap R^{\circ-1}(z+)$. Choose
$\\{U_{z-},U_{z},U_{z+}\\}$ a thickening of $\\{z-,z,z+\\}$ so that for every
$x-\in U_{z-},x+\in U_{z+}$ we have $U_{z}\subset R^{\circ}(x-)\cap
R^{\circ-1}(x+)$. By intersecting with $U$ we may assume that
$U_{z-},U_{z+}\subset U$.

Let $\\{U_{z_{j}}:j=1,\dots,k\\}$ be a subcover of $X$. If $x_{j}-\in
U_{z_{j}-},x_{j}+\in U_{z_{j}+},$ then
$\\{x_{j}-:j=1,\dots,k\\}\cup\\{x_{j}+:j=1,\dots,k\\}$ is a spanning set with
cardinality $2k$.

Now assume that $X$ has no isolated points. We can then choose for each $j$
sequences of distinct points $\\{x_{ij}-:i\in\mathbb{N}\\}$ in $U_{z_{j}-}$
and $\\{x_{ij}+:i\in\mathbb{N}\\}$ in $U_{z_{j}+}$. Since there are no
isolated points, every open set is uncountable and so we can inductively make
the choices so that for each $j$ none the points of
$\\{x_{ij}-:i\in\mathbb{N}\\}\cup\\{x_{ij}+:i\in\mathbb{N}\\}$ are contained
in $\bigcup_{k<j}\
\\{x_{ik}-:i\in\mathbb{N}\\}\cup\\{x_{ik}+:i\in\mathbb{N}\\}$.

Let $H_{i}=\\{x_{ij}-:j=1,\dots,k\\}\cup\\{x_{ij}+:j=1,\dots,k\\}$ to define
the pairwise disjoint sequence of spanning sets each with cardinality $2k$.

∎

An $n-$fold _partition_ $\\{C_{1},\dots,C_{n}\\}$ of a space $X$ is a cover by
$n$ pair-wise disjoint clopen sets. It is called _proper_ when no $C_{i}$ is
empty.

A _spanning set partition_ is a $2-$fold partition $\\{E,F\\}$ of $X$ by a
pair of complementary spanning sets.

######  Proposition 12.12.

Let $(X,R)$ be a topological tournament and $\\{E,F\\}$ be a $2-$fold
partition of $X$. The following conditions are equivalent.

  * (i)

The restriction $R\cap(E\times F)$ is a surjective relation from $E$ to $F$.

  * (ii)

For every $a\in E,\ R(a)\cap F\not=\emptyset$ and for every $b\in F,\
R^{-1}(b)\cap E\not=\emptyset$.

  * (iii)

The images $R(E)\supset F$ and $R^{-1}(F)\supset E$.

  * (iv)

For every $a\in E,$ and $b\in F$, there exists $a^{\prime}\in E,b^{\prime}\in
F$ such that $a\rightharpoonup b^{\prime},$ and $a^{\prime}\rightharpoonup b$.

These conditions imply that neither $E$ nor $F$ is empty.

Furthermore, the following conditions are equivalent.

  * (i)

$R\cap[(E\times F)\cup(F\times E)]$ is a surjective relation on $X$.

  * (ii)

$R\cap(E\times F)$ is a surjective relation from $E$ to $F$ and $R\cap(F\times
E)$ is a surjective relation from $F$ to $E$.

  * (iii)

$R(E)\cap R^{-1}(E)\supset F$ and $R(F)\cap R^{-1}(F)\supset E$.

If $\\{E,F\\}$ is a spanning set partition, then the restriction $R|[(E\times
F)\cup(F\times E)]$ is a surjective relation on $X$. Conversely, if $(X,R)$ is
balanced and $R|[(E\times F)\cup(F\times E)]$ is a surjective, then $(E,F)$ is
a spanning set partition.

###### Proof.

The equivalences are easy to check. Definition 12.9(iii) shows that a spanning
set partition satisfies $R(E)\cap R^{-1}(E)\supset F$ and $R(F)\cap
R^{-1}(F)\supset E$. The converse holds when $(X,R)$ is balanced because if a
point $x$ is balanced, then $R^{\circ}(x)$ and $R^{\circ-1}(x)$ meet $E$
whenever $x$ is in the interior of $E$.

∎

Now we develop the _attachment construction_. We begin with two examples.

######  Proposition 12.13.

Let $(X,R)$ be a topological tournament and $\\{E,F\\}$ be a $2-$fold
partition of $X$.

Given a point $u$ not in $X$ let $X^{\prime}\ =\ X\cup\\{u\\}$ with $u$
isolated and define the topological tournament $R^{\prime}$ on $X^{\prime}$ by

(12.11) $R^{\prime}|X\ =\ R,\qquad\text{and}\quad R^{\prime\circ}(u)\ =\ F,\ \
R^{\prime\circ-1}(u)\ =\ E.$

When $X$ is compact, $X^{\prime}$ is compact. If $(X,R)$ is wac, then
$(X^{\prime},R^{\prime})$ is wac.

Assume that the relation $R\cap(F\times E)$ is surjective.

If $(X,R)$ is arc cyclic, then $(X^{\prime},R^{\prime})$ is arc cyclic.

If $(X,R)$ is prime, then $(X^{\prime},R^{\prime})$ is prime.

###### Proof.

The compactness and wac results are obvious.

If $b\in F$, there exists $a\in E$ with $b\rightharpoonup a$ and if $a\in E$,
there exists $b\in F$ with $b\rightharpoonup a$. In each case, $\\{a,u,b\\}$
is a $3-$cycle.

It follows that $(X^{\prime},R^{\prime})$ is arc cyclic when $(X,R)$ is.

Now assume that $(X,R)$ is prime and that $U$ is a non-trivial, closed $Q$
invariant subset of $X^{\prime}$.

If any pair in $X$ is contained in $U$, then all of $X$ is contained in $U$.
Using the above $3-$cycles we see that $u\in U$ as well.

If $\\{x,u\\}\subset U$ for some $x\in X$, then the above $3-$cycles show that
there exists $x^{\prime}$ in the complementary member of the pair $\\{E,F\\}$
with $x^{\prime}\in U$. Again since $(X,R)$ is prime, $X\subset U$.

∎

######  Proposition 12.14.

Let $(X,R)$ be a topological tournament and $\\{E,F\\}$ be a $2-$fold
partition of $X$.

Given distinct points $u,v$ not in $X$ let $X^{\prime\prime}\ =\
X\cup\\{u,v\\}$ with $u,v$ isolated and define the tournament
$R^{\prime\prime}$ on $X^{\prime\prime}$ by

(12.12) $\begin{split}R^{\prime\prime}|X\ =\ R,\qquad\text{and}\quad
R^{\prime\prime\circ}(u)\ =\ F\cup\\{v\\},\ \ R^{\prime\prime\circ}(v)\ =\
E,\\\ \text{so that}\qquad R^{\prime\prime\circ-1}(u)\ =\ E,\ \
R^{\prime\prime\circ-1}(v)\ =\ F\cup\\{u\\}.\hskip 28.45274pt\end{split}$

When $X$ is compact, $X^{\prime\prime}$ is compact. If $(X,R)$ is wac, then
$(X^{\prime\prime},R^{\prime\prime})$ is wac.

Assume that $R(E)\supset F$ and $R^{-1}(E)\supset F$, that is, for all $b\in
F$, there exist $a,a^{\prime}\in A$ such that $a\rightharpoonup
b\rightharpoonup a^{\prime}$. This assumption includes the possibility that
$F=\emptyset$ and so $EA=X$.

If $(X,R)$ is arc cyclic, then $(X^{\prime\prime},R^{\prime\prime})$ is arc
cyclic.

Assume, in addition, that $F$ is nonempty.

If $(X,R)$ is prime, then $(X^{\prime\prime},R^{\prime\prime})$ is prime.

###### Proof.

The compactness and wac results are again obvious.

Now assume that $R(E)\supset F$ and $R^{-1}(E)\supset F$.

If $a\in E$, then $\\{a,u,v\\}$ is a $3-$cycle.

If $b\in F$, we may choose $a,a^{\prime}\in E$ such that $a\rightharpoonup
b\rightharpoonup a^{\prime}$, so that $\\{a,b,v\\}$ and $\\{b,a^{\prime},u\\}$
are $3-$cycles.

It follows that $(X^{\prime\prime},R^{\prime\prime})$ is arc cyclic if $(X,R)$
is.

Observe that if $F=\emptyset$, then $E=X$ is a closed $Q$ invariant subset of
$X^{\prime\prime}$. Smashing it to a point we see that
$(X^{\prime\prime},R^{\prime\prime})$ has a $3-$cycle quotient.

Now assume that $F$ is nonempty and that $(X,R)$ is prime. Notice that since
$F$ is nonempty, there are at least two points in $E$ and so at least three
points in $X$. It follows that $(X,R)$ is not an arc. Let $U$ be a non-
trivial, closed $Q$ invariant subset of $X^{\prime\prime}$.

If $U$ contains any pair in $X$, then it contains all of $X$ because $(X,R)$
is prime. Then since $F$ is nonempty, the three cycles $\\{a,b,v\\}$ and
$\\{b,a^{\prime},u\\}$ imply that $u,v\in U$.

If $\\{b,v\\}\subset U$ with $b\in B$, then $a\in U$ and if or
$\\{b,u\\}\subset U$ then $a^{\prime}\in U$. Since two points of $X$ are in
$U$, all of $X^{\prime\prime}$ is contained in $U$, again.

Now assume $\\{a,u\\}\subset U$ or $\\{a,v\\}\subset U$ with $a\in A$, then
the $3-$cycle $\\{a,u,v\\}$ for all $a\in E$ first implies first that both $u$
and $v$ are in $U$ and then that all of $E\subset U$. If $b\in B$, then the
cycle $\\{a,b,v\\}$ implies that $b\in U$. Thus, $F\subset U$ and so
$X^{\prime\prime}=U$.

Thus, $(X^{\prime\prime},R^{\prime\prime})$ is prime.

∎

Let $(Y,S)$ be a tournament containing isolated points $u,v$ with $(u,v)\in
S$. Assume that there does not exist $y\in Y\setminus\\{u,v\\}$ such that
either $y\in S(u)\cap S(v)$ or $y\in S^{-1}(u)\cap S^{-1}(v)$. With
$E=S^{\circ-1}(u)=S^{\circ}(v)$ and so
$F=S^{\circ}(u)\setminus\\{v\\}=S^{\circ-1}(v)\setminus\\{u\\}$, we see that
$(Y,S)$ is isomorphic to $(X^{\prime\prime},R^{\prime\prime})$ with
$X=Y\setminus\\{u,v\\}$ and $R=S|X$. Following [2] we then call $(Y,S)$
_reducible_ via $\\{u,v\\}$. If $(Y,S)$ is finite and no such pair $u,v$
exists, then, as above, we call $(Y,S)$ _irreducible_.

For a topological tournament $(X,R)$ we will call two $2-$fold partitions
$(E_{1},F_{1})$ and $(E_{2},F_{2})$ distinct when $E_{1}\not=E_{2}$. Note that

(12.13) $E_{1}\ \not=\ E_{2}\quad\Leftrightarrow\quad F_{1}\ \not=\
F_{2}\quad\Leftrightarrow\quad[(E_{1}\cap F_{2})\cup(E_{2}\cap F_{1})]\ \not=\
\emptyset.$

This allow the possibility, which we will frequently use, that $E_{1}=F_{2}$
and $F_{1}=E_{2}$.

For the attachment construction we begin with two topological tournaments
$(Y,S)$ and $(X,R)$. In the resulting tournament, $(Y,S)$ and $(X,R)$ play
symmetric roles, but it is convenient to use an asymmetric construction
method. We assume that $X$ and $Y$ are disjoint.

Let $\\{C_{i}:i=1,\dots,n\\}$ be a proper $n-$fold partition of $Y$. Let
$\\{(E_{i},F_{i}):i=1,\dots,n\\}$ be a list of $n$ pairwise distinct $2-$fold
partitions of $X$. Define $Z=X\cup Y$ with the topology on $Z$ so that $X$ and
$Y$ are clopen subsets of $Z$ with their initial topologies the relative
topologies from $Z$. Define the topological tournament

$T\subset Z\times Z=(Y\times Y)\cup(X\times X)\cup[(X\times Y)\cup(Y\times
X)]$

by

(12.14) $T\ =\ S\cup R\cup(\bigcup_{i}[(E_{i}\times C_{i})\cup(C_{i}\times
F_{i})]).$

We call $(Z,T)$ the _attachment_ of $(Y,S)$ to $(X,R)$ via
$\\{C_{i}:i=1,\dots,n\\}$ and $\\{(E_{i},F_{i}):i=1,\dots,n\\}$.

Clearly, if $(Y,S)$ and $(X,R)$ are both compact, wac or locally arc cyclic,
then $(Z,T)$ satisfies the corresponding property.

For example, the tournament $(X^{\prime\prime},R^{\prime\prime})$ of
Proposition 12.14 is the attachment of the arc $Y=\\{u,v\\}$ with
$u\rightharpoonup v$ to $(X,R)$ with $C_{1}=\\{u\\},C_{2}=\\{v\\}$ and
$(E_{1},F_{1})=(E,F),(E_{2},F_{2})=(F,E)$.

The tournament $(X^{\prime},R^{\prime})$ of Proposition 12.13 is the
attachment of the trivial tournament on $Y=\\{u\\}$ to $(X,R)$ with
$C_{1}=\\{u\\}$ and $(E_{1},F_{1})=(E,F)$.

######  Theorem 12.15.

Let $(Z,T)$ be the attachment of $(Y,S)$ to $(X,R)$ via
$\\{C_{i}:i=1,\dots,n\\}$ and $\\{(E_{i},F_{i}):i=1,\dots,n\\}$. Assume that
$n\geq 2$ and that for each $(E_{i},F_{i})$ the relation $R\cap(F_{i}\times
E_{i})$ is surjective (and so neither $E_{i}$ nor $F_{i}$ is empty), e.g. it
suffices that each $(E_{i},F_{i})$ be a spanning partition.

If $(Y,S)$ and $(X,R)$ are arc cyclic, then $(Z,T)$ is arc cyclic.

If $(Y,S)$ and $(X,R)$ are prime, then $(Z,T)$ is prime.

If $Y$ is finite with $|Y|=n$ so that each $C_{i}$ is a singleton, and $(X,R)$
is prime, then $(Z,T)$ is prime.

If $X=\bigcup_{i}E_{i}$, then $T\cap(X\times Y)$ is a surjective relation from
$X$ to $Y$.

If $X=\bigcup_{i}F_{i}$, then $T\cap(Y\times X)$ is a surjective relation from
$Y$ to $X$.

###### Proof.

If $c\in C_{i}$ and $b\in F_{i}$, then there exists $a\in E_{i}$ such that
$b\rightharpoonup a$. If $c\in C_{i}$ and $a\in E_{i}$, then there exists
$b\in F_{i}$ such that $b\rightharpoonup a$. In each case, $\\{c,b,a\\}$ is a
$3-$cycle.

It follows that $(Z,T)$ is arc cyclic if $(Y,S)$ and $(X,R)$ are arc cyclic.

Now assume that $(X,R)$ is prime and that $U$ is a nontrivial, closed $Q$
invariant subset of $Z$.

If any pair of $X$ is in $U$, then $X\subset U$ because $(X,R)$ is prime and
so the above $3-$cycles imply that every $c\in C_{i}$ is in $U$. Thus,
$Z\subset U$.

If some pair $\\{c,a\\}\subset U$ with $c\in C_{i}$ and $a\in E_{i}$, then the
above $3-$cycles show that there exists $b\in F_{i}\cap U$. Since some pair in
$X$ is contained in $U$ again $U=Z$. Similarly, if $\\{c,b\\}\subset U$ with
$c\in C_{i}$ and $b\in F_{i}$ we have $U=X$.

There remains the case when some pair of $Y$ is contained in $U$.

Case 1: Assume that $(Y,S)$ is prime. It then follows that $Y\subset U$. Let
$c_{1}\in C_{i},c_{2}\in C_{j}$ with $i\not=j$. Recall that the $C_{i}$’s are
nonempty and $n\geq 2.$ There exists $x\in(E_{i}\cap F_{j})\cup(E_{j}\cap
F_{i})$ because the $2-$fold partitions of $X$ are distinct. It follows that
$T^{\circ}(x)$ and $T^{\circ-1}(x)$ meet $U$ and so $x\in U$. Since
$\\{c_{2},x\\}\subset U$ it follows as above that $U=Z$.

Case 2: Assume that $|Y|=n$. If a pair $\\{c_{1},c_{2}\\}\subset U$ with
$c_{1}\not=c_{2}$ in $Y$, there exist $i\not=j$ such that $\\{c_{1}\\}=C_{i}$
and $\\{c_{2}\\}=C_{j}$. Since $i\not=j$ we may choose $x$ as in Case 1, and
so obtain that $U=Z$.

If $y\in C_{i}$, then with $x\in E_{i},x^{\prime}\in F_{i}$ we have
$x\rightharpoonup y\rightharpoonup x^{\prime}$. If $x\in X$ and
$\bigcup_{i}E_{i}=X$, then $x\in E_{i}$ for some $i$ and so $x\rightharpoonup
y$ for $y\in C_{i}$. Hence, $T\cap(X\times Y)$ is surjective. Similarly if
$\bigcup_{i}F_{i}=X$, then $T\cap(Y\times X)$ is surjective.

∎

Remark: A _Special Case_ which we will use repeatedly has $n=2$, with
$(E_{1},F_{1})$ a spanning partition and $(E_{2},F_{2})=(F_{1},E_{1})$, in
which case, of course, $E_{1}\cup E_{2}=F_{1}\cup F_{2}=X$.

######  Examples 9.

Constructing Generalized Reduced Doubles

Let $(X,R)$ be a compact tournament with no initial or terminal point. Assume
that $X$ totally disconnected with no isolated points. By Proposition 10.22
the assumption that $X$ be totally disconnected is redundant when $(X,R)$ is
prime.

Let $\\{E,F\\}$ be a spanning partition, which exists by Theorem 12.11. Define

(12.15) $C_{1}\ =\ E_{1}\ =\ F_{2}\ =\ E,\quad\text{and}\quad C_{2}\ =\ F_{1}\
=\ E_{2}\ =\ F.$

We let $X\pm=X\times\\{\pm 1\\}$, writing, as before, $x\pm$ for $(x,\pm 1)$
with $x\in X$. The tournaments $(X\pm,R\pm)$ are defined so that each is a
copy of $(X,R)$ via the isomorphisms $x\pm\mapsto x$.

We obtain a generalized reduced double
$2^{\prime}(X,R)=(2^{\prime}X,2^{\prime}R)$, as defined in Example 6 (b), by
using the attachment of $(X+,R+)$ to $(X-,R-)$ via $\\{C_{1},C_{2}\\}$ and
$\\{(E_{1},F_{1}),(E_{2},F_{2})\\}$. This is an example of the Special Case
mentioned in the Remark after Theorem 12.15. In particular,
$2^{\prime}R\cap[(X-\times X+)\cup(X+\times X-)]$ is a surjective relation on
$2^{\prime}X$.

It follows that if $(X,R)$ is wac, locally arc cyclic, arc cyclic or prime,
then $2^{\prime}(X,R)$ satisfies the corresponding property.

##  13\.  Prime Tournament Examples

We will show that $(\mathbb{Z}[2],\widehat{A})$, the standard tournament of
$2-$adics, is a prime tournament. This provides us with an example of a prime,
arc cyclic tournament on a Cantor set. We will use it to construct other such
examples. However, we require an invariant which will allow us to distinguish
among such examples. What we will use for such a tournament $(X,R)$ is the
collection of the almost wac tournaments which are the restrictions of $R$ to
the subsets $R(x)$ as $x$ varies over $X$. In particular, we will look at the
prime quotients of these restrictions.

Now recall that we regard the additive group of $2-$adic integers,
$\mathbb{Z}[2]$ as the product $\\{0.1\\}^{\mathbb{N}}$ with addition of two
sequences pointwise but with carrying to the right. We write
${\mathbf{0}}=000\dots$ for the identity element, instead of $e$, and we write
the group additively. Thus, $\mathbb{Z}[2]$ is a topological group on a Cantor
set. In fact, as it is the inverse limit of the finite rings