# Infinitary Noetherian Constructions II. Transfinite Words and the Regular
Subword Topology

Jean Goubault-Larrecq  Université Paris-Saclay, CNRS, ENS Paris-Saclay,
Laboratoire Méthodes Formelles, 91190, Gif-sur-Yvette, France<EMAIL_ADDRESS>,  Simon Halfon  Université Paris-Saclay, CNRS, ENS Paris-Saclay, Centre
Borelli, 91190, Gif-sur-Yvette, France<EMAIL_ADDRESS>and
Aliaume Lopez  Université Paris-Saclay, CNRS, ENS Paris-Saclay, Laboratoire
Méthodes Formelles, 91190, Gif-sur-Yvette, France  Université de Paris, IRIF,
CNRS, F-75013 Paris, France<EMAIL_ADDRESS>
###### Abstract.

We show that the spaces of transfinite words, namely ordinal-indexed words,
over a Noetherian space, is also Noetherian, under a natural topology which we
call the regular subword topology. We characterize its sobrification and its
specialization ordering, and we give an upper bound on its sobrification rank
and on its stature.

###### Key words and phrases:

Noetherian spaces, well-quasi-orders, infinite words

###### 2010 Mathematics Subject Classification:

Primary 54G99; Secondary 06A07, 06B30

The first and third authors were supported by grant ANR-17-CE40-0028 of the
French National Research Agency ANR (project BRAVAS)

##  1\. Introduction

Given a well-quasi-order $X$ (wqo, for short), the space of infinite words
$X^{\omega}$ (of length $\omega$) need not be wqo in the subword preordering.
One way of correcting this anomaly is to turn to the stronger notion of better
quasi-orderings [11]. Another one is to turn to the weaker notion of
Noetherian space. Noetherian spaces are a natural, topological generalization
of wqos with many similar properties [5, Section 9.7]. For example, there are
Noetherian analogues of Higman’s theorem and of Kruskal’s theorem.
Noetherianity is also preserved by some infinitary constructions such as
powerset.

In part I of this work [6] we have shown that, given a Noetherian space $X$,
$X^{\omega}$ is again Noetherian, with a natural topology, the subword
topology. The same works for the set of finite-or-infinite words
$X^{\leq\omega}$. The purpose of the present paper is to extend this to spaces
$X^{<\alpha}$ of transfinite words, namely words indexed by ordinals strictly
smaller than a fixed ordinal bound $\alpha$.

The topology we choose is a natural generalization of that of part I. The bulk
of the work consists in showing that if $X$ is Noetherian, then so is
$X^{<\alpha}$.

#### _Outline._

Section 2 recapitulates a few basic notions, and is also where we state our
basic tool for showing that $X^{<\alpha}$ is Noetherian, as Proposition 2.1.
We define the regular subword topology on $X^{<\alpha}$ in Section 3, based on
so-called transfinite products. After a few basic results on transfinite words
in Section 4, we show that the $\alpha$-products, a specific kind of
transfinite products, form an irredundant subbase of closed subsets of
$X^{<\alpha}$, where $\alpha$ is a special kind of ordinal which we call a
bound; this is one of the conditions of Proposition 2.1. We characterize
inclusion of transfinite products in Section 6, and we show that every
transfinite product has a canonical form in Section 7. This allows us to show
that the collection of $\alpha$-products is well-founded in Section 8; this is
another condition of Proposition 2.1. The final condition requires us to
express finite intersections of $\alpha$-products as finite unions of
$\alpha$-products, which we do in Section 9. In the process, we obtain an
upper bound on the stature and reduced sobrification rank of $X^{<\alpha}$.
Finally, we describe the specialization preordering of $X^{<\alpha}$ in
Section 10.

##  2\. Preliminaries

###  2.1. Topology.

Most of the following can be found in [5].

Every topological space has a _specialization_ preordering $\leq$, defined by
$x\leq y$ if and only if every open neighborhood of $x$ contains $y$. The
closure of $\\{x\\}$ is the _principal ideal_ $\mathop{\downarrow}\nolimits
x\mathrel{\buildrel\text{def}\over{=}}\\{y\in X\mid y\leq x\\}$. We silently
consider any topological space as a preordered set under $\leq$.

The _Alexandroff topology_ of a preordering $\leq$ is its family of upwards-
closed sets. Among the topologies with a given specialization preordering
$\leq$, it is the finest. The coarsest is the _upper topology_ , whose closed
sets are intersections of sets of the form $\mathop{\downarrow}\nolimits E$,
$E$ finite; we write $\mathop{\downarrow}\nolimits E$ for $\bigcup_{x\in
E}\mathop{\downarrow}\nolimits x$.

A _Noetherian_ space is a topological space in which every open set is
compact. We do not require compactness to imply separation.

A strict partial ordering $<$ is _well-founded_ if and only if there is no
infinite strictly descending chain. By a slight abuse of language, we say that
a preordering $\leq$ is well-founded if and only if its strict part ($x<y$ if
and only if $x\leq y$ and $y\not\leq x$) is well-founded. A space is
Noetherian if and only if its set of closed subsets is well-founded under
inclusion.

A closed subset $C$ is _irreducible_ if and only if $C\neq\emptyset$, and for
all closed sets $F_{1}$, $F_{2}$ such that $C\subseteq F_{1}\cup F_{2}$, $C$
is included in $F_{1}$ or in $F_{2}$. The closure
$\mathop{\downarrow}\nolimits x$ of every point $x$ is irreducible closed. The
product $C_{1}\times C_{2}$ of two irreducible closed subsets is irreducible
in the product topology. A space is _sober_ if and only if every irreducible
closed subset is the closure $\mathop{\downarrow}\nolimits x$ of a unique
point $x$.

An important property is that, in a Noetherian space $X$, every closed subset
is a _finite_ union of irreducible closed subsets.

The _sobrification_ $\mathcal{S}X$ of a topological space $X$ is its set of
irreducible closed subsets, with the sets $\diamond
U\mathrel{\buildrel\text{def}\over{=}}\\{C\in\mathcal{S}X\mid C\cap
U\neq\emptyset\\}$, $U$ open in $X$, as open sets. The specialization ordering
of $\mathcal{S}X$ is inclusion. $\mathcal{S}X$ is sober, and its lattice of
open subsets is order-isomorphic to that of $X$, through $U\mapsto\diamond U$.
In particular, $X$ is Noetherian if and only if $\mathcal{S}X$ is.

$\mathcal{S}$ defines a functor: for every continuous map $f\colon X\to Y$,
$\mathcal{S}f$ maps every $C\in\mathcal{S}X$ to the closure $cl(f[C])$ in $Y$
of $f[C]$. (We write $f[C]$ for the direct image $\\{f(x)\mid x\in C\\}$.) In
particular, $cl(f[C])$ is irreducible closed.

The sober Noetherian spaces can be characterized order-theoretically: they are
exactly the sets $X$ with a well-founded preordering $\leq$ such that every
finite intersection of principal ideals can be expressed as a finite union of
principal ideals; the topology of $X$ is the upper topology of $\leq$. Then
the closed subsets are exactly the sets $\mathop{\downarrow}\nolimits E$ with
$E$ finite.

The following proposition refines this, and will be the core of our
constructions. A family of sets $\mathcal{P}$ is _irredundant_ if and only if
every element of $\mathcal{P}$ is irreducible in $\mathcal{P}$, namely: no
element of $\mathcal{P}$ is empty, and for all $P,P_{1},P_{2}\in\mathcal{P}$,
if $P\subseteq P_{1}\cup P_{2}$ then $P\subseteq P_{1}$ or $P\subseteq P_{2}$.

######  Proposition 2.1.

Let $\mathcal{P}$ be a family of subsets of a set $X$, such that:

  1. (1)

$\mathcal{P}$ is well-founded under inclusion;

  2. (2)

$X$ can be written as a finite union of elements of $\mathcal{P}$;

  3. (3)

for all $P,Q\in\mathcal{P}$, $P\cap Q$ is a finite union of elements of
$\mathcal{P}$.

Then $X$, with the coarsest topology that makes every element of $\mathcal{P}$
closed, is Noetherian. If $\mathcal{P}$ is irredundant, its irreducible closed
subsets are exactly the elements of $\mathcal{P}$, and $\mathcal{S}X$ equals
$\mathcal{P}$ with the upper topology of inclusion.

###### Proof.

By assumption, $\mathcal{P}$ is sober Noetherian in the upper topology of
$\subseteq$.

For every $x\in X$, since $\mathcal{P}$ is well-founded (property (1)), there
is a minimal element $P$ of $\mathcal{P}$ that contains $x$. For every
$Q\in\mathcal{P}$ that contains $x$, we can write $P\cap Q$ as
$\bigcup_{i=1}^{n}P_{i}$ where each $P_{i}$ is in $\mathcal{P}$, by (3). Then
$x$ is in one $P_{i}$, and by minimality of $P$, $P=P_{i}$. It follows that
$P=P_{i}\subseteq P\cap Q$, so $P\subseteq Q$. Hence $P$ is the smallest
element of $\mathcal{P}$ that contains $x$. Let us write that element as
$\eta(x)$; for every $Q\in\mathcal{P}$, $\eta(x)\subseteq Q$ if and only if
$x\in Q$.

This defines a map $\eta\colon X\to\mathcal{P}$. For every finite subset
$E\mathrel{\buildrel\text{def}\over{=}}\\{P_{1},\cdots,P_{n}\\}$ of
$\mathcal{P}$, $\eta^{-1}(\mathop{\downarrow}\nolimits E)=\\{x\in X\mid\exists
i,\eta(x)\subseteq P_{i}\\}=\bigcup_{i=1}^{n}P_{i}$ is closed in $X$. Hence
$\eta$ is continuous. Taking $n\mathrel{\buildrel\text{def}\over{=}}1$, we
obtain that every element $P$ of $\mathcal{P}$ can be written as
$\eta^{-1}(\mathop{\downarrow}\nolimits\\{P\\})$. Since $\eta^{-1}$ commutes
with all intersections and (finite) unions, $\eta$ is a _full_ map, viz. every
closed subset of $X$ is the inverse image of some closed subset of
$\mathcal{P}$—this was called an _initial_ map in [6]. Lemma 8 of that paper
states that any space from which there is a full map to a Noetherian space is
itself Noetherian. Therefore $X$ is Noetherian.

By Lemma 9 of the same paper, every irreducible closed subset $C$ of $X$ is of
the form $\eta^{-1}(\mathcal{C})$ for some irreducible closed subset
$\mathcal{C}$ of $\mathcal{P}$. Since $\mathcal{P}$ is sober,
$\mathcal{C}=\mathop{\downarrow}\nolimits P$ for some unique
$P\in\mathcal{P}$, hence $C=P$.

Conversely, and assuming $\mathcal{P}$ irredundant, we claim that every
element $P$ of $\mathcal{P}$ is irreducible in $X$. By assumption, $P$ is non-
empty. Since $\eta$ is full, the closed subsets of $X$ are exactly the inverse
images of closed subsets of $\mathcal{P}$, namely the finite unions of
elements of $P$. Hence it suffices to show that if $P$ is included in a finite
union of elements of $\mathcal{P}$, then it is included in one of them. This
follows directly from the fact that $\mathcal{P}$ is irredundant. ∎

###  2.2. Ordinals.

An _indecomposable_ ordinal is one of the form $\omega^{\beta}$, where $\beta$
is an ordinal; equivalently, an ordinal that cannot be written as the sum of
two strictly smaller ordinals. The other ordinals are _decomposable_. For all
ordinals $\alpha$, $\beta$, $\gamma$, we have:  (1) if $\gamma$ is
indecomposable and $\alpha,\beta<\gamma$, then $\alpha+\beta<\gamma$;  (2) if
$\gamma$ is indecomposable, $\alpha<\gamma$, and $\beta\leq\gamma$, then
$\alpha+\beta\leq\gamma$;  (3) every ordinal $\alpha$ can be written in a
unique way as a finite sum of indecomposable ordinals
$\gamma_{1}+\cdots+\gamma_{m}$ (its _Cantor normal form_), where
$\alpha\geq\gamma_{1}\geq\cdots\geq\gamma_{m}$;  (4) the ordering on Cantor
normal forms is lexicographic:
$\gamma_{1}+\cdots+\gamma_{m}<\gamma^{\prime}_{1}+\cdots+\gamma^{\prime}_{n}$
(where both sides are in Cantor normal form) if and only if for some $i\geq
1$, $\gamma_{1}=\gamma^{\prime}_{1}$, …, $\gamma_{i-1}=\gamma^{\prime}_{i-1}$
and either $i=m+1\leq n$, or $i\leq m,n$ and $\gamma_{i}<\gamma^{\prime}_{i}$.

##  3\. Transfinite words

We call _transfinite word_ on a set $X$ any map from $\alpha$ to $X$, where
$\alpha$ is any ordinal. Such a word $w$ has _length_
$|w|\mathrel{\buildrel\text{def}\over{=}}\alpha$. Seen as the set of ordinals
$\beta$ strictly less than $\alpha$, $|w|$ is also the _domain_ of $w$, and we
write ${w}({\beta})$ for the letter at position $\beta$ in $w$, for every
$\beta\in|w|$ (equivalently, $\beta<|w|$).

When $X$ is preordered by $\leq$, the _subword preordering_ $\leq$ is defined
on transfinite words by $w\leq_{*}w^{\prime}$ if and only if there is a
strictly increasing map $f\colon|w|\to|w^{\prime}|$ such that
${w}({\beta})\leq{w^{\prime}}({f(\beta)})$ for every $\beta$,
$0\leq\beta<|w|$. We also say that $f$ _exhibits_ $w$ as a subword of
$w^{\prime}$.

We write $X^{<\alpha}$ for the set of transfinite words $w$ of length
$|w|<\alpha$. $X^{\leq\alpha}$ denotes $X^{<\alpha+1}$. For example,
$X^{<\omega}$ is the set $X^{*}$ of finite words on $X$, and
$X^{<\omega+1}=X^{\leq\omega}$ is the set of finite-or-infinite words studied
in Part I [6].

The _concatenation_ $ww^{\prime}$ of $w$ and $w^{\prime}$ is the transfinite
word of length $|w|+|w^{\prime}|$ such that
${(ww^{\prime})}({\beta})={w}({\beta})$ for every $\beta<|w|$, and
${(ww^{\prime})}({|w|+\beta})={w^{\prime}}({\beta})$ for every
$\beta<|w^{\prime}|$. We write $AB$ for the set $\\{ww^{\prime}\mid w\in
A,w^{\prime}\in B\\}$.

We are interested in the following topology. We cannot work on the (proper)
class of all transfinite words over $X$, for foundational reasons. Instead we
work on _sets_ $Y$ of transfinite words; this leads us to take intersections
with $Y$ here and there. Usually, $Y$ will be a set of the form $X^{<\alpha}$.

######  Definition 3.1 (Regular subword topology).

The _regular subword topology_ on any set of transfinite words $Y$ on a space
$X$ is the coarsest topology that makes the sets
$(F_{1}^{<\alpha_{1}}F_{2}^{<\alpha_{2}}\allowbreak\cdots\allowbreak
F_{n}^{<\alpha_{n}})\cap Y$ closed, where $n\in\mathbb{N}$, $F_{1}$, $F_{2}$,
…, $F_{n}$ are closed subsets of $X$, and $\alpha_{1}$, $\alpha_{2}$, …,
$\alpha_{n}$ are ordinals.

The following class of ordinals will be ubiquitous.

######  Definition 3.2 (Bound).

A _bound_ is an ordinal of the form $\omega^{\beta}$ or $\omega^{\beta}+1$,
$\beta\geq 0$. The _trivial bound_ is $\omega^{0}$ ($=1$), all others are
_non-trivial_. A _proper bound_ is one of the form $\omega^{\beta}$ or
$\omega^{\beta}+1$ with $\beta\geq 1$.

######  Definition 3.3 (Preatom, atom, product).

Let $X$ be a topological space. A _preatom_ is an expression of the form
$F^{<\gamma}$, where $F$ is a closed subset of $X$ and $\gamma$ is a bound. An
_atom_ is a preatom $F^{<\gamma}$ such that $\gamma$ is non-trivial, and if
$\gamma=\omega^{0}+1$ then $F$ is irreducible closed in $X$.

A _transfinite product_ $P$ is any set of the form $A_{1}A_{2}\cdots A_{n}$,
where $n\in\mathbb{N}$ and each $A_{i}$ is an atom. We write $\varepsilon$
when $n=0$, namely
$\varepsilon\mathrel{\buildrel\text{def}\over{=}}\\{\epsilon\\}$.

When $\beta=0$, $F^{<\omega^{0}}=\varepsilon=\\{\epsilon\\}$, and
$F^{<\omega^{0}+1}$ is sometimes written as $F^{?}$: that is the set of words
of length at most $1$, whose only letter if any is in $F$.

######  Lemma 3.4.

For every set $F$, for every ordinal $\alpha$, one can write $F^{<\alpha}$ as
$F^{<\gamma_{1}}F^{<\gamma_{2}}\allowbreak\cdots\allowbreak F^{<\gamma_{m}}$,
where $m\in\mathbb{N}$ and each $\gamma_{i}$ is a non-trivial bound.

###### Proof.

If $\alpha=0$, we take $m\mathrel{\buildrel\text{def}\over{=}}0$. Otherwise,
let us write $\alpha$ in Cantor normal form, as a finite sum
$\gamma_{1}+\cdots+\gamma_{m}$ of indecomposable ordinals, where $m\neq 0$ and
$\alpha\geq\gamma_{1}\geq\cdots\geq\gamma_{m}$. We claim that $F^{<\alpha}$ is
equal to $F^{\leq\gamma_{1}}F^{\leq\gamma_{2}}\cdots
F^{\leq\gamma_{m-1}}F^{<\gamma_{m}}$. The result will follow, since the latter
is equal to $F^{<\gamma_{1}+1}F^{<\gamma_{2}+1}\cdots
F^{<\gamma_{m-1}+1}F^{<\gamma_{m}}$: when $\gamma_{m}\neq 1$, all the
superscripts are non-trivial bounds; when $\gamma_{m}=1$, this simplifies to
$F^{<\gamma_{1}+1}F^{<\gamma_{2}+1}\allowbreak\cdots\allowbreak
F^{<\gamma_{m-1}+1}$, where all superscripts are non trivial bounds, too.

$F^{\leq\gamma_{1}}\cdots F^{\leq\gamma_{m-1}}F^{<\gamma_{m}}$ is included in
$F^{<\alpha}$. Conversely, let $w\in F^{<\alpha}$, and let us write $|w|$ in
Cantor normal form as $\gamma^{\prime}_{1}+\cdots+\gamma^{\prime}_{n}$. We can
write $w$ as a concatenation $w_{1}\cdots w_{n}$ where
$|w_{1}|=\gamma^{\prime}_{1}$, …, $|w_{n}|=\gamma^{\prime}_{n}$. Since
$|w|<\alpha$, there is a number $i\geq 1$ such that
$\gamma^{\prime}_{1}=\gamma_{1}$, …, $\gamma^{\prime}_{i-1}=\gamma_{i-1}$ and
either $i=n+1\leq m$, or $i\leq m,n$ and $\gamma^{\prime}_{i}<\gamma_{i}$. In
the first case, $w\in F^{\leq\gamma_{1}}\cdots F^{\leq\gamma_{n}}\subseteq
F^{\leq\gamma_{1}}\cdots F^{\leq\gamma_{m-1}}F^{<\gamma_{m}}$. In the second
case, since
$\gamma_{i}>\gamma^{\prime}_{i}\geq\gamma^{\prime}_{i+1}\geq\cdots\geq\gamma^{\prime}_{n}$
and $\gamma_{i}$ is indecomposable,
$\gamma^{\prime}_{i}+\cdots+\gamma^{\prime}_{n}<\gamma_{i}$. Then $w_{1}\in
F^{\leq\gamma_{1}}$, …, $w_{i-1}\in F^{\leq\gamma_{i-1}}$, and
$w_{i}w_{i+1}\cdots w_{n}$ is in $F^{<\gamma_{i}}$, hence in
$F^{\leq\gamma_{i}}\cdots F^{\leq\gamma_{m-1}}F^{<\gamma_{m}}$. ∎

######  Proposition 3.5.

The regular subword topology on any set of transfinite words $Y$ on a
Noetherian space $X$ is the coarsest topology that has the intersections of
$Y$ with transfinite products as closed sets.

###### Proof.

Let us consider a set of the form
$F_{1}^{<\alpha_{1}}F_{2}^{<\alpha_{2}}\cdots F_{n}^{<\alpha_{n}}$, where
$F_{1}$, …, $F_{n}$ are closed subsets of $X$, and $\alpha_{1}$, $\alpha_{2}$,
…, $\alpha_{n}$ are ordinals. We claim that we can rewrite it as a finite
union of transfinite products.

If some $\alpha_{i}$ equals $0$, then
$F_{1}^{<\alpha_{1}}F_{2}^{<\alpha_{2}}\cdots F_{n}^{<\alpha_{n}}$ is empty.
Otherwise, using Lemma 3.4, we may assume that every $\alpha_{i}$ is a non-
trivial bound. We may also remove the preatoms $F_{i}^{<\alpha_{i}}$ such that
$F_{i}=\emptyset$, since in that case $F_{i}^{\alpha_{i}}=\varepsilon$. Let
$I$ be the subset of those indices $i$, $1\leq i\leq n$, such that
$\alpha_{i}=\omega^{0}+1$. For each $i\in I$, we can write $F_{i}$ as a finite
union of irreducible closed subsets $C_{i1}$, …, $C_{ik_{i}}$ (and $k_{i}\neq
0$ since $F_{i}\neq\emptyset$), since $X$ is Noetherian. For every function
$f$ mapping each $i\in I$ to an element of $\\{1,\cdots,k_{i}\\}$, let $P_{f}$
be the transfinite product obtained from
$F_{1}^{<\alpha_{1}}F_{2}^{<\alpha_{2}}\cdots F_{n}^{<\alpha_{n}}$ by
replacing each preatom $F_{i}^{<\alpha_{i}}$, $i\in I$, by $C_{if(i)}^{?}$.
Then $F_{1}^{<\alpha_{1}}F_{2}^{<\alpha_{2}}\cdots F_{n}^{<\alpha_{n}}$ is the
finite union of the transfinite products $P_{f}$, when $f$ varies over the
finitely many possible functions. ∎

##  4\. Elementary combinatorics on transfinite words

######  Lemma 4.1.

Let $\gamma$, $\gamma^{\prime}$ be two bounds, and $u$ and $v$ be transfinite
words.

  1. (1)

If $\gamma$ is indecomposable and $|u|,|v|<\gamma$, then $|uv|<\gamma$.

  2. (2)

If $\gamma<\gamma^{\prime}$ and $|u|<\gamma$, $|v|<\gamma^{\prime}$, then
$|uv|<\gamma^{\prime}$.

  3. (3)

If $\gamma$ is proper and $|u|<\gamma$, then for every $x\in X$,
$|xw|<\gamma$.

###### Proof.

(1) By definition of indecomposability, using $|uv|=|u|+|v|$.

(2) follows from (1) if $\gamma^{\prime}$ is indecomposable. Otherwise, let us
write $\gamma^{\prime}$ as $\omega^{\beta}+1$. Then $\gamma<\gamma^{\prime}$
means $\gamma\leq\omega^{\beta}$. Since $|u|<\gamma$, $|u|<\omega^{\beta}$ and
since $|v|<\gamma^{\prime}$, $|v|\leq\omega^{\beta}$. Hence
$|uv|=|u|+|v|\leq\omega^{\beta}<\gamma^{\prime}$, because $\omega^{\beta}$ is
indecomposable.

(3) Since $\gamma$ is proper, we have $|x|=\omega^{0}<\gamma$; then (3)
follows from (2). ∎

######  Lemma 4.2.

Every closed set in the regular subword topology is downwards-closed with
respect to $\leq_{*}$.

###### Proof.

Given any two downwards-closed subsets $A$ and $B$ with respect to $\leq_{*}$,
their product $AB$ is, too: if $w\leq_{*}w^{\prime}$ and $w^{\prime}\in AB$,
then we can write $w^{\prime}$ as $u^{\prime}v^{\prime}$ where $u^{\prime}\in
A$ and $v^{\prime}\in B$, and it is then easy to show that $w=uv$ for some
$u\leq_{*}u^{\prime}$ and $v\leq_{*}v^{\prime}$. It is clear that every atom
is downwards-closed, as well as $\varepsilon$, so every transfinite product is
downwards-closed, hence also every intersection of finite unions of
transfinite products. ∎

For every indecomposable ordinal $\gamma$, there is a so-called _Hessenberg
pairing_ map $H\colon\gamma\times\gamma\to\gamma$, which is injective; for all
$\alpha<\beta<\gamma$ and $\delta<\gamma$, $H(\alpha,\delta)<H(\beta,\delta)$
and $H(\delta,\alpha)<H(\delta,\beta)$ [10, Exercise 2.23 (ii)], and it is
easy to see that $H(\alpha,0)\geq\alpha$ for every $\alpha<\gamma$.

######  Lemma 4.3.

Let $F$ be a non-empty subset of a set $X$. For every transfinite word $w$ on
$X$ such that $|w|$ is an indecomposable ordinal $\gamma$, and whose letters
are in $F$, there is a word $w^{\prime}$ of length $\gamma$, whose letters are
in $F$ again, such that for every way of writing $w^{\prime}$ as a
concatenation $uv$ with $|u|<\gamma$, $w\leq_{*}v$.

###### Proof.

Let us pick $x\in F$. We build $w^{\prime}$ as the following word of length
$\gamma$:
${w^{\prime}}({H(\alpha,\beta)})\mathrel{\buildrel\text{def}\over{=}}{w}({\beta})$
for all $\alpha,\beta<\gamma$, and
${w^{\prime}}({\delta})\mathrel{\buildrel\text{def}\over{=}}x$ for every
position $\delta<\gamma$ that is not in the range of $H$. Now let us write
$w^{\prime}$ as $uv$ with $|u|<\gamma$. Then $H(|u|,\\_)$ exhibits $w$ as a
subword of $v$, using the fact that $H(|u|,0)\geq|u|$. ∎

##  5\. Continuity and irredundancy

######  Lemma 5.1.

Let $X$ be a Noetherian space.

  1. (1)

For any set $Y$ of transfinite words on $X$ containing $X^{\leq 1}$, the
function $i\colon X\to Y$ mapping $x\in X$ to the one-letter word $x$ is
continuous.

  2. (2)

For every ordinal $\beta$, the concatenation map $cat\colon
X^{<\omega^{\beta}}\times X^{\leq\omega^{\beta}}\to X^{\leq\omega^{\beta}}$
(resp., $cat\colon X^{<\omega^{\beta}}\times X^{<\omega^{\beta}}\to
X^{<\omega^{\beta}}$) is continuous.

###### Proof.

(1) For every preatom $F^{<\gamma}$ with $\gamma$ non-trivial,
$i^{-1}(F^{<\gamma})=F$ is closed. Then, the inverse image of any transfinite
product $A_{1}A_{2}\cdots A_{n}\cap Y$ is $i^{-1}(A_{1})\cup\cdots\cup
i^{-1}(A_{n})$, which is closed.

(2) First, $cat$ is well-defined by Lemma 4.1 (1). As far as continuity is
concerned, let $P\mathrel{\buildrel\text{def}\over{=}}A_{1}A_{2}\cdots A_{n}$
be any transfinite product. We show that $cat^{-1}(P)$ (or rather,
$cat^{-1}(P\cap X^{\leq\omega^{\beta}})$) is (the intersection of
$X^{<\omega^{\beta}}\times X^{\leq\omega^{\beta}}$ with) a finite union
$\bigcup_{i=1}^{m}P_{i}\times Q_{i}$ of products of pairs of transfinite word
products $P_{i}$, $Q_{i}$, by induction on $n$. If $n=0$, then $cat^{-1}(P)$
only contains $(\epsilon,\epsilon)$, hence is equal to
$\varepsilon\times\varepsilon$. Otherwise, let
$P^{\prime}\mathrel{\buildrel\text{def}\over{=}}A_{2}\cdots A_{n}$, and let us
write $A_{1}$ as $F^{<\gamma}$, where $\gamma$ is a non-trivial bound, and $F$
is closed. By induction hypothesis, $cat^{-1}(P^{\prime})$ is a finite union
$\bigcup_{i=1}^{m}P_{i}\times Q_{i}$, where $P_{i}$ and $Q_{i}$ are
transfinite word products. The pairs of words $u$, $v$ whose concatenation are
in $P$ are those such that $u$ is of the form $u_{1}u_{2}$ with $u_{1}\in
A_{1}$ and $(u_{2},v)\in cat^{-1}(P^{\prime})$ (namely, the elements of
$\bigcup_{i=1}^{m}A_{1}P_{i}\times Q_{i}$, since, as one sees easily,
concatenation distributes over union) or such that $v$ is of the form
$v_{1}v_{2}$ with $uv_{1}\in A_{1}$ and $v_{2}\in P^{\prime}$. In order to
conclude, it therefore suffices to show that the set $A$ of pairs $(u,v)$ with
$v$ of the form $v_{1}v_{2}$, $uv_{1}\in A_{1}$ and $v_{2}\in P^{\prime}$, is
a finite union of products of pairs of transfinite products.

If $\gamma$ is of the form $\omega^{\beta}$, then $uv_{1}\in A_{1}$ if and
only if $u\in F^{<\gamma}$ and $v_{1}\in F^{<\gamma}$. The only if direction
is clear, and the if direction is by Lemma 4.1 (1). Hence $A=F^{<\gamma}\times
F^{<\gamma}P^{\prime}$ in this case.

If $\gamma$ is of the form $\omega^{\beta}+1$, then $uv_{1}\in A_{1}$ if and
only if $u\in F^{<\omega^{\beta}}$ and $v_{1}\in F^{\leq\omega^{\beta}}$, or
$u\in F^{\leq\omega^{\beta}}$ and $v_{1}=\epsilon$. In the only if direction,
we reason by cases, depending whether $|u|=\omega^{\beta}$ or not. In the if
direction, the case $v_{1}=\epsilon$ is obvious, while $u\in
F^{<\omega^{\beta}}$ and $v_{1}\in F^{\leq\omega^{\beta}}$ imply $uv_{1}\in
A_{1}=F^{<\gamma}$ by Lemma 4.1 (2). Hence $A=(F^{<\omega^{\beta}}\times
F^{<\gamma}P^{\prime})\cup(F^{<\gamma}\times P^{\prime})$ in this case. ∎

On spaces of the form $X^{<\alpha}$, the following refinement of the notion of
transfinite product will be the family $\mathcal{P}$ we will use Proposition
2.1 on.

######  Definition 5.2 ($\alpha$-product).

For a topological space $X$ and a bound $\alpha$, the _$\alpha$ -products_ are
the products of the form
${F_{1}}^{<\gamma_{1}}{F_{2}}^{<\gamma_{2}}\cdots{F_{n}}^{<\gamma_{n}}$ where
$n\in\mathbb{N}$,

  * •

$\gamma_{i}\leq\alpha$ and $F_{i}$ is non-empty for each $i$, $1\leq i\leq n$,

  * •

and if $\alpha$ is decomposable, then $\gamma_{i}<\alpha$ for every $i$,
$1\leq i<n$; namely, the only $\gamma_{i}$ that is equal to $\alpha$, if any,
is obtained with $i=n$.

######  Proposition 5.3.

For every topological space $X$ and every bound $\alpha$, the
$\alpha$-products are closed in $X^{<\alpha}$, and their complements form a
subbase of the regular subword topology.

###### Proof.

Let
$P\mathrel{\buildrel\text{def}\over{=}}{F_{1}}^{<\gamma_{1}}{F_{2}}^{<\gamma_{2}}\cdots{F_{n}}^{<\gamma_{n}}$
be an $\alpha$-product. Let us consider any transfinite word $w$ in $P$, and
let us write it as $u_{1}u_{2}\cdots u_{n}$, where
$u_{i}\in{F_{i}}^{<\gamma_{i}}$ for each $i$. If $\alpha$ is indecomposable,
then $|w|<\alpha$ because $|u_{i}|<\alpha$ for every $i$, and using Lemma 4.1
(1). Otherwise, let us write $\alpha$ as $\omega^{\beta}+1$. If
$\gamma_{i}\leq\omega^{\beta}$ for every $i$, by the same argument
$|w|<\omega^{\beta}$, hence $|w|<\alpha$. By the second item in the definition
of $\alpha$-products, the only remaining possibility is that
$\gamma_{i}\leq\omega^{\beta}$ for every $i$, $1\leq i<n$, that $n\geq 1$ and
that $\gamma_{n}=\omega^{\beta}+1=\alpha$. Then $|u_{n}|<\alpha$; since
$|u_{n-1}|<\gamma_{n-1}<\alpha$, we obtain $|u_{n-1}u_{n}|<\alpha$ by Lemma
4.1 (2); then $|u_{n-2}u_{n-1}u_{n}|<\alpha$, …, and eventually $|w|<\alpha$.

To show the second part of the proposition, we claim that the intersection of
every transfinite product
$P\mathrel{\buildrel\text{def}\over{=}}{F_{1}}^{\gamma_{1}}{F_{2}}^{<\gamma_{2}}\cdots{F_{n}}^{<\gamma_{n}}$
with $X^{<\alpha}$ is a finite union of $\alpha$-products. If $n=0$, $P$ is
already an $\alpha$-product, so let us assume $n\neq 0$. Given any $w\in P\cap
X^{<\alpha}$, we can write $w$ as $u_{1}u_{2}\cdots u_{n}$, where
$u_{i}\in{F_{i}}^{<\gamma_{i}}$ for each $i$. For each $i$, not only
$|u_{i}|<\gamma_{i}$ but also $|u_{i}|\leq|w|<\alpha$, so we can assume
without loss of generality that $\gamma_{i}\leq\alpha$ for every $i$, $1\leq
i\leq n$.

If $\alpha$ is indecomposable, then this makes $P$ an $\alpha$-product.
Henceforth let us assume that $\alpha$ is decomposable, say
$\alpha\mathrel{\buildrel\text{def}\over{=}}\omega^{\beta}+1$. Then we can
rewrite every transfinite product of the form $F^{<\alpha}Q$ included in
$X^{<\alpha}$ as $F^{<\omega^{\beta}}Q\cup F^{<\alpha}$. Indeed, every word
$w\mathrel{\buildrel\text{def}\over{=}}uv$ with $u\in F^{<\alpha}$ and $v\in
Q$ is either such that $|u|<\omega^{\beta}$ (then $w\in
F^{<\omega^{\beta}}Q$), or $|u|=\omega^{\beta}$. If $|u|=\omega^{\beta}$, then
$v=\epsilon$, otherwise $|w|>\omega^{\beta}$, which is impossible since $w\in
X^{<\alpha}$; so $w\in F^{<\alpha}$.

We can therefore rewrite $P$ as follows. Let $i_{1}<\cdots<i_{k}$ be the list
of indices $i$ between $1$ and $n-1$ such that $\gamma_{i}=\alpha$. Let
$\gamma^{\prime}_{i}\mathrel{\buildrel\text{def}\over{=}}\gamma_{i}$ for every
$i$ different from $i_{1}$, …, $i_{k}$, and
$\gamma^{\prime}_{i}\mathrel{\buildrel\text{def}\over{=}}\omega^{\beta}$
otherwise. (I.e., we replace the exponents equal to $\alpha=\omega^{\beta}+1$
by $\omega^{\beta}$.) Then $P$ is the union of the $\alpha$-products
${F_{1}}^{<\gamma^{\prime}_{1}}{F_{2}}^{<\gamma^{\prime}_{2}}\cdots{F_{i_{j}-1}}^{<\gamma^{\prime}_{i_{j}-1}}{F_{i_{j}}}^{<\alpha}$,
$1\leq j\leq k$, and
${F_{1}}^{<\gamma^{\prime}_{1}}{F_{2}}^{<\gamma^{\prime}_{2}}\cdots{F_{n}}^{<\gamma^{\prime}_{n}}$.
∎

######  Proposition 5.4.

Let $X$ be a Noetherian space. For every bound $\alpha$, every
$\alpha$-product is irreducible in $X^{<\alpha}$. Hence the family of
$\alpha$-products is irredundant.

###### Proof.

We first show that every atom $F^{<\gamma}$ is irreducible in $X^{<\alpha}$.

If $\gamma=\omega^{0}+1$ and $F$ is irreducible, then $i$ is continuous (Lemma
5.1 (1)), so $\mathcal{S}i(F)=cl(i[F])$ is irreducible closed in
$X^{<\alpha}$. Then $cl(i[F])$ is non-empty, and downwards-closed under
$\leq_{*}$ by Lemma 4.2, so it contains $\epsilon$. Clearly $i[F]\subseteq
cl(i[F])$, so $F^{<\gamma}=F^{?}=\\{\epsilon\\}\cup i[F]\subseteq cl(i[F])$.
Also, $i[F]\subseteq F^{<\gamma}$, and $F^{<\gamma}$ is closed. Therefore
$F^{<\gamma}=cl(i[F])$, so $F^{<\gamma}$ is irreducible.

Let now $\gamma$ be a proper bound. We show that $F^{<\gamma}$ is directed in
$\leq_{*}$, namely that $F^{<\gamma}\neq\emptyset$ and that any two elements
$u$, $v$ of $F^{<\gamma}$ have an upper bound $w$ in $F^{<\gamma}$. If
$\gamma$ is indecomposable, then $w\mathrel{\buildrel\text{def}\over{=}}uv$
fits, using Lemma 4.1 (1). Otherwise, let
$\gamma\mathrel{\buildrel\text{def}\over{=}}\omega^{\beta}+1$. If
$|u|<\omega^{\beta}$, then $uv$ fits again, by Lemma 4.1 (2). If
$|v|<\omega^{\beta}$, then we pick $vu$ instead. Finally, if
$|u|=|v|=\omega^{\beta}$, we define $w$ as the one-for-one interleaving of $u$
and $v$, namely as the word of length $\omega^{\beta}$ such that, for every
ordinal $\lambda+n<\omega^{\beta}$ (where $\lambda$ is $0$ or a limit ordinal,
and $n\in\mathbb{N}$), ${w}({\lambda+2n})={u}({\lambda+n})$ and
${w}({\lambda+2n+1})={v}({\lambda+n})$.

Since $F^{<\gamma}$ is directed, $F^{<\gamma}$ is irreducible. Indeed, if
$F^{<\gamma}\subseteq\mathcal{C}_{1}\cup\mathcal{C}_{2}$ where
$\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are closed in $X^{<\alpha}$, but
$F^{<\gamma}\not\subseteq\mathcal{C}_{1},\mathcal{C}_{2}$, then we can pick
$u\in F^{<\gamma}\smallsetminus\mathcal{C}_{1}$ and $v\in
F^{<\gamma}\smallsetminus\mathcal{C}_{2}$; let $w$ be an upper bound of $u$
and $v$ in $F^{<\gamma}$, then $w$ is neither in $\mathcal{C}_{1}$ nor in
$\mathcal{C}_{2}$, since those sets are downwards-closed under $\leq_{*}$
(Lemma 4.2), and therefore $w\in
F^{<\gamma}\smallsetminus(\mathcal{C}_{1}\cup\mathcal{C}_{2})$, which is
impossible.

We now prove that every $\alpha$-product $P$ is irreducible in $X^{<\alpha}$,
by induction on the number $n$ of atoms in $P$. If $n=0$, then
$P=\varepsilon$, and this is clear. If $n\geq 1$, we have just seen that
$P=F^{<\gamma}$ is irreducible. If $n\geq 2$, we write $P$ as $F^{<\gamma}Q$
where $Q$ is shorter, hence irreducible by induction hypothesis.

If $\alpha$ is indecomposable, then $F^{<\gamma}$ is irreducible in
$X^{<\alpha}$, so $F^{<\gamma}\times Q$ is irreducible in $X^{<\alpha}\times
X^{<\alpha}$. By Lemma 5.1 (2), $cat$ is continuous from the latter to
$X^{<\alpha}$; so $\mathcal{S}{cat}(F^{<\gamma}\times
Q)=cl(cat[F^{<\gamma}\times Q])$ is irreducible in $X^{<\alpha}$. Now
$cat[F^{<\gamma}\times Q]=F^{<\gamma}Q$ is closed, hence equal to its own
closure.

If $\alpha$ is decomposable, then let
$\alpha\mathrel{\buildrel\text{def}\over{=}}\omega^{\beta}+1$. By the second
item in the definition of $\alpha$-products, we must have $\gamma<\alpha$,
hence $\gamma\leq\omega^{\beta}$. Then $F^{<\gamma}$ is irreducible in
$X^{<\omega^{\beta}}$, while $Q$ is irreducible in $X^{\leq\omega^{\beta}}$.
We use Lemma 5.1 (2): $cat$ is continuous from $X^{<\omega^{\beta}}\times
X^{\leq\omega^{\beta}}$ to $X^{\leq\omega^{\beta}}$. Then we conclude as above
that $F^{<\gamma}Q=cat[F^{<\gamma}\times Q]=cl(cat[F^{<\gamma}\times Q])$ is
irreducible. ∎

##  6\. Inclusion of transfinite products

We start with necessary conditions for inclusion of transfinite products. We
abbreviate “$\gamma=\gamma^{\prime}$ and $\gamma$ is decomposable” as
“$\gamma=\gamma^{\prime}$ is decomposable”.

######  Lemma 6.1.

Let $\gamma$, $\gamma^{\prime}$ be two non-trivial bounds, and let
$F^{<\gamma}P$, ${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ be two transfinite
products. If $F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$,
then:

  1. (1)

$P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$.

  2. (2)

If $F\not\subseteq F^{\prime}$ then $F^{<\gamma}P\subseteq P^{\prime}$.

  3. (3)

If $F$ is non-empty and if $\gamma>\gamma^{\prime}$, then
$F^{<\gamma}P\subseteq P^{\prime}$.

  4. (4)

If $F\subseteq F^{\prime}$ and $F\neq\emptyset$, and if
$\gamma=\gamma^{\prime}$ is decomposable, then $P\subseteq P^{\prime}$.

###### Proof.

Let us assume that
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$.

(1) $P\subseteq F^{<\gamma}P$, because $\epsilon\in F^{<\gamma}$, and
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ by
assumption.

(2) Let us assume $F\not\subseteq F^{\prime}$. Then there is a letter $x$ in
$F\smallsetminus F^{\prime}$. Let $w\in F^{<\gamma}P$ be arbitrary, and let us
write $w$ as $uv$ where $u\in F^{<\gamma}$ and $v\in P$.

If $\gamma$ is proper, then $xu$ is in $F^{<\gamma}$ again, by Lemma 4.1 (3),
so $xw=xuv$ is in $F^{<\gamma}P$, hence in
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. Since $x\not\in F^{\prime}$,
$xuv$ is in $P^{\prime}$, so $w=uv$ is in $P^{\prime}$, by Lemma 4.2 and since
$w\leq_{*}xuv$. Since $w$ is arbitrary, $F^{<\gamma}P\subseteq P^{\prime}$.

It remains to deal with the case $\gamma=\omega^{0}+1$, $F$ irreducible. Let
$A\mathrel{\buildrel\text{def}\over{=}}\\{y\in X\mid\forall v\in P,yv\in
P^{\prime}\\}$. For every $y\in F\smallsetminus F^{\prime}$, $y$ is in $A$:
indeed, for every $v\in P$, $yv$ is in $F^{?}P=F^{<\gamma}P$ hence in
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, and since $y\not\in F^{\prime}$,
$yv$ must be in $P^{\prime}$. This means that $F\subseteq F^{\prime}\cup A$.
$A$ is also equal to $\bigcap_{v\in P}cat(i(\\_),v)^{-1}(P^{\prime})$, where
$cat(i(\\_),v)\colon y\mapsto cat(i(y),v)=yv$ is continuous by Lemma 5.1.
Hence $A$ is closed. Since $F$ is irreducible, and since $F\not\subseteq
F^{\prime}$, $F$ must be included in $A$; equivalently, $FP\subseteq
P^{\prime}$. Since $P^{\prime}$ is downwards-closed with respect to $\leq_{*}$
(Lemma 4.2), and since every element $u$ of $P$ is a subword of $xu\in FP$,
$F^{?}P=P\cup FP$ is also included in $P^{\prime}$—namely,
$F^{<\gamma}P\subseteq P^{\prime}$.

(3) Let us assume
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, with
$F\neq\emptyset$ and $\gamma>{\gamma^{\prime}}$. We pick $x\in F$. For every
$\alpha$, let $x^{\alpha}$ be the word of length $\alpha$ whose sole letter is
$x$.

Whether $\gamma^{\prime}$ is equal to $\omega^{\beta^{\prime}}$ or to
$\omega^{\beta^{\prime}+1}$, we define ${\gamma^{\prime}}^{-}$ as
$\omega^{\beta^{\prime}}$. For every transfinite word $v$, if
$x^{{\gamma^{\prime}}^{-}}v\in{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, then
$v\in P^{\prime}$. Indeed, assuming
$x^{{\gamma^{\prime}}^{-}}v\in{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, we
can write $x^{{\gamma^{\prime}}^{-}}v$ as $v_{1}v_{2}$ where
$v_{1}\in{F^{\prime}}^{<\gamma^{\prime}}$ and $v_{2}\in P^{\prime}$. If
$\gamma^{\prime}=\omega^{\beta^{\prime}}$, then
$|x^{{\gamma^{\prime}}^{-}}|=\omega^{\beta^{\prime}}>|v_{1}|$; if
$\gamma^{\prime}=\omega^{\beta^{\prime}+1}$, then
$|x^{{\gamma^{\prime}}^{-}}|=\omega^{\beta^{\prime}}\geq|v_{1}|$. In any case,
$|v_{1}|\leq|x^{{\gamma^{\prime}}^{-}}|$, so $v_{1}$ is a prefix of
$x^{{\gamma^{\prime}}^{-}}$, and $v_{2}$ is of the form $x^{\alpha}v$ for some
ordinal $\alpha$. Since $v_{2}\in P^{\prime}$, and
$v\leq_{*}x^{\alpha}v=v_{2}$, $v\in P^{\prime}$ by Lemma 4.2.

Let $w\in F^{<\gamma}P$ be arbitrary, and let us write $w$ as $uv$ where $u\in
F^{<\gamma}$ and $v\in P$. With the aim of showing that $w\in P^{\prime}$, we
form the transfinite word $x^{{\gamma^{\prime}}^{-}}u$. Its letters are in
$F$, and its length is ${\gamma^{\prime}}^{-}+|u|$.

If ${\gamma^{\prime}}^{-}+|u|<\gamma$, then $x^{{\gamma^{\prime}}^{-}}u$ is in
$F^{<\gamma}$, so $x^{{\gamma^{\prime}}^{-}}w=x^{{\gamma^{\prime}}^{-}}uv$ is
in $F^{<\gamma}P$, hence in ${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. We
have seen that this implies $w\in P^{\prime}$.

Henceforth, we assume that ${\gamma^{\prime}}^{-}+|u|\geq\gamma$. We recall
that ${\gamma^{\prime}}^{-}\leq\gamma^{\prime}<\gamma$ and that $|u|<\gamma$.
If $\gamma$ were indecomposable, then we would have
${\gamma^{\prime}}^{-}+|u|<\gamma$, contradicting our assumption. Hence
$\gamma=\omega^{\beta}+1$ for some ordinal $\beta$. Then
$\gamma^{\prime}\leq\omega^{\beta}$ and $|u|\leq\omega^{\beta}$. If
$\gamma^{\prime}<\omega^{\beta}$, then
${\gamma^{\prime}}^{-}+|u|\leq\omega^{\beta}<\gamma$, which is impossible
again. Therefore $\gamma^{\prime}=\omega^{\beta}$.

To sum up, $|u|<\gamma=\omega^{\beta}+1$, and
$\gamma^{\prime}=\omega^{\beta}$. Let $W$ be $u$ itself if
$|u|=\omega^{\beta}$, else $ux^{\omega^{\beta}}$. In each case, $W$ is a word
of length $\omega^{\beta}$ whose letters are all in $F$, so $W\in
F^{<\gamma}$. We use Lemma 4.3: let $W^{\prime}$ be a word of length
$\omega^{\beta}$, whose letters are all in $F$, and such that for every way of
writing $W^{\prime}$ as $UV$ with $|U|<\omega^{\beta}$, $W\leq_{*}V$. Then
$W^{\prime}v$ is in $F^{<\gamma}P$, hence in
$F^{\prime<\gamma^{\prime}}P^{\prime}$. Let us write $W^{\prime}v$ as
$Uv^{\prime}$ where $|U|<\gamma^{\prime}$ and $v^{\prime}\in P^{\prime}$.
Since $|U|<\gamma^{\prime}=\omega^{\beta}$, $U$ is a prefix of $W^{\prime}$,
and we can therefore write $W^{\prime}$ as $UV$ for some transfinite word $V$,
and $v^{\prime}$ as $Vv$. By construction, $W\leq_{*}V$. Therefore
$u\leq_{*}W\leq_{*}V$, and hence $w=uv\leq_{*}Vv=v^{\prime}$. Since
$v^{\prime}$ is in $P^{\prime}$, so is $w$, by Lemma 4.2.

(4) Since $F\neq\emptyset$, let us pick $x\in F$. We write
$\gamma=\gamma^{\prime}$ as $\omega^{\beta}+1$. For every $w\in P$,
$x^{\omega^{\beta}}w$ is in $F^{<\gamma}P$, hence in
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. Hence we can write
$x^{\omega^{\beta}}w$ as $uv$ where $|u|<\gamma^{\prime}$ (namely,
$|u|\leq\omega^{\beta}$) and $v\in P^{\prime}$. Since
$|u|\leq\omega^{\beta}=|x^{\omega^{\beta}}|$, we can write $v$ as
$x^{\alpha}w$ for some ordinal $\alpha$. In particular, $w\leq_{*}v$, and
since $v\in P^{\prime}$, $w$ is in $P^{\prime}$, by Lemma 4.2. ∎

We turn to sufficient conditions. There are three cases, depending on the
relative positions and indecomposability statuses of $\gamma$ and
$\gamma^{\prime}$.

######  Lemma 6.2.

Let $\gamma$, $\gamma^{\prime}$ be two non-trivial bounds, and let
$F^{<\gamma}P$, ${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ be two transfinite
products. Assuming that $\gamma<\gamma^{\prime}$, or that
$\gamma=\gamma^{\prime}$ is indecomposable,
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ if and only
if:

  1. (1)

$F\subseteq F^{\prime}$ and
$P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$,

  2. (2)

or $F\not\subseteq F^{\prime}$ and $F^{<\gamma}P\subseteq P^{\prime}$.

###### Proof.

The ‘only if’ direction is by Lemma 6.1 (1) and (2). We deal with the ‘if’
direction. Note that $\gamma\leq\gamma^{\prime}$; also, if $\gamma^{\prime}$
is decomposable, then $\gamma<\gamma^{\prime}$.

(1) For every $w\in F^{<\gamma}P$, let us write $w$ as $uv$ with $u\in
F^{<\gamma}$ and $v\in P$. Since
$P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, $v$ is in
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. Let us write $v$ as $v_{1}v_{2}$
with $v_{1}\in{F^{\prime}}^{<\gamma^{\prime}}$ and $v_{2}\in P^{\prime}$. Then
the letters of $uv_{1}$ are all in $F^{\prime}$, and
$|uv_{1}|<\gamma^{\prime}$ by Lemma 4.1 (2). It follows that $uv_{1}$ is in
${F^{\prime}}^{<\gamma^{\prime}}$, so $w=uv_{1}v_{2}$ is in
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$.

(2) If $F^{<\gamma}P\subseteq P^{\prime}$, then $F^{<\gamma}P\subseteq
P^{\prime}\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, where the last
inequality is because every $w\in P^{\prime}$ can be written as $\epsilon
w\in{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. ∎

######  Lemma 6.3.

Let $\gamma$, $\gamma^{\prime}$ be two non-trivial bounds, and $F^{<\gamma}P$,
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ be two transfinite products.
Assuming that $\gamma>\gamma^{\prime}$,
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ if and only
if:

  1. (1)

$F$ is empty and $P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$,

  2. (2)

or $F$ is non-empty and $F^{<\gamma}P\subseteq P^{\prime}$.

###### Proof.

(1) If $F=\emptyset$ then $F^{<\gamma}P=P$, and the equivalence is clear.

(2) If $F^{<\gamma}P\subseteq P^{\prime}$, then
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ since
$P^{\prime}$ is trivially included in
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. The ‘only if’ direction is by
Lemma 6.1 (3). ∎

######  Lemma 6.4.

Let $\gamma$, $\gamma^{\prime}$ be two non-trivial bounds, and let
$F^{<\gamma}P$, ${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ be two transfinite
products. Assuming that $\gamma=\gamma^{\prime}$ is decomposable,
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ if and only
if:

  1. (1)

$F$ is empty and $P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$,

  2. (2)

or $F$ is non-empty, $F\subseteq F^{\prime}$, and $P\subseteq P^{\prime}$,

  3. (3)

or $F\not\subseteq F^{\prime}$ and $F^{<\gamma}P\subseteq P^{\prime}$.

###### Proof.

If $F$ is empty then $F^{<\gamma}P=P$, so
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ is equivalent
to $P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. Henceforth, we
assume $F$ non-empty.

‘If’ direction. If $F\subseteq F^{\prime}$ and $P\subseteq P^{\prime}$, then
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ is obvious
(recall that $\gamma=\gamma^{\prime}$). If $F^{\gamma}P\subseteq P^{\prime}$,
then $F^{\gamma}P\subseteq
P^{\prime}\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$.

‘Only if’. Let us assume
$F^{<\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. If
$F\subseteq F^{\prime}$, then $P\subseteq P^{\prime}$ by Lemma 6.1 (4).
Otherwise, $F^{<\gamma}P\subseteq P^{\prime}$ by Lemma 6.1 (2). ∎

##  7\. Reduced products

We can write transfinite products in many equivalent ways. For example,
$\emptyset^{<\gamma}=\varepsilon$ for every non-trivial bound $\gamma$. Here
are a few other cases.

######  Lemma 7.1.

Let $F$, $F^{\prime}$ be non-empty closed subsets of a topological space $X$,
and $\gamma$, $\gamma^{\prime}$ be non-trivial bounds. If $F\subseteq
F^{\prime}$, then:

  1. (1)

If $\gamma<\gamma^{\prime}$ or if $\gamma=\gamma^{\prime}$ is indecomposable,
then $F^{<\gamma}{F^{\prime}}^{<\gamma^{\prime}}=F^{\prime<\gamma^{\prime}}$.

  2. (2)

If $\gamma^{\prime}$ is indecomposable, and $\gamma\leq\gamma^{\prime}$, then
${F^{\prime}}^{<\gamma^{\prime}}F^{<\gamma}={F^{\prime}}^{<\gamma^{\prime}}$.

###### Proof.

The right-hand sides are always included in the left-hand sides.

(1) By Lemma 6.2, and remembering that $F\subseteq F^{\prime}$,
$F^{<\gamma}{F^{\prime}}^{<\gamma^{\prime}}\subseteq
F^{\prime<\gamma^{\prime}}$ if and only if
${F^{\prime}}^{<\gamma^{\prime}}\subseteq F^{\prime<\gamma^{\prime}}$, which
is simply true.

(2) Since $\gamma^{\prime}$ is indecomposable, by the same lemma,
${F^{\prime}}^{<\gamma^{\prime}}F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$
if and only if $F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$, which
holds since $F\subseteq F^{\prime}$ and $\gamma\leq\gamma^{\prime}$. ∎

We will see that this leads to canonical forms for transfinite products. As in
[7, Theorem 4.22], and to reduce excessive pedantry related to the difference
between syntax and semantics, we write $\mathtt{A}$, $\mathtt{B}$ (resp.,
$\mathtt{P}$, $\mathtt{Q}$) to denote atoms, resp. sequences of atoms
(syntax), and $A$, $B$, $P$, $Q$ for their respective semantics. Hence if
$\mathtt{P}=\mathtt{A}_{1}\mathtt{A}_{2}\cdots\mathtt{A}_{n}$ (as a sequence),
then $P=A_{1}A_{2}\cdots A_{n}$ (as a product). The (syntactic) atoms
$\mathtt{A}$ are pairs $(F,\gamma)$ of a closed set $F$ and a non-trivial
bound $\gamma$ (with $F$ irreducible if $\gamma=\omega^{0}+1$), and then
$A=F^{<\gamma}$. Note that $\mathtt{P}=\mathtt{Q}$ implies $P=Q$, but the
converse may fail.

######  Definition 7.2 (Reduced).

A sequence of atoms
$\mathtt{P}\mathrel{\buildrel\text{def}\over{=}}\mathtt{A}_{1},\mathtt{A}_{2}\cdots\mathtt{A}_{n}$
on $X$, where
$\mathtt{A}_{i}\mathrel{\buildrel\text{def}\over{=}}(F_{i},\gamma_{i})$ for
each $i$, is _reduced_ if and only if:

  1. (1)

$F_{i}$ is a non-empty closed subset of $X$ ($1\leq i\leq n$);

  2. (2)

for every $i$, $1\leq i<n$, such that $\gamma_{i}<\gamma_{i+1}$, $F_{i}$ is
not included in $F_{i+1}$;

  3. (3)

for every $i$, $1\leq i<n$, such that $\gamma_{i}=\gamma_{i+1}$ is
indecomposable, $F_{i}$ and $F_{i+1}$ are incomparable;

  4. (4)

and for every $i$, $1\leq i<n$, such that $\gamma_{i}$ is indecomposable and
$\gamma_{i}>\gamma_{i+1}$, $F_{i}$ does not contain $F_{i+1}$.

######  Lemma 7.3.

For all non-empty closed subsets $F$ and $F^{\prime}$ of a space $X$, for all
non-trivial bounds $\gamma$, $\gamma^{\prime}$,

  1. (1)

$F^{<\gamma}\neq\varepsilon$;

  2. (2)

$F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$ if and only if
$F\subseteq F^{\prime}$ and $\gamma\leq\gamma^{\prime}$.

  3. (3)

$F^{<\gamma}={F^{\prime}}^{<\gamma^{\prime}}$ if and only if $F=F^{\prime}$
and $\gamma=\gamma^{\prime}$.

###### Proof.

(1) Since $F$ is non-empty, let us pick $x$ in $F$. Since $\gamma$ is non-
trivial, the one-letter word $x$ is in $F^{<\gamma}$, whence the conclusion.

(2) If $\gamma<\gamma^{\prime}$ or if $\gamma=\gamma^{\prime}$ is
indecomposable, then by Lemma 6.2,
$F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$ if and only if
$F\subseteq F^{\prime}$ and
$\varepsilon\subseteq{F^{\prime}}^{<\gamma^{\prime}}$ (true), or
$F\not\subseteq F^{\prime}$ and $F^{<\gamma}\subseteq\varepsilon$ (false, by
(1)). Hence $F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$ if and only
if $F\subseteq F^{\prime}$ in this case. If $\gamma>\gamma^{\prime}$, by Lemma
6.3 $F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$ reduces to
$F^{<\gamma}\subseteq\varepsilon$, which is false by (1). If
$\gamma=\gamma^{\prime}$ is decomposable, then by Lemma 6.4,
$F^{<\gamma}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$ if and only if
$F\subseteq F^{\prime}$ and $\varepsilon\subseteq\varepsilon$, or
$F\not\subseteq F^{\prime}$ and $F^{<\gamma}\subseteq\varepsilon$;
equivalently, if $F\subseteq F^{\prime}$.

(3) follows immediately from (2). ∎

######  Lemma 7.4.

The only reduced sequence of atoms with semantics $\\{\epsilon\\}$ is
$\varepsilon$.

###### Proof.

Let $\mathtt{P}$ be a reduced sequence of atoms of length $n\geq 1$, say
$\mathtt{A}_{1}\cdots\mathtt{A}_{n}$. By Lemma 7.3 (1), each $A_{i}$ contains
a non-empty word, and their concatenation is a non-empty word in $P$. ∎

######  Lemma 7.5.

For all atoms $A\mathrel{\buildrel\text{def}\over{=}}F^{<\gamma}$,
$B\mathrel{\buildrel\text{def}\over{=}}{F^{\prime}}^{<\gamma^{\prime}}$, with
$F,F^{\prime}\neq\emptyset$, and all transfinite products $P$, $Q$, if
$AP\subseteq BQ$ and $A\not\subseteq B$, then $AP\subseteq Q$.

###### Proof.

Since $A\not\subseteq B$, we have $F\not\subseteq F^{\prime}$ or
$\gamma>\gamma^{\prime}$, by Lemma 7.3 (2).

If $\gamma>\gamma^{\prime}$, then Lemma 6.3 and
$F^{\gamma}P\subseteq{F^{\prime}}^{<\gamma^{\prime}}Q$ imply
$F^{<\gamma}P\subseteq Q$. Let us therefore assume
$\gamma\leq\gamma^{\prime}$, and $F\not\subseteq F^{\prime}$. If
$\gamma<\gamma^{\prime}$ or $\gamma=\gamma^{\prime}$ is indecomposable, then
by Lemma 6.2, $F^{<\gamma}P\subseteq Q$ again. If $\gamma=\gamma^{\prime}$ is
decomposable, then we reach the same conclusion by using Lemma 6.4. ∎

######  Lemma 7.6.

For any reduced sequence of atoms $\mathtt{A}\mathtt{P}$, $AP=A$ implies
$\mathtt{P}=\varepsilon$.

###### Proof.

By contradiction, let us assume that
$\mathtt{P}=\mathtt{A}^{\prime}\mathtt{Q}$, and let us write $A$ as
$F^{<\gamma}$ and $A^{\prime}$ as ${F^{\prime}}^{<\gamma^{\prime}}$. Since
$AP\subseteq A$, either $\gamma$ is indecomposable and $P\subseteq A$ by Lemma
6.2, or $\gamma$ is decomposable and $P\subseteq\varepsilon$ by Lemma 6.4. The
latter is impossible by Lemma 7.4. Hence $\gamma$ is indecomposable, and
$P=A^{\prime}Q\subseteq A$.

If $\gamma^{\prime}\leq\gamma$, then by Lemma 6.2 $F^{\prime}\subseteq F$ and
$Q\subseteq A$, or $F^{\prime}\not\subseteq F$ and $P\subseteq\varepsilon$.
The latter is impossible, as above. The former is impossible, too, because
$\mathtt{A}\mathtt{A}^{\prime}\mathtt{Q}$ is reduced, using Definition 7.2
(3), (4). If $\gamma^{\prime}>\gamma$, then
$P=A^{\prime}Q\subseteq\varepsilon$ by Lemma 6.3, which is impossible by Lemma
7.4. ∎

######  Lemma 7.7.

For every reduced sequence of atoms $\mathtt{A}\mathtt{P}$ and for every atom
$\mathtt{B}\mathrel{\buildrel\text{def}\over{=}}{F^{\prime}}^{<\gamma^{\prime}}$
with $F^{\prime}\neq\emptyset$, if $AP=B$ then $\mathtt{P}=\varepsilon$.

###### Proof.

By induction on the number $n\geq 1$ of atoms in $\mathtt{A}\mathtt{P}$. If
$n=1$, this is vacuous, so let $n\geq 2$. If $B\subseteq A$, then $B\subseteq
A\subseteq AP=B$, so $A=B$. Then $AP=A$, so $\mathtt{P}=\varepsilon$ by Lemma
7.6.

We now assume $B\not\subseteq A$, and we will show that this is impossible. By
Lemma 7.5 applied to $B\subseteq AP$, we have $B\subseteq P$. Since
$P\subseteq AP=B$, $B=P$. By induction hypothesis, if we write $\mathtt{P}$ as
$\mathtt{A}^{\prime}\mathtt{Q}$ where $\mathtt{A}^{\prime}$ is an atom, then
$\mathtt{Q}=\varepsilon$, so $\mathtt{P}=\mathtt{A}^{\prime}$. Then $P=B$
entails $\mathtt{P}=\mathtt{A}^{\prime}=\mathtt{B}$ by Lemma 7.3 (3). Let us
write $\mathtt{A}$ as $(F,\gamma)$ and $\mathtt{B}$ as
$(F^{\prime},\gamma^{\prime})$. Since $A\subseteq AP=B$, Lemma 7.3 (2) entails
that $F\subseteq F^{\prime}$ and $\gamma\leq\gamma^{\prime}$. But
$\mathtt{A}\mathtt{P}=(F,\gamma)(F^{\prime},\gamma^{\prime})$ is reduced, and
this contradicts Definition 7.2 (2) if $\gamma<\gamma^{\prime}$, (3) if
$\gamma=\gamma^{\prime}$ is indecomposable. Hence $\gamma=\gamma^{\prime}$ is
decomposable. Then $AP=F^{<\gamma}{F^{\prime}}^{<\gamma}\subseteq
B=F^{\prime<\gamma}$ implies ${F^{\prime}}^{<\gamma}\subseteq\varepsilon$ by
Lemma 6.4, and that is impossible by Lemma 7.3 (1). ∎

######  Proposition 7.8.

For all reduced sequences of atoms $\mathtt{P}$ and $\mathtt{Q}$, $P=Q$ if and
only if $\mathtt{P}=\mathtt{Q}$.

###### Proof.

The ‘if’ direction is trivial. We show that $P=Q$ implies
$\mathtt{P}=\mathtt{Q}$ by induction on the sum $|\mathtt{P}|+|\mathtt{Q}|$ of
the sizes of $\mathtt{P}$ and $\mathtt{Q}$, where by size we mean number of
atoms. Henceforth we assume $P=Q$.

If $|\mathtt{P}|=0$, then $P=Q=\\{\epsilon\\}$. By Lemma 7.4,
$\mathtt{Q}=\varepsilon=\mathtt{P}$. The situation is symmetric if
$|\mathtt{Q}|=0$. Let us assume $|\mathtt{P}|,|\mathtt{Q}|\geq 1$. If
$|\mathtt{Q}|=1$, $\mathtt{Q}$ is an atom $\mathtt{B}$. By Lemma 7.7, we must
have $|\mathtt{P}|=1$. Then $P=Q$ implies $\mathtt{P}=\mathtt{Q}$ by Lemma 7.3
(3). Similarly if $|\mathtt{P}|=1$. The interesting case is the remaining one:
$|\mathtt{P}|,|\mathtt{Q}|\geq 2$. Let us write $\mathtt{P}$ as
$\mathtt{A}_{1}\mathtt{A}_{2}\mathtt{P}^{\prime}$ and $\mathtt{Q}$ as
$\mathtt{B}_{1}\mathtt{B}_{2}\mathtt{Q}^{\prime}$.

We first claim that if $A_{1}\not\subseteq B_{1}$, then
$\mathtt{P}=\mathtt{B}_{2}\mathtt{Q}^{\prime}$. Indeed, under that assumption,
and since $P=A_{1}A_{2}P^{\prime}\subseteq Q=B_{1}B_{2}Q^{\prime}$, we obtain
$P\subseteq B_{2}Q^{\prime}$ by Lemma 7.5. In turn, $B_{2}Q^{\prime}\subseteq
B_{1}B_{2}Q^{\prime}=Q=P$, so $P=B_{2}Q^{\prime}$. The induction hypothesis
then yields $\mathtt{P}=\mathtt{B}_{2}\mathtt{Q}^{\prime}$. Similarly, if
$B_{1}\not\subseteq A_{1}$ then
$\mathtt{Q}=\mathtt{A}_{2}\mathtt{P}^{\prime}$.

It follows that we cannot have $A_{1}\not\subseteq B_{1}$ and
$B_{1}\not\subseteq A_{1}$. Otherwise,
$\mathtt{Q}=\mathtt{B}_{1}\mathtt{B}_{2}\mathtt{Q}^{\prime}=\mathtt{B}_{1}\mathtt{P}=\mathtt{B}_{1}\mathtt{A}_{1}\mathtt{A}_{2}\mathtt{P}^{\prime}=\mathtt{B}_{1}\mathtt{A}_{1}\mathtt{Q}$,
which is impossible since
$|\mathtt{B}_{1}\mathtt{A}_{1}\mathtt{Q}|\neq|\mathtt{Q}|$.

Hence $A_{1}$ is included in $B_{1}$, or conversely. Without loss of
generality, let us assume $B_{1}\subseteq A_{1}$. We claim that, in fact,
$A_{1}=B_{1}$. We reason by contradiction, and we assume $A_{1}\not\subseteq
B_{1}$. Then we have seen that $\mathtt{P}=\mathtt{B}_{2}\mathtt{Q}^{\prime}$.
By syntactic matching, $\mathtt{A}_{1}=\mathtt{B}_{2}$ (and
$\mathtt{Q}^{\prime}=\mathtt{A}_{2}\mathtt{P}^{\prime}$). Let us write
$\mathtt{B}_{1}$ as $(F,\gamma)$ and $\mathtt{B}_{2}$ as
$(F^{\prime},\gamma^{\prime})$. Since $B_{1}\subseteq A_{1}=B_{2}$,
$F\subseteq F^{\prime}$ and $\gamma\leq\gamma^{\prime}$ by Lemma 7.3 (2). By
Definition 7.2 (2) and (3) applied to
$\mathtt{B}_{1}\mathtt{B}_{2}\mathtt{Q}^{\prime}$, it is impossible that
$\gamma<\gamma^{\prime}$, or that $\gamma=\gamma^{\prime}$ is indecomposable.
Hence $\gamma=\gamma^{\prime}$ is decomposable. Now
$Q=B_{1}B_{2}Q^{\prime}=F^{<\gamma}B_{2}Q^{\prime}$ is included in
$P=A_{1}A_{2}P^{\prime}=B_{2}A_{2}P^{\prime}={F^{\prime}}^{<\gamma}A_{2}P^{\prime}$
(since $\mathtt{A}_{1}=\mathtt{B}_{2}$ and $\gamma=\gamma^{\prime}$), so
$B_{2}Q^{\prime}\subseteq A_{2}P^{\prime}$ by Lemma 6.4 (2). Since
$A_{2}P^{\prime}\subseteq P=B_{2}Q^{\prime}$ (because
$\mathtt{P}=\mathtt{B}_{2}\mathtt{Q}^{\prime}$),
$A_{2}P^{\prime}=B_{2}Q^{\prime}$. By induction hypothesis,
$\mathtt{A}_{2}\mathtt{P}^{\prime}=\mathtt{B}_{2}\mathtt{Q}^{\prime}$, so
$\mathtt{A}_{2}\mathtt{P}^{\prime}=\mathtt{P}$. This is impossible since
$\mathtt{P}=\mathtt{A}_{1}\mathtt{A}_{2}\mathtt{P}^{\prime}$. Having reached a
contradiction, we conclude that $A_{1}=B_{1}$, so
$\mathtt{A}_{1}=\mathtt{B}_{1}$ by Lemma 7.3 (3).

We now claim that $A_{2}P^{\prime}\subseteq B_{2}Q^{\prime}$. We know that
$P=A_{1}A_{2}P^{\prime}$ is included in
$Q=B_{1}B_{2}Q^{\prime}=A_{1}B_{2}Q^{\prime}$ (since $A_{1}=B_{1}$). Let us
write $\mathtt{A}_{1}$ as $(F,\gamma)$ and $\mathtt{A}_{2}$ as
$(F^{\prime},\gamma^{\prime})$. If $\gamma$ is decomposable, then
$A_{2}P^{\prime}\subseteq B_{2}Q^{\prime}$ by Lemma 6.4 (2). Let therefore
$\gamma$ be indecomposable. Since
$\mathtt{A}_{1}\mathtt{A}_{2}\mathtt{P}^{\prime}$ is reduced, we cannot have
$F\supseteq F^{\prime}$ and $\gamma\geq\gamma^{\prime}$, by Definition 7.2 (3)
and (4), so $A_{2}\not\subseteq A_{1}$, by Lemma 7.3 (2). Then
$A_{2}P^{\prime}\subseteq A_{1}A_{2}P^{\prime}\subseteq A_{1}B_{2}Q^{\prime}$,
so $A_{2}P^{\prime}\subseteq B_{2}Q^{\prime}$ by Lemma 7.5.

Symmetrically, $B_{2}Q^{\prime}$ is included in $A_{2}P^{\prime}$, so
$A_{2}P^{\prime}=B_{2}Q^{\prime}$. By the induction hypothesis
$\mathtt{A}_{2}\mathtt{P}^{\prime}=\mathtt{B}_{2}\mathtt{Q}^{\prime}$. We
remember that $\mathtt{A}_{1}=\mathtt{B}_{1}$, so $\mathtt{P}=\mathtt{Q}$. ∎

We can always rewrite any transfinite product into a reduced product with the
same semantics, using Lemma 7.1, whence the following.

######  Corollary 7.9.

Every transfinite product is equal to $P$ for some unique reduced sequence of
atoms $\mathtt{P}$.

Corollary 7.9 allows us to conflate the notions of transfinite product and of
reduced sequence of atoms. By abuse of language, we will call _reduced
product_ any transfinite product $P$ written in such a way that $\mathtt{P}$
is reduced. A _reduced $\alpha$-product_ is an $\alpha$-product that is
reduced in this sense.

##  8\. Well-foundedness

The _rank_ (or _height_) of an element $x$ in a well-founded poset $P$ is
defined by well-founded induction as the least ordinal strictly larger than
the ranks of all elements $y<x$. We write $||F||$ for the rank of $F$ in the
lattice of closed subsets of a Noetherian space $X$. For any Noetherian space
$F$, $||F||$ is the _stature_ of $F$ [8], generalizing the notion of the same
name on wqos [1].

Given any two ordinals
$\alpha\mathrel{\buildrel\text{def}\over{=}}\omega^{\alpha_{1}}+\cdots+\omega^{\alpha_{m}}$
and
$\beta\mathrel{\buildrel\text{def}\over{=}}\omega^{\beta_{1}}+\cdots+\omega^{\beta_{n}}$
in Cantor normal form, their _natural sum_ $\alpha\oplus\beta$ is defined as
$\omega^{\gamma_{1}}+\cdots+\omega^{\gamma_{m+n}}$, where
$\gamma_{1}\geq\cdots\geq\gamma_{m+n}$ is the list obtained by sorting the
list $\alpha_{1},\cdots,\alpha_{m},\beta_{1},\cdots,\beta_{n}$ in decreasing
order. This operation is associative and commutative, and strictly monotonic
in both arguments.

An ordinal $\delta$ is _critical_ if and only if $\omega^{\delta}=\delta$. For
every ordinal $\alpha$, let ${\alpha}^{\circ}$ be $\alpha+1$ if
$\alpha=\delta+n$ for some critical ordinal $\delta$ and some
$n\in\mathbb{N}$, and $\alpha$ otherwise. Then
$\alpha<\omega^{{\alpha}^{\circ}}$, and $\alpha\mapsto{\alpha}^{\circ}$ is
strictly monotonic [8, Lemmata 12.3 and 12.4]. For every proper bound
$\gamma$, we define $[{\gamma}]$ as the rank of $\gamma$ in the poset of all
proper bounds less than or equal to $\gamma$. Explicitly, and writing $n$ for
a natural number and $\lambda$ for a limit ordinal, $[{\omega^{n}}]=2n-2$ and
$[{\omega^{n}+1}]=2n-1$ if $n\geq 1$, $[{\omega^{\lambda+n}}]=\lambda+2n$, and
$[{\omega^{\lambda+n}+1}]=\lambda+2n+1$.

######  Definition 8.1.

Let $X$ be a Noetherian space. For every atom $F^{<\gamma}$, let
$\varphi(F^{<\gamma})$ be $||F||$ if $\gamma=\omega^{0}+1$, and
$\omega^{{(||F||\oplus[{\gamma}])}^{\circ}}$ otherwise. For every reduced
product $P\mathrel{\buildrel\text{def}\over{=}}A_{1}\cdots A_{n}$, let
$\varphi(P)\mathrel{\buildrel\text{def}\over{=}}\bigoplus_{i=1}^{n}\varphi(A_{i})$.

######  Proposition 8.2.

Let $X$ be a Noetherian space. For all reduced products $P$ and $P^{\prime}$,
$P\subseteq P^{\prime}$ implies $\varphi(P)\leq\varphi(P^{\prime})$, and
$P\subsetneq P^{\prime}$ implies $\varphi(P)<\varphi(P^{\prime})$.

###### Proof.

We first claim that $\varphi$ is strictly monotonic on reduced atoms: for all
atoms $F^{<\gamma}$ and ${F^{\prime}}^{<\gamma^{\prime}}$ with
$F,F^{\prime}\neq\emptyset$ and
$F^{<\gamma}\subsetneq{F^{\prime}}^{<\gamma^{\prime}}$,
$\varphi(F^{<\gamma})<\varphi({F^{\prime}}^{<\gamma^{\prime}})$. By Lemma 7.3,
$F\subseteq F^{\prime}$ and $\gamma\leq\gamma^{\prime}$, and not both are
equalities. If $\gamma=\gamma^{\prime}=\omega^{0}+1$ and $F\subsetneq
F^{\prime}$, then
$\varphi(F^{<\gamma})=||F||<||F^{\prime}||=\varphi({F^{\prime}}^{<\gamma^{\prime}})$.
If $\gamma=\omega^{0}+1<\gamma^{\prime}$, then
$\varphi(F^{<\gamma})=||F||\leq||F||\oplus[{\gamma}]<||F^{\prime}||\oplus[{\gamma^{\prime}}]<\omega^{{(||F^{\prime}||\oplus[{\gamma^{\prime}}])}^{\circ}}=\varphi({F^{\prime}}^{<\gamma^{\prime}})$.
If both $\gamma$ and $\gamma^{\prime}$ are proper, then
$\varphi(F^{<\gamma})=\omega^{{(||F||\oplus[{\gamma}])}^{\circ}}<\omega^{{(||F^{\prime}||\oplus[{\gamma^{\prime}}])}^{\circ}}=\varphi({F^{\prime}}^{<\gamma^{\prime}})$
since $\oplus$, $[{\\_}]$, ${\\_}^{\circ}$ and $||\\_||$ are strictly
monotonic.

We prove the proposition by induction on the sum of the lengths of $P$ and
$P^{\prime}$. Let $P\subseteq P^{\prime}$. If $P=\varepsilon$, then
$\varphi(P)=0\leq\varphi(P^{\prime})$. If additionally
$P^{\prime}\neq\varepsilon$, then $\varphi(P^{\prime})$ is the natural sum of
at least one term, and all those terms are non-zero: they are of the form
$\varphi(F^{<\gamma})$, and when $\gamma=\omega^{0}+1$,
$\varphi(F^{<\gamma})=||F||\neq 0$ since $F\neq\emptyset$, otherwise
$\varphi(F^{<\gamma})$ is a power of $\omega$.

We now assume that $P\neq\varepsilon$. We write $P$ as $A_{1}\cdots A_{n}$,
where $A_{1}$, …, $A_{n}$ are atoms and $n\geq 1$. Let also
$Q\mathrel{\buildrel\text{def}\over{=}}A_{2}\cdots A_{n}$, so $P=A_{1}Q$.
Since $P\subseteq P^{\prime}$, $P^{\prime}\neq\varepsilon$, so
$P^{\prime}=A^{\prime}_{1}Q^{\prime}$ for some atom $A^{\prime}_{1}$ and some
(reduced) product $Q^{\prime}$. We write $A^{\prime}_{1}$ as
${F^{\prime}}^{<\gamma^{\prime}}$, and $A_{i}$ as ${F_{i}}^{<\gamma_{i}}$ for
each $i$, $1\leq i\leq n$.

If $A_{1}\not\subseteq A^{\prime}_{1}$, then Lemma 7.5 entails that
$P=A_{1}Q\subseteq Q^{\prime}$. By induction hypothesis,
$\varphi(P)\leq\varphi(Q^{\prime})$. Now
$\varphi(P^{\prime})=\varphi(A^{\prime}_{1})\oplus\varphi(Q^{\prime})>\varphi(Q^{\prime})$,
so $\varphi(P)<\varphi(P^{\prime})$. We turn to the other cases: from now on,
$A_{1}\subseteq A^{\prime}_{1}$.

_If $\gamma^{\prime}$ is indecomposable._ We say that $A_{i}$ is _small_ if
$A_{i}\subseteq A^{\prime}_{1}$, equivalently $F_{i}\subseteq F^{\prime}$ and
$\gamma_{i}\leq\gamma^{\prime}$, by Lemma 7.3 (2). $A_{1}$ is small. Let $k$
be largest such that $A_{1}$, …, $A_{k}$ are small, and
$R\mathrel{\buildrel\text{def}\over{=}}A_{k+1}\cdots A_{n}$. Then $R\subseteq
Q^{\prime}$. This is clear if $R=\varepsilon$; otherwise $k<n$,
$A_{k+1}\not\subseteq A^{\prime}_{1}$, then $R\subseteq P\subseteq
P^{\prime}=A^{\prime}_{1}Q^{\prime}$ implies $R\subseteq Q^{\prime}$ by Lemma
7.5. By the induction hypothesis, $\varphi(R)\leq\varphi(Q^{\prime})$.

If some $A_{i}$ with $1\leq i\leq k$ is equal to
$A^{\prime}_{1}={F^{\prime}}^{<\gamma^{\prime}}$, then
$\gamma_{i}=\gamma^{\prime}$ is indecomposable, and $F_{i}=F^{\prime}$. We
cannot have $i\geq 2$, since that would contradict Definition 7.2 (2) or (3)
at positions $i-1$ and $i$; similarly, $i\leq k-1$ would contradict Definition
7.2 (4) or (3) at positions $i$ and $i+1$. Hence $k=1$. Then
$\varphi(P)=\varphi(A_{1})\oplus\varphi(R)\leq\varphi(A^{\prime}_{1})\oplus\varphi(Q^{\prime})=\varphi(P^{\prime})$.
Additionally, if $P\neq P^{\prime}$, then (since $k=1$ and
$A_{1}=A^{\prime}_{1}$) $R$ is strictly included in $Q^{\prime}$, so
$\varphi(R)<\varphi(Q^{\prime})$ by induction hypothesis, from which
$\varphi(P)<\varphi(P^{\prime})$ follows.

Otherwise, $k\geq 1$ and $A_{1}$, …, $A_{k}$ are all strictly included in
$A^{\prime}_{1}$. Then $\varphi(A_{i})<\varphi(A^{\prime}_{1})$ for every $i$
with $1\leq i\leq k$. Since $\gamma^{\prime}$ is indecomposable,
$\gamma^{\prime}\neq\omega^{0}+1$, so
$\varphi(A^{\prime}_{1})=\omega^{{(||F^{\prime}||\oplus[{\gamma^{\prime}}])}^{\circ}}$.
The latter is indecomposable, so
$\varphi(A_{1})\oplus\cdots\oplus\varphi(A_{k})<\varphi(A^{\prime}_{1})$.
Together with $\varphi(R)\leq\varphi(Q^{\prime})$, this implies that
$\varphi(P)=\varphi(A_{1})\oplus\cdots\oplus\varphi(A_{k})\oplus\varphi(R)<\varphi(A^{\prime}_{1})\oplus\varphi(Q^{\prime})=\varphi(P^{\prime})$.

_If $\gamma^{\prime}$ is decomposable._ In that case, we say that $A_{i}$ is
_small_ if and only if $F_{i}\subseteq F^{\prime}$ and
$\gamma_{i}<\gamma^{\prime}$ (not $\leq$). Let $k$ be largest such that
$A_{1}$, …, $A_{k}$ are small, and
$R\mathrel{\buildrel\text{def}\over{=}}A_{k+1}\cdots A_{n}$. $A_{1}$, …,
$A_{k}$ are all strictly included in $A^{\prime}_{1}$, so
$\varphi(A_{i})<\varphi(A^{\prime}_{1})$ for every $i$ with $1\leq i\leq k$.
If $k\geq 1$, then $\gamma_{1}<\gamma^{\prime}$, so
$\gamma^{\prime}\neq\omega^{0}+1$, and therefore $\varphi(A^{\prime}_{1})$ is
indecomposable. Hence
$\varphi(A_{1})\oplus\cdots\oplus\varphi(A_{k})<\varphi(A^{\prime}_{1})$. If
$k=0$, the same inequality holds, vacuously.

If $R=\varepsilon$, then
$\varphi(P)=\varphi(A_{1})\oplus\cdots\oplus\varphi(A_{k})<\varphi(A^{\prime}_{1})\leq\varphi(P^{\prime})$.
We now assume $R\neq\varepsilon$. Then, $k<n$, and $A_{k+1}$ is not small.

If $A_{k+1}\not\subseteq A^{\prime}_{1}$, then $R\subseteq P\subseteq
P^{\prime}=A^{\prime}_{1}Q^{\prime}$ implies $R\subseteq Q^{\prime}$ by Lemma
7.5, hence $\varphi(R)\leq\varphi(Q^{\prime})$ by induction hypothesis. Then
$\varphi(P)=\varphi(A_{1})\oplus\cdots\oplus\varphi(A_{k})\oplus\varphi(R)<\varphi(A^{\prime}_{1})\oplus\varphi(Q^{\prime})=\varphi(P^{\prime})$.

There remains one case, where $A_{k+1}\subseteq A^{\prime}_{1}$ and $A_{k+1}$
is not small. Then $F_{k+1}\subseteq F^{\prime}$ and
$\gamma_{k+1}=\gamma^{\prime}$, which is decomposable. Let
$R^{\prime}\mathrel{\buildrel\text{def}\over{=}}A_{k+2}\cdots A_{n}$.
$R=A_{k+1}R^{\prime}\subseteq P\subseteq P^{\prime}=A^{\prime}_{1}Q^{\prime}$
implies $R^{\prime}\subseteq Q^{\prime}$ by Lemma 6.4. By induction
hypothesis, $\varphi(R^{\prime})\leq\varphi(Q^{\prime})$. If $k=0$, then
$P=A_{1}R^{\prime}$, $P^{\prime}=A^{\prime}_{1}Q^{\prime}$, $A_{1}\subseteq
A^{\prime}_{1}$, and $R^{\prime}\subseteq Q^{\prime}$, so
$\varphi(P)=\varphi(A_{1})\oplus\varphi(R^{\prime})\leq\varphi(A^{\prime}_{1})\oplus\varphi(Q^{\prime})=\varphi(P^{\prime})$;
additionally, if $P\subsetneq P^{\prime}$ then $A_{1}\subsetneq
A^{\prime}_{1}$ or $R^{\prime}\subsetneq Q^{\prime}$, which implies
$\varphi(P)<\varphi(P^{\prime})$. Let us now assume $k\geq 1$. We claim that
the inclusion $A_{k+1}\subseteq A^{\prime}_{1}$ is strict: if
$A_{k+1}=A^{\prime}_{1}$, then $A_{k}\subseteq A^{\prime}_{1}=A_{k+1}$ and
$\gamma_{k}<\gamma^{\prime}=\gamma_{k+1}$, contradicting Definition 7.2 (2).
Hence $A_{1}$, …, $A_{k}$, and also $A_{k+1}$, are strictly included in
$A^{\prime}_{1}$. We recall that, since $k\geq 1$, $\varphi(A^{\prime}_{1})$
is indecomposable, so
$\varphi(P)=\varphi(A_{1})\oplus\cdots\oplus\varphi(A_{k})\oplus\varphi(A_{k+1})\oplus\varphi(R^{\prime})<\varphi(A^{\prime}_{1})\oplus\varphi(Q^{\prime})=\varphi(P^{\prime})$.
∎

######  Corollary 8.3.

Let $X$ be a Noetherian space. The inclusion ordering on transfinite products
on $X$ is well-founded. Additionally, the ordinal rank of $X^{<\alpha}$ in the
poset of all $\alpha$-products is at most
$\omega^{{(||X||\oplus[{\alpha}])}^{\circ}}$, for every bound $\alpha$.

##  9\. Intersections of transfinite products

######  Lemma 9.1.

Let $X$ be a topological space. The intersection of two transfinite products
satisfies the following properties:

  1. (1)

$\varepsilon\cap P^{\prime}=P\cap\varepsilon=\varepsilon$.

  2. (2)

If $\gamma<\gamma^{\prime}$ or if $\gamma=\gamma^{\prime}$ is indecomposable,
then $F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ is equal to
the union of $(F\cap
F^{\prime})^{<\gamma}(P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime})$ and of
$(F\cap F^{\prime})^{<\gamma}(F^{<\gamma}P\cap P^{\prime})$.

  3. (3)

If $\gamma=\gamma^{\prime}$ is decomposable, say
$\gamma=\gamma^{\prime}=\omega^{\beta}+1$, then
$F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ is the union of
$(F\cap F^{\prime})^{<\gamma}(P\cap P^{\prime})$, of $(F\cap
F^{\prime})^{<\omega^{\beta}}(P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime})$,
and of $(F\cap F^{\prime})^{<\omega^{\beta}}(F^{<\gamma}P\cap P^{\prime})$.

###### Proof.

(1) is clear. Let us deal with the left to right inclusions for the other
cases, as a first step. For every $w\in
F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, let us write $w$
as $uv$ where $u\in F^{<\gamma}$, $v\in P$ and also as $u^{\prime}v^{\prime}$
where $u^{\prime}\in{F^{\prime}}^{<\gamma^{\prime}}$, $v^{\prime}\in
P^{\prime}$.

(2) If $|u|\leq|u^{\prime}|$, $u$ is a prefix of $u^{\prime}$, so its letters
are not just in $F$, but also in $F^{\prime}$. Therefore $u\in(F\cap
F^{\prime})^{<\gamma}$. Also, $v$ is in $P$, and is a suffix of $w$. In
particular, $v\leq_{*}w$, so $v\in{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$
by Lemma 4.2. If $|u^{\prime}|\leq|u|$, then symmetrically $u^{\prime}$ is in
$(F\cap F^{\prime})^{<\gamma}$ and $v^{\prime}$ is in $P^{\prime}$ and in
$F^{<\gamma}P$.

(3) If $|u|=|u^{\prime}|=\omega^{\beta}$, then $u=u^{\prime}$ is in $(F\cap
F^{\prime})^{<\gamma}$, and $v=v^{\prime}$ is in $P\cap P^{\prime}$, so
$w\in(F\cap F^{\prime})^{<\gamma}(P\cap P^{\prime})$. Otherwise, $w$ is in
$F^{<\omega^{\beta}}P$ and in ${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$, or
in $F^{<\gamma}P$ and in ${F^{\prime}}^{<\omega^{\beta}}P^{\prime}$. In the
first case, by (2) with $\omega^{\beta}$ in lieu of $\gamma$, $w$ is in
$(F\cap
F^{\prime})^{<\omega^{\beta}}(P\cap{F^{\prime}}^{\gamma^{\prime}}P^{\prime})\cup(F\cap
F^{\prime})^{<\omega^{\beta}}(F^{<\omega^{\beta}}P\cap P^{\prime})$, hence in
$(F\cap
F^{\prime})^{<\omega^{\beta}}(P\cap{F^{\prime}}^{\gamma^{\prime}}P^{\prime})\cup(F\cap
F^{\prime})^{<\omega^{\beta}}(F^{<\gamma}P\cap P^{\prime})$. In the second
case, a similar argument leads to the same result.

We now deal with the right to left inclusions.

(2) $(F\cap
F^{\prime})^{<\gamma}(P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime})$ is
included both in $F^{<\gamma}P$ (because $F\cap F^{\prime}\subseteq F$) and in
$(F\cap
F^{\prime})^{<\gamma}{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}={F^{\prime}}^{\gamma^{\prime}}P^{\prime}$,
by Lemma 7.1 (1). Similarly for $(F\cap F^{\prime})^{<\gamma}(F^{<\gamma}P\cap
P^{\prime})$.

(3) $(F\cap F^{\prime})^{<\gamma}(P\cap P^{\prime})$ is included both in
$F^{<\gamma}P$ and in
${F^{\prime}}^{<\gamma}P^{\prime}={F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$.
$(F\cap
F^{\prime})^{<\omega^{\beta}}(P\cap{F^{\prime}}^{\gamma^{\prime}}P^{\prime})$
is included both in $(F\cap F^{\prime})^{<\omega^{\beta}}P\subseteq
F^{<\omega^{\beta}}P\subseteq F^{<\gamma}P$ and in $(F\cap
F^{\prime})^{<\omega^{\beta}}{F^{\prime}}^{\gamma^{\prime}}P^{\prime}={F^{\prime}}^{\gamma^{\prime}}P^{\prime}$
(by Lemma 7.1 (1)). Finally, $(F\cap
F^{\prime})^{<\omega^{\beta}}(F^{<\gamma}P\cap P^{\prime})$ is included both
in $(F\cap F^{\prime})^{<\omega^{\beta}}F^{<\gamma}P=F^{<\gamma}P$ (by Lemma
7.1 (1)) and in $(F\cap
F^{\prime})^{<\omega^{\beta}}P^{\prime}\subseteq{F^{\prime}}^{<\omega^{\beta}}P^{\prime}\subseteq{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$.
∎

######  Corollary 9.2.

Let $X$ be a Noetherian space. The intersection of any two transfinite
products is a finite union of transfinite products.

###### Proof.

By induction on the sum of their sizes, using Lemma 9.1. The case where one of
them is $\varepsilon$ is obvious, so we deal with the intersection of two
transfinite products $F^{<\gamma}P$ and
${F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$. Without loss of generality,
$\gamma\leq\gamma^{\prime}$.

By induction hypothesis, $P\cap{F^{\prime}}^{\gamma^{\prime}}P^{\prime}$,
$F^{<\gamma}P\cap P^{\prime}$ and $P\cap P^{\prime}$ are finite unions of
transfinite products, say $\bigcup_{i}P_{i}$, $\bigcup_{j}Q_{j}$, and
$\bigcup_{k}R_{k}$ respectively.

The intersection $F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$
can then be expressed as $\bigcup_{i}(F\cap
F^{\prime})^{<\gamma}P_{i}\cup\bigcup_{j}(F\cap F^{\prime})^{<\gamma}Q_{j}$
when $\gamma<\gamma^{\prime}$ or if $\gamma=\gamma^{\prime}$ is
indecomposable. If $\gamma=\gamma^{\prime}$ is of the form $\omega^{\beta}+1$,
then it can be expressed as the union $\bigcup_{k}(F\cap
F^{\prime})^{<\gamma}R_{k}\cup\bigcup_{i}(F\cap
F^{\prime})^{<\omega^{\beta}}P_{i}\cup\bigcup_{j}(F\cap
F^{\prime})^{<\omega^{\beta}}Q_{j}$.

This is a finite union of transfinite products, except when
$\gamma=\omega^{0}+1$. In that case, we need to refine the expressions above.
If $F\cap F^{\prime}$ is empty, then $(F\cap
F^{\prime})^{<\gamma}=\varepsilon$, so
$F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}=\bigcup_{i}P_{i}\cup\bigcup_{j}Q_{j}$
when $\omega^{0}+1<\gamma^{\prime}$, and
$F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}=\bigcup_{k}R_{k}\cup\bigcup_{i}P_{i}\cup\bigcup_{j}Q_{j}$
otherwise.

We now assume that $F\cap F^{\prime}\neq\emptyset$. $F$ and $F^{\prime}$ are
irreducible closed, and since $X$ is Noetherian, $F\cap F^{\prime}$ is a
finite union $\bigcup_{\ell=1}^{n}C_{\ell}$ of irreducible closed subsets.
Since $F\cap F^{\prime}\neq\emptyset$, $n$ is non-zero. Then $(F\cap
F^{\prime})^{<\gamma}=(F\cap F^{\prime})^{?}=\bigcup_{\ell=1}^{n}C_{\ell}^{?}$
(an equality that would fail if $n$ were zero), and $(F\cap
F^{\prime})^{<\omega^{0}}=\varepsilon$. Therefore
$F^{<\gamma}P\cap{F^{\prime}}^{<\gamma^{\prime}}P^{\prime}$ is equal to
$\bigcup_{i,\ell}C_{\ell}^{?}P_{i}\cup\bigcup_{j,\ell}C_{\ell}^{?}Q_{j}$ when
$\omega^{0}+1<\gamma^{\prime}$, or to
$\bigcup_{k,\ell}C_{\ell}^{?}R_{k}\cup\bigcup_{i}P_{i}\cup\bigcup_{j}Q_{j}$ if
$\gamma^{\prime}=\omega^{0}+1$. ∎

We can now use Proposition 2.1. If $X$ is Noetherian, the class $\mathcal{P}$
of $\alpha$-products is well-founded under inclusion by Corollary 8.3,
$X^{<\alpha}$ is a finite union of elements of $\mathcal{P}$ since it is, in
fact, an $\alpha$-product. The intersection of any two elements of
$\mathcal{P}$ is a finite union of elements of $\mathcal{P}$ by Corollary 9.2,
and $\mathcal{P}$ is irredundant by Proposition 5.4. Moreover, the regular
subword topology on $X^{<\alpha}$ is the coarsest that makes every element of
$\mathcal{P}$ a closed set, by Proposition 5.3. Since every set of transfinite
words on $X$ is included in $X^{<\alpha}$ for some bound $\alpha$, we obtain
the following.

######  Theorem 9.3.

For every Noetherian space $X$, every space $Y$ of transfinite words on $X$ is
Noetherian in the regular subword topology. For every bound $\alpha$, the
irreducible closed subsets of $X^{<\alpha}$ are the $\alpha$-products.

Given any non-empty Noetherian space $Z$, $Z$ has finitely many maximal
irreducible closed subsets $C_{1},\cdots,C_{n}\in\mathcal{S}Z$ ($n\geq 1$).
The _reduced sobrification rank_ $\mathrm{rsob}\;Z$ is the maximum of the
ranks of $C_{1}$, …, $C_{n}$ in $\mathcal{S}Z$ [8, Lemma 4.2]; then
$||Z||\leq\omega^{\mathrm{rsob}\;Z}\otimes n$ [8, Proposition 4.5]. For
$Z\mathrel{\buildrel\text{def}\over{=}}X^{<\alpha}$, $n=1$ since $X^{<\alpha}$
is the unique largest element of $\mathcal{S}{(X^{<\alpha})}$, so the $\otimes
n$ part vanishes. Corollary 8.3 then gives us the following quantitative
information.

######  Proposition 9.4.

For every Noetherian space $X$, for every bound $\alpha$, we have
$\mathrm{rsob}\;{X^{<\alpha}}\leq\omega^{{(||X||\oplus[{\alpha}])}^{\circ}}$
and $||X^{<\alpha}||\leq\omega^{\omega^{{(||X||\oplus[{\alpha}])}^{\circ}}}$.

It is not our purpose to give an exact formula for
$\mathrm{rsob}\;{X^{<\alpha}}$ and $||X^{<\alpha}||$ here. When
$\alpha=\omega^{1}$, $X^{<\alpha}=X^{*}$, $[{\alpha}]=0$, and Proposition 9.4
implies $\mathrm{rsob}\;{X^{*}}\leq\omega^{{||X||}^{\circ}}$ and
$||X^{*}||\leq\omega^{\omega^{{||X||}^{\circ}}}$. The former is essentially
optimal, since if $X\neq\emptyset$, then
$\mathrm{rsob}\;{X^{*}}=\omega^{{||X||}^{\circ}}$ [8, Theorem 12.13], while
the latter is close to optimal: if $||X||$ is infinite,
$||X^{*}||=\omega^{\omega^{{||X||}^{\circ}}}$ [8, Theorem 12.23], generalizing
Schmidt’s formula on well-partial-orders [13, Theorem 9]. But when $||X||$ is
finite and non-empty,
$||X^{*}||=\omega^{\omega^{||X||-1}}<\omega^{\omega^{{||X||}^{\circ}}}$.

##  10\. The specialization ordering

A non-empty transfinite word $w$ is indecomposable if and only if for every
way of writing $w$ as $uv$ where $v\neq\epsilon$, we have $w\leq_{*}v$. Pouzet
[12], confirming a conjecture of Jullien [9], shows that $\leq$ is a better-
quasi-ordering if and only if every transfinite word (of countable length) is
a concatenation of finitely many indecomposable words.

Given a topological space $X$ and an ordinal $\alpha$, let us call $w\in
X^{<\alpha}$ _topologically indecomposable_ if and only $w\neq\epsilon$ and,
for every way of writing $w$ as $uv$ where $v\neq\epsilon$, $w$ is in the
closure $\overline{v}$ of $v$ in $X^{<\alpha}$. (We write $\overline{v}$
instead of the more cumbersome notation $cl(\\{v\\})$.) When $w=uv$, we have
$v\leq_{*}w$, so $v\in\overline{w}$ by Lemma 4.2. Hence $w$ is topologically
indecomposable if and only if for every way of writing $w$ as $uv$ where
$v\neq\epsilon$, $\overline{v}=\overline{w}$.

######  Lemma 10.1.

Let $X$ be a topological space. Every indecomposable transfinite word $w$ on
$X$ is topologically indecomposable.

###### Proof.

Whenever $w=uv$ with $v\neq\epsilon$, $w\leq_{*}v$; so $w\in\overline{v}$ by
Lemma 4.2. ∎

######  Lemma 10.2.

Every transfinite word $w\in X^{<\alpha}$, where $X$ is Noetherian and
$\alpha$ is a bound, can be written as a finite concatenation of topologically
indecomposable transfinite words.

###### Proof.

By induction on $|w|$. The claim is clear if $w=\epsilon$. Otherwise, for
every $\beta<|w|$, we write $w$ as $u_{\beta}v_{\beta}$ where $u_{\beta}$ is
its prefix of length $\beta$, and $v_{\beta}\neq\epsilon$. For all ordinals
$\beta<\gamma<|w|$, $v_{\gamma}\leq_{*}v_{\beta}$, so
$v_{\gamma}\in\overline{v_{\beta}}$ by Lemma 4.2, and therefore
$\overline{v_{\gamma}}\subseteq\overline{v_{\beta}}$. Hence the closed sets
$\overline{v_{\beta}}$ with $\beta<|w|$ form a chain. Since $X^{<\alpha}$ is
Noetherian (Theorem 9.3), inclusion is well-founded on its closed subsets, so
there is a non-empty suffix $v_{\beta}$ of $w$ whose closure
$\overline{v_{\beta}}$ is smallest.

We claim that $v_{\beta}$ is topologically indecomposable. Let us write
$v_{\beta}$ as $uv$ with $v\neq\epsilon$. Then $v=v_{\gamma}$, where
$\gamma\mathrel{\buildrel\text{def}\over{=}}\beta+|u|$, and
$u_{\gamma}=u_{\beta}u$. Since $\overline{v_{\beta}}$ is least and
$\overline{v_{\gamma}}\subseteq\overline{v_{\beta}}$, owing to the fact that
$\gamma\geq\beta$, we deduce that
$\overline{v}=\overline{v_{\gamma}}=\overline{v_{\beta}}$.

By induction hypothesis (indeed, $|u_{\beta}|=\beta<|w|$), $u_{\beta}$ is a
finite concatenation of topologically indecomposable words; hence so is
$w=u_{\beta}v_{\beta}$. ∎

######  Lemma 10.3.

The length of a topologically indecomposable transfinite word is
indecomposable.

###### Proof.

Let $w$ be topologically indecomposable in $X^{<\alpha}$, and let us assume
that $|w|$ is decomposable. We write $|w|$ as $\beta+\gamma$, where
$0<\beta,\gamma<|w|$, and then $w$ as $uv$ where $|u|=\beta$. We have
$|v|=\gamma\neq 0$, so $v\neq\epsilon$. Since $w$ is topologically
indecomposable, $w$ is in $\overline{v}$. Now $v$ is in $X^{<\gamma+1}$, which
is closed, so $\overline{v}\subseteq X^{<\gamma+1}$. Thus $w$ is in
$X^{<\gamma+1}$, which is impossible since $\gamma<|w|$. ∎

######  Lemma 10.4.

Let $X$ be a Noetherian space and $\alpha$ be a bound. For every non-empty
transfinite word $w\in X^{<\alpha}$, $w$ is topologically indecomposable if
and only if for every closed subset of $X^{<\alpha}$ that is the concatenation
$\mathcal{C}_{1}\mathcal{C}_{2}$ of two closed subsets of $X^{\alpha}$, if
$w\in\mathcal{C}_{1}\mathcal{C}_{2}$ then $w$ is in $\mathcal{C}_{1}$ or in
$\mathcal{C}_{2}$.

###### Proof.

Let $w$ be topologically indecomposable. Since
$w\in\mathcal{C}_{1}\mathcal{C}_{2}$, we can write $w$ as $uv$ where
$u\in\mathcal{C}_{1}$ and $v\in\mathcal{C}_{2}$. If $v=\epsilon$, then $w=u$
is in $\mathcal{C}_{1}$. Otherwise, since $w$ is topologically indecomposable,
$\overline{v}=\overline{w}$. But $\overline{v}\subseteq\mathcal{C}_{2}$, so
$w\in\overline{w}\subseteq\mathcal{C}_{2}$.

Conversely, let us assume that $w\in\mathcal{C}_{1}\mathcal{C}_{2}$ implies
$w\in\mathcal{C}_{1}$ or $w\in\mathcal{C}_{2}$ for every concatenation of two
closed sets $\mathcal{C}_{1}\mathcal{C}_{2}$. For any way of writing $w$ as
$uv$ with $v\neq\emptyset$, we let
$\mathcal{C}_{1}\mathrel{\buildrel\text{def}\over{=}}\overline{u}$ and
$\mathcal{C}_{2}\mathrel{\buildrel\text{def}\over{=}}\overline{v}$. Then
$w\in\overline{u}$ or $w\in\overline{v}$. Let
$\beta\mathrel{\buildrel\text{def}\over{=}}|u|$. We note that $\beta<|w|$,
since $|w|=\beta+|v|$ and $v\neq\epsilon$. Clearly, $u$ is in the closed set
$X^{<\beta+1}$, so $\overline{u}\subseteq X^{<\beta+1}$. If
$w\in\overline{u}$, then $w$ is in $X^{<\beta+1}$, so $|w|\leq\beta<|w|$,
which is impossible. Therefore $w$ is in $\overline{v}$. ∎

Let us write $\mathop{\mathrm{Im}}\nolimits w$ for the set of letters in a
transfinite word $w$. We call _support_ $\mathrm{supp}\;{w}$ of $w$ the
closure $cl(\mathop{\mathrm{Im}}\nolimits w)$ of
$\mathop{\mathrm{Im}}\nolimits w$.

######  Lemma 10.5.

Let $X$ be Noetherian, and $\alpha$ be a bound. For every topologically
indecomposable word $w\in X^{<\alpha}$,
$\overline{w}={(\mathrm{supp}\;{w})}^{<\gamma+1}$, where
$\gamma\mathrel{\buildrel\text{def}\over{=}}|w|$.

###### Proof.

Let $F\mathrel{\buildrel\text{def}\over{=}}\mathrm{supp}\;{w}$. Clearly, $w$
is in $F^{<\gamma+1}$, so $\overline{w}\subseteq F^{<\gamma+1}$. Conversely,
$\overline{w}$ is irreducible closed, hence is an $\alpha$-product
$P\mathrel{\buildrel\text{def}\over{=}}A_{1}\cdots A_{n}$, by Theorem 9.3.
Since $w\in P$ is topologically indecomposable hence non-empty, $n\geq 1$.
Using Lemma 10.4, $w$ is in some $A_{i}$. Let us write $A_{i}$ as
${F^{\prime}}^{<\gamma^{\prime}}$. Since $|w|=\gamma$,
$\gamma<\gamma^{\prime}$. Every letter of $w$ is in $F^{\prime}$, so
$\mathop{\mathrm{Im}}\nolimits w\subseteq F^{\prime}$, and taking closures,
$F\subseteq F^{\prime}$. Then
$F^{<\gamma+1}\subseteq{F^{\prime}}^{<\gamma^{\prime}}$, so
$F^{<\gamma+1}\subseteq P=\overline{w}$. ∎

######  Lemma 10.6.

Let $X$ be a Noetherian space, $\beta$ be an ordinal, and $\alpha$ be
$\omega^{\beta}$ or $\omega^{\beta}+1$. The map $cat$ is closed and continuous
from $X^{<\omega^{\beta}}\times X^{<\alpha}$ to $X^{<\alpha}$.

###### Proof.

It is well-defined and continuous by Lemma 5.1. By Theorem 9.3, the closed
subsets of $X^{<\omega^{\beta}}$ (resp., $X^{<\alpha}$) are the finite unions
of $\omega^{\beta}$-products (resp., $\alpha$-products). Given any two such
closed sets expressed as finite unions of such products
$\mathcal{C}_{1}\mathrel{\buildrel\text{def}\over{=}}\bigcup_{i=1}^{m}P_{i}$
and
$\mathcal{C}_{2}\mathrel{\buildrel\text{def}\over{=}}\bigcup_{j=1}^{n}Q_{j}$,
the image $\mathcal{C}_{1}\mathcal{C}_{2}$ of
$\mathcal{C}_{1}\times\mathcal{C}_{2}$ by $cat$ is the (finite) union over all
$i$ and $j$ of the products $P_{i}Q_{j}$. ∎

######  Lemma 10.7.

Let $X$ be a Noetherian space, and $\alpha$ be a bound. For all sets of
transfinite words $A$ and $B$ such that $AB\subseteq X^{<\alpha}$,
$cl(AB)=cl(A)cl(B)$.

###### Proof.

We use Lemma 10.6. If $\alpha$ is indecomposable, $cat$ is continuous from
$X^{<\alpha}\times X^{<\alpha}$ to $X^{<\alpha}$, so $cl(A)cl(B)\subseteq
cl(AB)$; $cat$ is closed, so $cl(A)cl(B)$ is closed, and contains $AB$, so it
contains $cl(AB)$.

If $\alpha=\omega^{\beta}+1$, then either $B\subseteq\\{\epsilon\\}$, in which
case $cl(AB)=cl(A)=cl(A)cl(B)$; or $A\subseteq X^{<\omega^{\beta}}$ (else we
could pick $u\in A$ of length $\omega^{\beta}$, $v\neq\epsilon$ in $B$, and
then $uv\in AB$ would not be in $X^{<\alpha}$), then we reason as above, using
the fact that $cat$ is closed and continuous from $X^{<\omega^{\beta}}\times
X^{<\alpha}$ to $X^{<\alpha}$. ∎

######  Proposition 10.8.

For every Noetherian space $X$ and every bound $\alpha$, the regular subword
topology on $X^{<\alpha}$ is the coarsest one such that $F^{<\gamma}$ is
closed for every closed subset $F$ of $X$ and every ordinal
$\gamma\leq\alpha$, and such that $\mathcal{C}_{1}\mathcal{C}_{2}$ is closed
for all closed subsets $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ such that
$\mathcal{C}_{1}\mathcal{C}_{2}\subseteq X^{<\alpha}$.

###### Proof.

Let us call admissible any topology $\tau$ containing the sets $F^{<\gamma}$
as closed sets, and closed under concatenations
$\mathcal{C}_{1}\mathcal{C}_{2}$, as described above. The regular subword
topology is admissible, since
$\mathcal{C}_{1}\mathcal{C}_{2}=cl(\mathcal{C}_{1})cl(\mathcal{C}_{2})=cl(\mathcal{C}_{1}\mathcal{C}_{2})$,
by Lemma 10.7. Given any admissible topology $\tau$, we see that the
$\alpha$-products are closed in $\tau$, so $\tau$ is finer than the regular
subword topology. ∎

There is a full, continuous map $\eta_{Y}\colon
y\mapsto\mathop{\downarrow}\nolimits y$ from any space $Y$ to $\mathcal{S}Y$.
The specialization preordering $\leq$ on $Y$ is characterized by $y\leq
y^{\prime}$ if and only if $y\in\eta_{Y}(y^{\prime})$. We note that
$\eta_{X^{<\alpha}}(w)$ is simply $\overline{w}$.

######  Theorem 10.9.

Let $X$ be a Noetherian space, $\alpha$ be a bound. Let
$\leq^{\mathrm{top}}_{*}$ be the specialization preordering of $X^{<\alpha}$.

  1. (1)

For every $w\in X^{<\alpha}$, one can write $w$ as a finite concatenation of
topologically indecomposable words $w_{1}\cdots w_{n}$, and then
$\eta_{X^{<\alpha}}(w)=\overline{w}=\overline{w_{1}}\cdots\overline{w_{n}}$,
and $\overline{w_{i}}={(\mathrm{supp}\;{w_{i}})}^{<|w_{i}|+1}$ for every $i$,
$1\leq i\leq n$.

  2. (2)

For all transfinite words $w\mathrel{\buildrel\text{def}\over{=}}w_{1}\cdots
w_{m}$ and
$w^{\prime}\mathrel{\buildrel\text{def}\over{=}}w^{\prime}_{1}\cdots
w^{\prime}_{n}$ written as finite concatenations of topologically
indecomposable words, $w\leq^{\mathrm{top}}_{*}w^{\prime}$ if and only if
there are indices $0=i_{0}\leq i_{1}\leq\cdots\leq i_{n-1}\leq i_{n}=m$ such
that for every $j$ with $1\leq j\leq n$,
$\bigcup_{i=i_{j-1}+1}^{i_{j}}\mathrm{supp}\;{w}_{i}\subseteq\mathrm{supp}\;{w^{\prime}_{j}}$
and $|w_{i_{j-1}+1}|+\cdots+|w_{i_{j}}|\leq|w^{\prime}_{j}|$.

###### Proof.

(1) We write $w$ as $w_{1}\cdots w_{n}$ where each $w_{i}$ is topologically
indecomposable by Lemma 10.2. By Lemma 10.7 (and since
$\overline{\epsilon}=\varepsilon$ in the base case $n=0$),
$\overline{w}=\overline{w_{1}}\cdots\overline{w_{n}}$. Finally,
$\overline{w_{i}}={(\mathrm{supp}\;{w_{i}})}^{<|w_{i}|+1}$ by Lemma 10.5.

(2) Lemma 10.4 has the following consequence. Given any closed set of the form
$\mathcal{C}_{1}\mathcal{C}_{2}$ with $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$
closed, for every transfinite word
$w\mathrel{\buildrel\text{def}\over{=}}w_{1}\cdots w_{m}$ in
$\mathcal{C}_{1}\mathcal{C}_{2}$, written as a finite concatenation of
topologically indecomposable words, there is an index $i$ with $0\leq i\leq m$
such that $w_{1}\cdots w_{i}\in\mathcal{C}_{1}$ and $w_{i+1}\cdots
w_{m}\in\mathcal{C}_{2}$. Indeed, since $w\in\mathcal{C}_{1}\mathcal{C}_{2}$,
there is an index $j$ with $1\leq j\leq m$ such that one can write $w_{j}$ as
$uv$, and $w_{1}\cdots w_{j-1}u\in\mathcal{C}_{1}$, $vw_{j}\cdots
w_{m}\in\mathcal{C}_{2}$. Since $cat$ is continuous (Lemma 5.1 (2)),
$\mathcal{C}^{\prime}_{1}\mathrel{\buildrel\text{def}\over{=}}cat(w_{1}\cdots
w_{j-1},\\_)^{-1}(\mathcal{C}_{1})$ and
$\mathcal{C}^{\prime}_{2}\mathrel{\buildrel\text{def}\over{=}}cat(\\_,w_{j}\cdots
w_{m})^{-1}(\mathcal{C}_{2})$ are closed. Now
$w_{j}=uv\in\mathcal{C}^{\prime}_{1}\mathcal{C}^{\prime}_{2}$, so $w_{j}$ is
in $\mathcal{C}^{\prime}_{1}$ or in $\mathcal{C}^{\prime}_{2}$; we define $i$
as $j$ in the first case, as $j-1$ in the second case.

By induction on $n$, if $w$ belongs to a finite product
$\mathcal{C}_{1}\cdots\mathcal{C}_{n}$ of closed subsets of $X^{<\alpha}$
included in $X^{<\alpha}$, we can find indices $0=i_{0}\leq
i_{1}\leq\cdots\leq i_{n-1}\leq i_{n}=m$ such that $w_{i_{j-1}+1}\cdots
w_{i_{j}}\in\mathcal{C}_{j}$ for every $j$ with $1\leq j\leq n$.

Let us assume that $w\leq^{\mathrm{top}}_{*}w^{\prime}$, namely
$\overline{w}\subseteq\overline{w^{\prime}}$. By (1),
$\overline{w_{1}}\cdots\overline{w_{m}}\subseteq\overline{w^{\prime}_{1}}\cdots\overline{w^{\prime}_{n}}$.
Letting
$\mathcal{C}_{j}\mathrel{\buildrel\text{def}\over{=}}\overline{w^{\prime}_{j}}$,
we obtain indices $0=i_{0}\leq i_{1}\leq\cdots\leq i_{n-1}\leq i_{n}=m$ such
that $w_{i_{j-1}+1}\cdots w_{i_{j}}\in\overline{w^{\prime}_{j}}$ for each $j$.
Since $w^{\prime}_{j}\in X^{<|w^{\prime}_{j}|+1}$,
$\overline{w^{\prime}_{j}}\subseteq X^{<|w^{\prime}_{j}|+1}$, so
$|w_{i_{j-1}+1}|+\cdots+|w_{i_{j}}|=|w_{i_{j-1}+1}\cdots
w_{i_{j}}|\leq|w^{\prime}_{j}|$. Also, for every $i$ with $i_{j-1}+1\leq i\leq
i_{j}$, $w_{i}\leq_{*}w_{i_{j-1}+1}\cdots w_{i_{j}}$, so $w_{i}$ is in
$\overline{w^{\prime}_{j}}$ by Lemma 4.2; hence
$\mathrm{supp}\;{w_{i}}\subseteq\mathrm{supp}\;{w^{\prime}_{j}}$, using Lemma
10.5 and Lemma 7.3 (2).

Conversely, if
$\bigcup_{i=i_{j-1}+1}^{i_{j}}\mathrm{supp}\;{w}_{i}\subseteq\mathrm{supp}\;{w^{\prime}_{j}}$
and $|w_{i_{j-1}+1}|+\cdots+|w_{i_{j}}|\leq|w^{\prime}_{j}|$ for every $j$,
then $w_{i_{j-1}+1}\cdots w_{i_{j}}$ is in
${(\mathrm{supp}\;{w^{\prime}_{j}})}^{<|w^{\prime}_{j}|+1}=\overline{w^{\prime}_{j}}$
(by Lemma 10.5); so $w=w_{1}\cdots w_{m}$ is in
$\overline{w^{\prime}_{1}}\cdots\overline{w^{\prime}_{n}}=\overline{w^{\prime}}$,
by (1). ∎

How does this compare to the subword ordering $\leq_{*}$? By Lemma 4.2,
$w\leq_{*}w^{\prime}$ implies $w\leq^{\mathrm{top}}_{*}w^{\prime}$. The
converse holds on $X^{*}=X^{<\omega}$ [5, Exercise 9.7.29], and on
$X^{<\omega+1}$ if $X$ is a wqo in its Alexandroff topology [6, Proposition
5.16]. Otherwise, the result may fail, as the following demonstrates.

######  Example 10.10.

Let $X$ be $\mathbb{N}$ with the cofinite topology. Since its non-trivial
closed subsets are finite, $X$ is Noetherian. In order to see that
$\leq^{\mathrm{top}}_{*}$ and $\leq_{*}$ differ, let
$w\mathrel{\buildrel\text{def}\over{=}}0\;1\;2\;\cdots$. For every non-empty
suffix $v$ of $w$, $\mathop{\mathrm{Im}}\nolimits v$ is infinite, so
$\mathrm{supp}\;{v}=\mathbb{N}$. Hence $w$ is topologically indecomposable. By
Theorem 10.9 (1), $\overline{w}=X^{<\omega+1}$, so
$w^{\prime}\leq^{\mathrm{top}}_{*}w$ for every $w^{\prime}\in X^{<\omega+1}$.
Hence, for example, $0^{\omega}\leq^{\mathrm{top}}_{*}w$, but
$0^{\omega}\not\leq_{*}w$.

For spaces $X^{<\alpha}$ with $\alpha\geq\omega^{2}$, we have the following.

######  Example 10.11.

Let $X$ be any Noetherian space with two elements $a$ and $b$ such that
$a\not\leq b$ (say, $\\{a,b\\}$ with the discrete topology). We claim that
$\leq^{\mathrm{top}}_{*}$ and $\leq_{*}$ differ on $X^{<\alpha}$, for any
bound $\alpha\geq\omega^{2}$. Let
$w^{\prime}\mathrel{\buildrel\text{def}\over{=}}(ab^{\omega})^{\omega}$
(explicitly, $w^{\prime}\colon\omega^{2}\to X$,
$w^{\prime}(\omega.m+n)\mathrel{\buildrel\text{def}\over{=}}a$ if $n=0$, $b$
otherwise); $w^{\prime}$ is indecomposable, hence topologically indecomposable
by Lemma 10.1, and
$\mathrm{supp}\;{w^{\prime}}=\mathop{\downarrow}\nolimits\\{a,b\\}$. Let
$w\mathrel{\buildrel\text{def}\over{=}}a^{\omega^{2}}$. This is also
indecomposable, and $\mathrm{supp}\;{w}=\mathop{\downarrow}\nolimits a$.
Hence, by Theorem 10.9, $w\leq^{\mathrm{top}}_{*}w^{\prime}$. However,
$w\not\leq_{*}w^{\prime}$, since there is no strictly increasing map from
$\omega^{2}$ into $\omega$, the subset of positions of $w^{\prime}$ where $a$
occurs.

##  11\. Conclusion and open problems

We have described a Noetherian topology on spaces of transfinite words over a
Noetherian space $X$, in particular on spaces $X^{<\alpha}$, where $\alpha$ is
a bound. In the latter situation, we have characterized its irreducible closed
subsets, and given upper bounds on the stature and reduced sobrification rank
of $X^{<\alpha}$. We have also characterized the specialization preordering
$\leq^{\mathrm{top}}_{*}$ of $X^{<\alpha}$.

Although we have not stressed it, the syntax of $\alpha$-products naturally
yields an S-representation of $X^{<\alpha}$, in the sense of [3, 6], provided
we restrict our bounds to lie in some class of ordinals with a computable
representation, decidable ordering, and decidable equality. S-representations
are important in forward analysis procedures for well-structured transition
systems [2].

We finish with two questions. First, it is frustrating that
$\leq^{\mathrm{top}}_{*}$ and $\leq_{*}$ differ in general. Is there a
natural, finer Noetherian topology on $X^{<\alpha}$ that would have $\leq_{*}$
as specialization preordering? The specialization preordering of a Noetherian
space is necessarily well-founded. Hence, if the desired topology exists,
$X^{<\alpha}$ is well-founded under $\leq_{*}$. By an argument similar to
Lemma 10.2, every non-empty word $w\in X^{<\alpha}$ would have an
indecomposable suffix, in other words $X$ would have to be $\beta$-better-
quasi-ordered in the sense of [12, Definition IV-2] or of [4, Chapter 8, 5.1],
for every $\beta<\alpha$.

Second, what is the exact stature of $X^{<\alpha}$? its reduced sobrification
rank?

## References

  * [1] A. Blass and Y. Gurevich.  Program termination and well partial orderings.  ACM Transactions on Computational Logic, 9(3), 2008.  Article 18.
  * [2] A. Finkel and J. Goubault-Larrecq.  Forward analysis for WSTS, part II: Complete WSTS.  Logical Methods in Computer Science, 8(3:28), 2012.
  * [3] A. Finkel and J. Goubault-Larrecq.  Forward analysis for WSTS, part I: Completions.  Mathematical Structures in Computer Science, 30(7):752–832, 2020\.
  * [4] R. Fraïssé.  Theory of Relations, volume 118 of Studies in Logic and the Foundation of Mathematics.  North-Holland, 1986.
  * [5] J. Goubault-Larrecq.  Non-Hausdorff Topology and Domain Theory—Selected Topics in Point-Set Topology, volume 22 of New Mathematical Monographs.  Cambridge University Press, 2013.
  * [6] J. Goubault-Larrecq.  Infinitary Noetherian constructions I. Infinite words.  Colloquium Mathematicum, 2022.  Published online, January 3rd, 2022.
  * [7] J. Goubault-Larrecq, S. Halfon, P. Karandikar, and K. N. Kumar.  The ideal approach to computing closed subsets in well-quasi-orderings.  In P. Schuster, M. Seisenberger, and A. Weiermann, editors, Well Quasi-orders in Computation, Logic, Language and Reasoning, volume 53 of Trends in Logic (Studia Logica Library), pages 55–105. Springer, Cham, 2019\.
  * [8] J. Goubault-Larrecq and B. Laboureix.  Statures and sobrification ranks of Noetherian spaces.  Houston Journal of Mathematics, 2023.  To appear, available as arXiv report arXiv:2112.06828 [math.GN].
  * [9] P. Jullien.  Contribution à l’Étude des Types d’Ordres Dispersés.  PhD thesis, Faculté des Sciences de l’Université de Marseille (Saint-Charles), 1969.
  * [10] A. Levy.  Basic Set Theory.  Dover Books on Mathematics, 2002.  Original Springer Verlag edition, Perspectives in Mathematical Logic, 1979\.
  * [11] C. S.-J. A. Nash-Williams.  On better-quasi-ordering transfinite sequences.  Mathematical Proceedings of the Cambridge Philosophical Society, 64(2):273–290, 1968.
  * [12] M. Pouzet.  Sur les prémeilleurordres.  Annales de l’institut Fourier, 22(2):1–19, 1972.
  * [13] D. Schmidt.  Well-Partial Orderings and their Maximal Order Types.  Habilitationsschrift, Universität Heidelberg, 1979.