# Non-commutative Stone duality

Mark V. Lawson  Mark V. Lawson, Department of Mathematics and the Maxwell
Institute for Mathematical Sciences, Heriot-Watt University, Riccarton,
Edinburgh EH14 4AS, UNITED KINGDOM<EMAIL_ADDRESS>This paper is
dedicated to the memory of my friend and colleague Iain Currie

###### Abstract.

We show explicitly that Boolean inverse semigroups are in duality with what we
term Boolean groupoids. This generalizes classical Stone duality, which we
refer to as commutative Stone duality, between generalized Boolean algebras
and locally compact Hausdorff $0$-dimensional spaces.

##  1\. Introduction

The theory of what we term non-commutative Stone duality grew out of the work
of a number of authors [18, 46, 43], [13, 14, 36] and [31, 47]. In this paper,
I shall concentrate on one aspect of that duality: namely, how Boolean inverse
semigroups are in duality with a class of étale groupoids called Boolean
groupoids. Specifically, we shall prove the following theorem and discuss some
special cases (proved as Theorem 7.12):

######  Theorem 1.1 (Non-commutative Stone duality).

The category of Boolean inverse semigroups and callitic morphisms is dually
equivalent to the category of Boolean groupoids and coherent, continuous,
covering functors.

This theorem generalizes what you will find in [24] since we shall not assume
that our topological groupoids are Hausdorff. Although this theorem can be
gleaned from our papers [24, 25, 26, 27, 30], what I describe here has not
been reported in one place before.

You might think that this duality is of merely parochial interest. It isn’t.
The work of Matui [39, 40] deals with étale groupoids of just the kind that
figure in our duality theorem. In addition, Matui refers constantly to
‘compact open $\mathcal{G}$-sets’. These are precisely what we call ‘compact-
open local bisections’ and are elements of the inverse semigroup associated
with the étale groupoid. Thus, even though Matui is not explicitly interested
in Boolean inverse semigroups, they are there implicitly.

Sections 3 to 7 are devoted to proving the above theorem whereas in Section 2
classical Stone duality is decribed since this sets the scene for our
generalization. Thus the reader familiar with classical Stone duality can
start reading at Section 3. In Section 3, we describe the theory of Boolean
inverse semigroups needed: these should be regarded as the non-commutative
generalized Boolean algebras. In Section 4, we describe Boolean groupoids;
these are a class of étale topological groupoids and play the role of ‘non-
commutative topological spaces’. In Section 5, we show how to pass from
Boolean groupoids to Boolean inverse semigroups, where a key role is played by
the compact-open local bisections of a Boolean groupoid. This is a
comparatively easy construction. In Section 6, we show how to pass from
Boolean inverse semigroups to Boolean groupoids by using ultrafilters. This is
quite technical. In Section 7, we show, amongst other things, that the above
two constructions are the inverse of each other. We also bring on board
morphisms and establish our main duality theorem which generalizes the
classical theory described in Section 2. Section 8 is dedicated to special
cases; for example, those Boolean inverse semigroups that have all binary
meets correspond to Hausdorff Boolean groupoids. There is one new application,
in Section 9, which shows how our theory can be used to give an account of
unitization first described in [59, Definition 6.6.1]. In Section 10, I shall
show how the above theory can be derived within a more general framework using
pseudogroups and arbitrary étale groupoids.

I shall assume you are conversant with the theory of inverse semigroups [20];
I will simply remind the reader of notation and terminology as we go along.
Observe that whenever I refer to an order on an inverse semigroup, I mean the
natural partial order defined on every inverse semigroup. I shall develop the
theory of Boolean inverse semigroups from scratch but this is no subsitute for
a proper development of that theory as in [59].

We shall also need some terminology from the theory of posets throughout. Let
$P$ be a poset with a minimum (or bottom) element denoted by zero $0$. In this
context, singleton sets such as $\\{x\\}$ will be written simply as $x$. If
$X\subseteq P$, define

$X^{\downarrow}=\\{y\in P\colon y\leq x\text{ for some }x\in X\\}\text{ and
}X^{\uparrow}=\\{y\in P\colon x\leq y\text{ for some }x\in X\\}.$

If $X=X^{\downarrow}$ we say that $X$ is an order-ideal. An order-ideal of the
form $a^{\downarrow}$ is said to be principal. If for any $x,y\in X$ there
exists $z\in X$ such that $z\leq x,y$, we say that $X$ is downwardly directed.
If $X=X^{\uparrow}$ we say that $X$ is upwardly closed.

Let $X$ and $Y$ be posets. A function $\theta\colon X\rightarrow Y$ is said to
be order-preserving if $x_{1}\leq x_{2}$ in $X$ implies that
$\theta(x_{1})\leq\theta(x_{2})$. An order-isomorphism is a bijection which is
order-preserving and whose inverse is order-preserving. Let $F$ be a non-
empty111Throughout this paper, filters will be assumed non-empty. subset of
$X$. We say that it is a filter if it is downwardly directed and upwardly
closed. We say that it is proper if it does not contain the zero. A maximal
proper filter is called an ultrafilter. Ultrafilters play an important role in
this paper.

A meet semilattice is a poset in which each pair of elements has a greatest
lower bound (or meet); we write $x\wedge y$ for the meet of $x$ and $y$. The
following was proved as [6, Lemma 12.3] and is very useful.

######  Lemma 1.2.

Let $P$ be a meet semilattice with bottom element $0$, and let $A$ be a proper
filter in $P$. Then $A$ is an ultrafilter if and only if the following holds:
if $x\wedge y\neq 0$ for all $x\in A$ then $y\in A$.

Let $Y$ be a meet semilattice with bottom element $0$. Let $X\subseteq Y$ be
any subset. Define $X^{\wedge}$ to be the set of all finite meets of elements
of $X$; we say that $X$ has the finite intersection property if $0\notin
X^{\wedge}$.

Our references for topology are [53] and [60]. The following contains the main
results we need. For the proof of (1), see [53, Section 23, Theorem A]; for
(2), see [53, Section 26, Theorem D]; for (3), see [53, Section 21, Theorem
A]; for (4), see [53, Section 21, Theorem D]; for (5), see [60, Theorem 18.2]

######  Lemma 1.3.

  1. (1)

The product of any non-empty set of compact spaces is compact.

  2. (2)

Every compact subspace of a Hausdorff space is closed.

  3. (3)

Every closed subspace of a compact space is compact.

  4. (4)

A topological space is compact if and only if every set of closed sets with
the finite intersection property has a non-empty intersection

  5. (5)

A Hausdorff space is locally compact if for every point $x$ there is a compact
set $U$ such that $x\in U^{\circ}$, the union of all the open sets contained
in $U$.

Let $\mathscr{B}$ be any set of subsets of a set $X$. We say that
$\mathscr{B}$ is a base if the union of all elements in $\mathscr{B}$ is $X$
and if $x\in B\cap C$ where $B,C\in\mathscr{B}$ then there is a
$D\in\mathscr{B}$ such that $x\in D\subseteq B\cap C$. Given a base
$\mathscr{B}$, we may define as a topology all those sets which are unions of
subsets of $\mathscr{B}$; where the empty set is the union of the empty subset
of $\mathscr{B}$. See [60, Theorem 5.3]. A space with a countable base is said
to be second countable. A topological space is discrete if every susbet is
open. A subset of a topological space is said to be clopen if it is both open
and closed. A topological space is said to be $0$-dimensional if it has a base
of clopen sets. If $X$ is a topological space we denote its set of open
subsets by $\Omega(X)$.

Acknowledgements. Some of the work for this paper was carried out at LaBRI,
Université de Bordeaux during April 2018 whilst visiting David Janin. I am
also grateful to Phil Scott for alerting me to typos. None of this work would
have been possible without my collaboration with Daniel Lenz, and some very
timely conversations with Pedro Resende.

##  2\. Classical Stone duality

In this section, we shall describe classical Stone duality which relates
generalized Boolean algebras to locally compact Hausdorff $0$-dimensional
spaces. There are no new results in this section — the theory is classical —
but it provides essential motivation for what we do when we come to study
Boolean inverse semigroups. In Section 2.1, we shall recall the definition and
first properties of Boolean algebras; in Section 2.2, we shall describe the
structure of finite Boolean algebras; in Section 2.3, we describe classical
Stone duality which deals with the relationship between Boolean algebras and
compact Boolean spaces — this is the version of Stone duality that you will
find well represented in the textbooks; finally, in Section 2.4, we describe
the extension of classical Stone duality to generalized Boolean algebras and
locally compact Boolean spaces.

###  2.1. Boolean algebras

Boolean algebras may not rank highly in the pantheon of algebraic structures
but they are, in fact, both mathematically interesting and remarkably useful;
for example, they form the basis of measure theory.

Formally, a Boolean algebra is a $6$-tuple
$(B,\vee,\wedge,\,^{\prime}\,\,,0,1)$ consisting of a set $B$, two binary
operations $\vee$, called join, and $\wedge$, called meet, one unary operation
$a\mapsto a^{\prime}$, called complementation, and two constants $0$ and $1$
satisfying the following ten axioms:

$\begin{array}[]{ll}{\rm(B1)}\,(x\vee y)\vee z=x\vee(y\vee
z).&{\rm(B6)}\,x\wedge 1=x.\\\ {\rm(B2)}\,x\vee y=y\vee
x.&{\rm(B7)}\,x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z).\\\ {\rm(B3)}\,x\vee
0=x.&{\rm(B8)}\,x\vee(y\wedge z)=(x\vee y)\wedge(x\vee z).\\\
{\rm(B4)}\,(x\wedge y)\wedge z=x\wedge(y\wedge z).&{\rm(B9)}\,x\vee
x^{\prime}=1.\\\ {\rm(B5)}\,x\wedge y=y\wedge x.&{\rm(B10)}\,x\wedge
x^{\prime}=0.\\\ \end{array}$

The element $0$ is called the bottom and $1$ is called the top. A function
$\theta\colon B\rightarrow C$ between two Boolean algebras is called a
homomorphism if it preserves the two binary operations, the unary operation
and maps constants to corresponding constants. Boolean algebras and their
homomorphisms form a category. Observe that for a function to be a
homomorphism of Boolean algebras it is enough that it preserves the constants
and maps meets and joins to meets and joins, respectively; this is because the
complement of $x$ is the unique element $y$ such that $1=x\vee y$ and
$0=x\wedge y$. The following lemma summarizes some important properties of
Boolean algebras that readily follow from these axioms.

######  Lemma 2.1.

In a Boolean algebra $B$, the following hold for all $x,y\in B$:

$\begin{array}[]{ll}(1)\,x\vee x=x\text{ and }x\wedge x=x.&(5)\,x\vee
y=x\vee(y\wedge x^{\prime}).\\\ (2)\,x\wedge 0=0.&(6)\,x^{\prime\prime}=x.\\\
(3)\,1\vee x=1.&(7)\,(x\vee y)^{\prime}=x^{\prime}\wedge y^{\prime}.\\\
(4)\,x=x\vee(x\wedge y).&(8)\,(x\wedge y)^{\prime}=x^{\prime}\vee
y^{\prime}.\\\ \end{array}$

The theory of Boolean algebras is described in an elementary fashion in [9]
and from a more advanced standpoint in [15] and [51]. The first two chapters
of [12] approach the subject of Stone duality from the perspective of frame
theory.

######  Example 2.2.

  1. (1)

The basic example of a Boolean algebra is the power set Boolean algebra which
consists of the set of all subsets, $\mathsf{P}(X)$, of the non-empty set $X$
with the operations $\cup$, $\cap$ and $\overline{A}=X\setminus A$ and the two
constants $\varnothing$, $X$.

  2. (2)

We denote by $\mathbb{B}$ the unique two-element Boolean algebra.

  3. (3)

Let $A$ be any finite non-empty set. We call $A$ an alphabet. Denote the free
monoid on $A$ by $A^{\ast}$; observe that $A^{\ast}$ is countably infinite.
Elements of the free monoid are called strings. The length of a string $x$ is
denoted by $|x|$. The empty string is denoted by $\varepsilon$. A subset of
$A^{\ast}$ is called a language over $A$. Recall that a language over an
alphabet $A$ is said to be recognizable if there is a finite-state automaton
that accepts it. By Kleene’s theorem [21], the set of recognizable languages
over $A$ is equal to the set of regular languages over $A$. Denote the set of
regular languages over $A$ by $\mbox{\rm Reg}(A)$. This set is a Boolean
algebra (with extra operations). This Boolean structure can be exploited to
provide a sophisticated way of studying families of regular languages [45, 8].

Boolean algebras have their roots in the work of George Boole, though the
definition of Boolean algebras seems to have been inspired by his work rather
than originating there [1, 11]. Until the 1930s, research on Boolean algebras
was essentially about axiomatics. However, with Stone’s paper [55], stability
in the definition of Boolean algebras emerges because he showed that each
Boolean algebra could be regarded as a (unital) ring in which each element was
idempotent; rings such as this are called Boolean rings. In the language of
category theory, his result shows that the category of unital Boolean algebras
is isomorphic to the category of unital Boolean rings (where I have stressed
the fact that the Boolean algebra has a top element and the ring has an
identity). The following result describes how the correspondence between
Boolean algebras and Boolean rings works at the algebraic level.

######  Theorem 2.3.

  1. (1)

Let $B$ be a Boolean algebra. Define $a+b=(a\wedge
b^{\prime})\vee(a^{\prime}\wedge b)$ (the symmetric difference) and $a\cdot
b=a\wedge b$. Then $(B,+,\cdot,1)$ is a Boolean ring.

  2. (2)

Let $(R,+,\cdot,1)$ be a Boolean ring. Define $a\vee b=a+b+a\cdot b$, $a\wedge
b=a\cdot b$ and $a^{\prime}=1-a$. Then $(R,\vee,\wedge,\,^{\prime}\,,0,1)$ is
a Boolean algebra

  3. (3)

The constructions (1) and (2) above are mutually inverse.

The above result is satisfying since the definition of Boolean rings could
hardly be simpler but also raises the interesting question of why Marshall H.
Stone (1903–1989), a functional analyst, should have been interested in
Boolean algebras in the first place. The reason is that Stone worked on the
spectral theory of symmetric operators and this led to an interest in algebras
of commuting projections. Such algebras are naturally Boolean algebras. The
connection between Boolean inverse semigroups and $C^{\ast}$-algebras
continues this tradition. The following theorem [7] puts this connection in a
slightly wider context. If $R$ is a ring denote its set of idempotents by
$\mathsf{E}(R)$.

######  Theorem 2.4.

Let $R$ be a unital commutative ring. Then the set $\mathsf{E}(R)$ is a
Boolean algebra when we define $e\vee f=e+f-ef$, $e\wedge f=e\cdot f$ and
$e^{\prime}=1-e$.

By Theorem 2.3 and Theorem 2.4, it is immediate that each Boolean algebra
arises as the Boolean algebra of idempotents of a commutative ring.

So far we have viewed Boolean algebras as purely algebraic objects, but in
fact they come equipped with a partial order that underpins this algebraic
structure. Let $B$ be a Boolean algebra. For $x,y\in B$, define $x\leq y$ if
and only if $x=x\wedge y$; we say that $y$ lies above $x$. The proofs of the
following are routine.

######  Lemma 2.5.

With the above definition, we have the following:

  1. (1)

$\leq$ is a partial order on $B$.

  2. (2)

$x\leq y$ if and only if $y=x\vee y$.

  3. (3)

$a\wedge b=\mbox{\rm glb}\\{a,b\\}$ and $a\vee b=\mbox{\rm lub}\\{a,b\\}$.

###  2.2. Finite Boolean algebras

In this section, we describe the structure of all finite Boolean algebras. It
is only included to provide motivation for the sections that follow. The
crucial idea is that of an atom. A non-zero element $x\in B$ of a Boolean
algebra is called an atom if $y\leq x$ implies that either $x=y$ or $y=0$. We
denote the set of atoms of the Boolean algebra $B$ by $\mathsf{at}(B)$. The
proof of the following is immediate from the definition.

######  Lemma 2.6.

Let $x$ and $y$ be atoms. Then either $x=y$ or $x\wedge y=0$.

The proof of the following is immediate.

######  Lemma 2.7.

let $B$ be a finite Boolean algebra. Then every non-zero element lies above an
atom.

The following lemma will be useful.

######  Lemma 2.8.

Let $B$ be a Boolean algebra. If $a\nleq b$ then $a\wedge b^{\prime}\neq 0$.

###### Proof.

Suppose that $a\wedge b^{\prime}=0$. Then $a=a\wedge 1=a\wedge(b\vee
b^{\prime})=a\wedge b$ from which we get that $a\leq b$, which is a
contradiction. ∎

Let $B$ be a finite Boolean algebra. For each $a\in B$, denote by
$\mathscr{U}_{a}$ the set of all atoms in $B$ below $a$. By Lemma 2.7, it
follows that $a\neq 0$ implies that $\mathscr{U}_{a}\neq\varnothing$. The
important properties of the sets $\mathscr{U}_{a}$ we shall need are listed
below.

######  Lemma 2.9.

Let $B$ be a finite Boolean algebra.

  1. (1)

$\mathscr{U}_{0}=\varnothing$.

  2. (2)

$\mathscr{U}_{1}=\mathsf{at}(B)$.

  3. (3)

$\mathscr{U}_{a}\cap\mathscr{U}_{b}=\mathscr{U}_{a\wedge b}$.

  4. (4)

$\mathscr{U}_{a}\cup\mathscr{U}_{b}=\mathscr{U}_{a\vee b}$.

  5. (5)

$\mathscr{U}_{a^{\prime}}=\overline{\mathscr{U}_{a}}$.

  6. (6)

$\mathscr{U}_{a}\subseteq\mathscr{U}_{b}$ if and only if $a\leq b$.

  7. (7)

$\mathscr{U}_{a}=\mathscr{U}_{b}$ if and only if $a=b$.

###### Proof.

(1) Immediate.

(2) Immediate.

(3) Let $x\in\mathscr{U}_{a}\cap\mathscr{U}_{b}$. Then $x\leq a$ and $x\leq
b$. It follows that $x\leq a\wedge b$ and so $x\in\mathscr{U}_{a\wedge b}$.
Now suppose that $x\in\mathscr{U}_{a\wedge b}$. Then $x\leq a\wedge b$. But
$a\wedge b\leq a,b$. It follows that $x\in\mathscr{U}_{a}\cap\mathscr{U}_{b}$.

(4) Let $x\in\mathscr{U}_{a}\cup\mathscr{U}_{b}$. Then $x\leq a$ or $x\leq b$.
In either case, $x\leq a\vee b$. It follows that $x\in\mathscr{U}_{a\vee b}$.
Conversely, suppose that $x\in\mathscr{U}_{a\vee b}$. Then $x\leq a\vee b$. It
follows that $x=(a\wedge x)\vee(b\wedge x)$. But $x$ is an atom. So, either
$a\wedge x=x$ or $b\wedge x=x$; that is either $x\leq a$ or $x\leq b$. Hence
$x\in\mathscr{U}_{a}\cup\mathscr{U}_{b}$.

(5) Suppose that $x\in\overline{\mathscr{U}_{a}}$. Then $x\nleq a$. It follows
that $x\wedge a^{\prime}\neq 0$ by Lemma 2.8. But $x$ is an atom, and so
$x\leq a^{\prime}$. The proof of the reverse inclusion follows by part (6) of
Lemma 2.1,

(6) Suppose that $\mathscr{U}_{a}\subseteq\mathscr{U}_{b}$. If $a\nleq b$ then
$a\wedge b^{\prime}\neq 0$ by Lemma 2.8. By Lemma 2.7, there is an atom $x\leq
a\wedge b^{\prime}$. It follows that $x\leq a$ and $x\leq b^{\prime}$ and so
$x\nleq b$ by part (5) above. This contradicts our assumption. It follows that
$a\leq b$. The proof of the converse is immediate.

(7) This is immediate by part (6). ∎

######  Proposition 2.10.

Every finite Boolean algebra is isomorphic to the Boolean algebra of subsets
of a finite set.

###### Proof.

Let $B$ be a finite Boolean algebra. As pour set, we take $\mathsf{at}(B))$,
the set of atoms of $B$. Define a function
$B\rightarrow\mathsf{P}(\mathsf{at}(B))$, the set of all subsets of
$\mathsf{at}(B)$, by $a\mapsto\mathscr{U}_{a}$. By part (7) of Lemma 2.9, this
is an injective morphism of Boolean algebras. It remains only to prove that it
is surjective. Let $A=\\{x_{1},\ldots,x_{n}\\}$ be any non-empty set of atoms.
Put $a=\bigvee_{i=1}^{n}x_{i}$. We shall prove that $\mathscr{U}_{a}=A$. Let
$x$ be any atom such that $x\leq a$. Then $x=(x\wedge
x_{1})\vee\ldots\vee(x\wedge x_{n})$. By Lemma 2.6, it follows that $x=x_{i}$
for some $i$. ∎

We now place the above result in its proper categorical context.

######  Theorem 2.11 (Stone duality for finite Boolean algebras).

The category of finite Boolean algebras and homomorphisms between them is
dually equivalent to the category of finite sets and functions between them.

###### Proof.

Let $\theta\colon B\rightarrow C$ be a homomorphism between finite Boolean
algebras. Define
$\theta^{\sharp}\colon\mathsf{at}(C)\rightarrow\mathsf{at}(B)$ by
$\theta^{\sharp}(f)=e$ if $f\leq\theta(e)$ and $e$ is an atom. We shall prove
that $\theta^{\sharp}$ really is a function. Suppose that $e$ and $e^{\prime}$
are distinct atoms such that $\theta(e)\geq f$ and $\theta(e^{\prime})\geq f$.
Then $\theta(e\wedge e^{\prime})\geq f$ but $e\wedge e^{\prime}=0$, since $e$
and $e$ are distinct atoms, which implies that $f=0$. This is a contradiction.
Therefore, $\theta^{\sharp}$ is a partial function. We have that
$1=\bigvee_{e\in\mathsf{at}(B)}\theta(e)$ inside $B$. Thus
$1=\theta(1)=\bigvee_{\theta(e)\in\mathsf{at}(B)}\theta(e)$. But $1\geq f$. It
follows that $f=\bigvee_{\theta(e)\in\mathsf{at}(B)}\theta(e)\wedge f$. But
$f$ is an atom, so that either $\theta(e)\wedge f=0$ or $\theta(e)\wedge f=f$.
In the latter case, $f\leq\theta(e)$. We have therefore proved that
$\theta^{\sharp}$ is a function.

Let $\alpha\colon\mathsf{at}(C)\rightarrow\mathsf{at}(B)$ be any function.
Define $\alpha^{\flat}\colon B\rightarrow C$ by

$\alpha^{\flat}(e)=\bigvee_{\alpha(f)\leq e}f.$

We prove that $\alpha^{\flat}$ is a homomorphism of Boolean algebras. The join
of the empty set is $0$ and so $\alpha^{\flat}(0)=0$. Similarly,
$\alpha^{\flat}(1)=1$. The fact that $\alpha^{\flat}$ preserves joins follows
from Lemma 2.9. The fact that $\alpha^{\flat}$ preserves meets is
straightforward. It follows that $\alpha^{\flat}$ is a homomorphism of Boolean
algebras.

Let $\theta\colon B\rightarrow C$ be a homomorphism of Boolean algebras. Then
$\theta^{\sharp}\colon\mathsf{at}(C)\rightarrow\mathsf{at}(B)$ is a well-
defined function. We calculate the effect of $(\theta^{\sharp})^{\flat}$ on
atoms. Let $e$ be an atom of $B$. Then by definition

$(\theta^{\sharp})^{\flat}(e)=\bigvee_{\theta^{\sharp}(f)\leq e}f.$

But $e$ is an atom and so $\theta^{\sharp}(f)=e$. It follows that
$f\leq\theta(e)$. We are therefore looking at the join of all atoms below
$\theta(e)$, which is exactly $\theta(e)$. We have therefore proved that
$(\theta^{\sharp})^{\flat}=\theta$ on atoms. It follows that
$(\theta^{\sharp})^{\flat}=\theta$ as functions.

Let $\alpha\colon\mathsf{at}(C)\rightarrow\mathsf{at}(B)$ be any function. Let
$f$ be any atom of $C$. Then $(\alpha^{\flat})^{\sharp}(f)=e$ if and only if
$f\leq\alpha^{\flat}(e)$ and $e$ is an atom. By definition

$\alpha^{\flat}(e)=\bigvee_{\alpha(i)\leq e}i=\bigvee_{\alpha(i)=e}i.$

It follows that $f=i$ for some $i$. Thus $\alpha(f)=e$. It follows that
$(\alpha^{\flat})^{\sharp}=\alpha$.

It is now routine to check that we have defined a dual equivalence of
categories. ∎

###  2.3. Arbitrary Boolean algebras

In the light of Proposition 2.10, it is tempting to conjecture that every
Boolean algebra is isomorphic to a powerset Boolean algebra. However, this
turns out to be false; powerset Boolean algebras always have atoms but there
are Boolean algebras that have no atoms at all (the atomless Boolean algebras.
See [9, page 118, Chapter 4]. The Lindenbaum-Tarski Boolean algebra
constructed from classical propositional logic is another example). To
describe arbitrary Boolean algebras, we have to adopt a different approach,
and this was just what Stone did [56]. You will find classical Stone duality
described in the following references [2, 9, 12, 15]. The approach is
symbolized below and can be viewed as a generalization of the finite case
described in the previous section:

$\begin{array}[]{ccc}\text{atom}&\stackrel{{\scriptstyle\text{\tiny replaced
by}}}{{\xrightarrow{\hskip 56.9055pt}}}&\text{ultrafilter}\\\
\text{powerset}&\stackrel{{\scriptstyle\text{\tiny replaced
by}}}{{\xrightarrow{\hskip 56.9055pt}}}&\text{topological space}\\\ \text{atom
}x\leq a&\stackrel{{\scriptstyle\text{\tiny replaced
by}}}{{\xrightarrow{\hskip 56.9055pt}}}&a\in F\text{ ultrafilter}\\\
\end{array}$

Let $B$ be an arbitrary Boolean algebra. There is no reason for $B$ to have
atoms, so we have to find ‘atom substitutes’ that do always exist. This is the
role of the ultrafilters. A non-empty subset $F$ of a Boolean algebra $B$ is
called a filter if it is closed under meets and upwardly closed. The filter
$F$ is said to be proper if $0\notin F$.

######  Example 2.12.

Let $a\in B$. Then $a^{\uparrow}$ is a filter called the principal filter
generated by $a$ .

A proper filter $F$ is said to be prime if $a\vee b\in F$ implies that $a\in
F$ or $b\in F$. A maximal proper filter is called an ultrafilter.

######  Lemma 2.13.

The following are equivalent for a proper filter $F$ in a Boolean algebra $B$.

  1. (1)

$F$ is an ultrafilter.

  2. (2)

For each non-zero $a\in B$ either $a\in F$ or $a^{\prime}\in F$.

  3. (3)

$F$ is a prime filter.

###### Proof.

(1)$\Rightarrow$(2). Let $F$ be an ultrafilter. Suppose that $a\notin F$. Then
by Lemma 1.2, there exists $b\in F$ such that $a\wedge b=0$. Now $1=a\vee
a^{\prime}$. Thus $b=(b\wedge a)\vee(b\wedge a^{\prime})$. So, $b=b\wedge
a^{\prime}$. It follows that $b\leq a^{\prime}$ giving $a^{\prime}\in F$, as
required.

(2)$\Rightarrow$(3). We prove that $F$ is a prime filter. Let $a\vee b\in F$.
Suppose that $a\notin F$ and $b\notin F$. Then, by assumption, $a^{\prime}\in
F$ and $b^{\prime}\in F$ so that $a^{\prime}\wedge b^{\prime}\in F$. Thus
$(a\vee b)^{\prime}\in F$ which is a contradiction.

(3)$\Rightarrow$(1). Let $F$ be a prime filter. We prove that $F$ is an
ultrafilter. We shall use Lemma 1.2. Assume that $a\in B$ is such that
$a\wedge b\neq 0$ for all $b\in F$. We shall prove that $a\in A$. Suppose for
the sake of argument, that $a\notin F$. Now, $1=a\vee a^{\prime}\in F$ and so
either $a\in F$ or $a^{\prime}\in F$. It follows that $a^{\prime}\in F$ but
$a\wedge a^{\prime}=0$ which contradicts our assumption about $a$. Thus $a\in
F$. We have therefore proved that $F$ is an ultrafilter. ∎

######  Remark 2.14.

In the light of the above result, the terms ‘prime filter’ and ‘ultrafilter’
are interchangeable in a Boolean algebra.

Ultrafilters are connected with homomorphisms to the two-element Boolean
algebra, $\mathbb{B}$. The following is the ultrafilter version of [12,
Proposition 2.2].

######  Lemma 2.15.

Let $B$ be a Boolean algebra.

  1. (1)

Let $\theta\colon B\rightarrow\mathbb{B}$ be a homomorphism of Boolean
algebras. Then $F=\theta^{-1}(1)$ is an ultrafilter.

  2. (2)

Let $F$ be an ultrafilter. Then the characteristic function $\chi_{F}\colon
B\rightarrow\mathbb{B}$ is a homomorphism of Boolean algebras.

###### Proof.

These are both straightforward to prove using Lemma 2.13. ∎

The above lemma actually establishes a bijection between ultrafilters in $B$
and homomorphisms from $B$ to the two-element Boolean algebra $\mathbb{B}$. We
can connect atoms with special kinds of ultrafilters; this enables us to link
what we are doing in this section with what we did previously.

######  Lemma 2.16.

Let $B$ be a Boolean algebra. The principal filter $F=a^{\uparrow}$ is a prime
filter if and only if $a$ is an atom.

###### Proof.

Let $a$ be an atom. From Example 2.12, we know that $a^{\uparrow}$ is a
filter. We shall prove that it is a prime filter. Suppose that $b\vee c\in F$.
Then $a\leq b\vee c$. Thus $a=(a\wedge b)\vee(a\wedge c)$. It cannot happen
that both $a\wedge b=0$ and $a\wedge c=0$. Also $a\wedge b\leq a$ and $a\wedge
c\leq a$. But $a$ is an atom. If $a\wedge b=a$ then $a\leq b$ and $b\in F$; if
$a\wedge b=0$ then $a\wedge c=a$ implying that $a\leq c$ and so $c\in F$. This
proves that $a^{\uparrow}$ is a prime filter. We now prove the converse.
Suppose that $F$ is an ultrafilter. We prove that $a$ is an atom. Suppose not.
Then there is $0<b<a$. Then $b^{\uparrow}$ is a filter and $F\subset
b^{\uparrow}$. But this contradicts the assumption that $F$ is an ultrafilter.
It follows that $a$ must be an atom. ∎

The above lemma is only interesting in the light of the following result. The
routine proof uses Zorn’s Lemma222The fairy godmother of mathematics. or see
[15, Chapter 1, Proposition 2.16].

######  Lemma 2.17 (Boolean Prime Ideal Theorem).

A non-empty subset of a Boolean algebra is contained in an ultrafilter if and
only if it has the finite intersection property.

The first corollary below is the analogue of the result for finite Boolean
algebras that every non-zero element is above an atom.

######  Corollary 2.18.

Every non-zero element of a Boolean algebra is contained in an ultrafilter.

The second corollary says that there are enough ultrafilters to separate
points; this is the analogue of the result that says in a finite Boolean
algebra each element is a join of the atoms below it.

######  Corollary 2.19.

Let $a$ and $b$ be distinct non-zero elements of a Boolean algebra. Then there
is an ultrafilter that contains one of the elements and omits the other.

###### Proof.

Since $a\neq b$ then either $a\nleq b$ or $b\nleq a$. Suppose that $a\nleq b$.
Then $a\wedge b^{\prime}\neq 0$ by Lemma 2.8. Thus by Corollary 2.18 there is
an ultrafilter $F$ that contains $a\wedge b^{\prime}$. It follows that $a\in
F$ and $b\notin F$. ∎

Ultrafilters are the first step in generalizing the theory of finite Boolean
algebras to arbitrary Boolean algebras. The second is to introduce topological
spaces to repace powersets. A compact Hausdorff space which is $0$-dimensional
is called a Boolean space; for emphasis, these will also be referred to in
this paper as compact Boolean spaces.

######  Lemma 2.20.

The clopen subsets of a Boolean space form a Boolean algebra.

###### Proof.

Let $X$ be a Boolean space and denote by $\mathsf{B}(X)$ the set of all clopen
subsets of $X$. Observe that $\varnothing,X\in\mathsf{B}(X)$. If
$A,B\in\mathsf{B}(X)$ then $A\cap B,A\cup B\in\mathsf{B}(X)$. Finally, if
$A\in\mathsf{B}(X)$ then $\overline{A}\in\mathsf{B}(X)$. ∎

Let $B$ be a Boolean algebra. Define $\mathsf{X}(B)$ to be the set of
ultrafilters on $B$. If $a\in B$ denote by $\mathscr{U}_{a}$ the set of
ultrafilters containing $a$.

######  Remark 2.21.

By Lemma 2.16, the above notation is consistent with that introduced in
Section 2.2.

######  Lemma 2.22.

Let $B$ be a Boolean algebra.

  1. (1)

$\mathscr{U}_{0}=\varnothing$.

  2. (2)

$\mathscr{U}_{1}=\mathsf{X}(B)$.

  3. (3)

$\mathscr{U}_{a}\cap\mathscr{U}_{b}=\mathscr{U}_{a\wedge b}$.

  4. (4)

$\mathscr{U}_{a}\cup\mathscr{U}_{b}=\mathscr{U}_{a\vee b}$.

  5. (5)

$\mathscr{U}_{a^{\prime}}=\overline{\mathscr{U}_{a}}$.

###### Proof.

The proofs of (1) and (2) are straightforward. The proof of (3) follows fom
the fact that filters are closed under meets. The proof of (4) follows from
the fact that ultrafilters are prime filters. The proof of (5) follows from
the fact that $a\wedge a^{\prime}=0$ and the fact that ultrafilters are proper
filters, and part (2) of Lemma 2.13. ∎

The above lemma tells us that the collection of sets $\mathscr{U}_{a}$, where
$a\in B$, forms the base for a topology on $\mathsf{X}(B)$. We shall first of
all determine the salient properties of this topological space.

######  Lemma 2.23.

For each Boolean algebra $B$, the topological space $\mathsf{X}(B)$ is
Boolean.

###### Proof.

The base of the topology consists of sets of the form $\mathscr{U}_{a}$. These
are open by fiat. But by part (5) of Lemma 2.22 they are also closed. It
follows that $\mathsf{X}(B)$ is $0$-dimensional. We prove that this space is
Hausdorff. Let $A$ and $B$ be distinct ultrafilters. Then there exists $a\in
A\setminus B$, and so $a\notin B$. We now use Lemma 2.13 to deduce that
$a^{\prime}\in B$. It follows that $A\in\mathscr{U}_{a}$ and
$B\in\mathscr{U}_{a^{\prime}}$. By part (3) of Lemma 2.22, we have that
$\mathscr{U}_{a}\cap\mathscr{U}_{a^{\prime}}=\varnothing$. Thus, the space
$\mathsf{X}(B)$ is Hausdorff. Finally, we prove that the space $\mathsf{X}(B)$
is compact. Let $\mathscr{C}=\\{\mathscr{U}_{a}\colon a\in I\\}$ be a cover of
$\mathsf{X}(B)$. Suppose that no finite subset of $\mathscr{C}$ covers
$\mathsf{X}(B)$. Then for any $a_{1},\ldots,a_{n}\in I$ we have that
$\mathscr{U}_{a_{1}}\cup\ldots\cup\mathscr{U}_{a_{n}}\neq\mathsf{X}(B)$. It
follows that $a_{1}\vee\ldots\vee a_{n}\neq 1$ and so
$a_{1}^{\prime}\wedge\ldots\wedge a_{n}^{\prime}\neq 0$. Thus the set
$I^{\prime}=\\{a^{\prime}\colon a\in I\\}$ has the finite intersection
property. By Lemma 2.17, there is an ultrafilter $F$ such that
$I^{\prime}\subseteq F$. By assumption, $F\in\mathscr{U}_{a}$ for some $a\in
I$ and so $a,a^{\prime}\in F$, which is a contradiction. ∎

The topological space $\mathsf{X}(B)$ is called the Stone space of the Boolean
algebra $B$.

We can now assemble Lemma 2.20 and Lemma 2.23 into the first main result.

######  Proposition 2.24.

  1. (1)

Let $B$ be a Boolean algebra. Then $B\cong\mathsf{B}(\mathsf{X}(B))$, where
here $\cong$ means an isomorphism of Boolean algebras.

  2. (2)

Let $X$ be a Boolean space. Then $X\cong\mathsf{X}(\mathsf{B}(X))$, where here
$\cong$ means a homeomorphism of topological spaces.

###### Proof.

(1) Define $\alpha\colon B\rightarrow\mathsf{B}\mathsf{X}(B)$ by
$a\mapsto\mathscr{U}_{a}$. By Lemma 2.22 this is a homomorphism of Boolean
algebras. It is injective by Corollary 2.19. We prove that it is surjective.
An element of $\mathsf{B}\mathsf{X}(B)$ is a clopen subset of $\mathsf{X}(B)$.
Since it is open, it is a union of open sets of the form $\mathscr{U}_{a}$,
but closed subsets of compact spaces are compact by part (3) of Lemma 1.3. It
follows that it is a union of a finite number of sets of the form
$\mathscr{U}_{a}$ and so must itself be of that form.

(2) Let $x\in X$. Define $O_{x}$ to be the set of all clopen sets that contain
$x$. It is easy to check that this is a prime filter in $\mathsf{B}(X)$ and so
$O_{x}\in\mathsf{X}(\mathsf{B}(X))$. Define $\beta\colon
X\rightarrow\mathsf{X}(\mathsf{B}(X))$ by $x\mapsto O_{x}$. Since both domain
and codomain spaces are compact and Hausdorff, to prove that $\beta$ is a
homeomorphism it is enough to prove that it is bijective and continuous.
Suppose that $O_{x}=O_{y}$. If $x\neq y$ then by the fact that $X$ is
Hausdorff we can find disjoint open sets $U$ and $V$ such that $x\in U$ and
$y\in V$. But $X$ is $0$-dimensional and so we can assume, without loss of
generality, that $U$ and $V$ are clopen from which we deduce that $O_{x}\neq
O_{y}$. It follows that $\beta$ is injective. Next, let $F$ be any ultrafilter
in $\mathsf{B}(X)$. Then this is an ultrafilter consisting of clopen subsets
of a compact space. Since $F$ is a filter, it has the finite intersection
property. By part (4) of Lemma 1.3, it follows that there is an element $x$ in
the intersection of all the elements of $F$. Thus $F\subseteq O_{x}$. But $F$
is an ultrafilter and so $F=O_{x}$. We have therefore proved our function is a
bijection. Finally, we prove continuity. Let $U$ be an open subset of
$\mathsf{X}(\mathsf{B}(X))$. Then $U$ is a union of the basic open sets which
are clopen. These have the form $\mathscr{U}_{A}$ where $A$ is a clopen subset
of $X$. Thus it is enough to calculate $\beta^{-1}(\mathscr{U}_{A})$. But
$O_{x}\in\mathscr{U}_{A}$ if and only if $x\in A$. Thus
$\beta^{-1}(\mathscr{U}_{A})=A$. ∎

We can extend the above result to maps to obtain the following:

######  Theorem 2.25 (Classical Stone duality I).

The category of Boolean algebras and their homomorphisms is dually equivalent
to the category of Boolean spaces and continuous functions between them.

###### Proof.

In Lemma 2.15, we proved that there is a bijective map between the
ultrafilters in $B$ and the Boolean algebra homomorphisms from $B$ to
$\mathbb{B}$, the $2$-element Boolean algebra. This bijection associates with
the ultrafilter $F$ its characteristic function $\chi_{F}$. Let $\theta\colon
B_{1}\rightarrow B_{2}$ be a homomorphism between Boolean algebras. Let
$F\in\mathsf{X}(B_{2})$ be an ultrafilter. Then $\chi_{F}\theta$ is the
characteristic function of an ultrafilter in $B_{1}$. In this way, we can map
homomomorphisms $B_{1}\rightarrow B_{2}$ to continuous functions
$\mathsf{X}(B_{1})\leftarrow\mathsf{X}(B_{2})$ with a consequent reversal of
arrows. In the other direction, let $\phi\colon X_{1}\rightarrow X_{2}$ be a
continuous function. Then $\phi^{-1}$ maps clopen sets to clopen sets. In this
way, we can map continuous functions $X_{1}\rightarrow X_{2}$ to homomorphisms
$\mathsf{B}(X_{1})\leftarrow\mathsf{B}(X_{2})$. The result now follows from
Proposition 2.24. ∎

######  Example 2.26.

Here are some examples of classical Stone duality.

  1. (1)

Let $B$ be a finite Boolean algebra. By Lemma 2.16, the ultrafilters of $B$
are in bijective correspondence with the atoms of $B$. We may therefore
identify the elements of $\mathsf{X}(B)$ with the set of atoms of $B$. Let
$a\in B$. We describe the set $\mathscr{U}_{a}$ in terms of atoms. The
ultrafilter $b^{\uparrow}\in\mathscr{U}_{a}$ if and only is $b\leq a$. So, the
set $\mathscr{U}_{a}$ is in bijective correspondence with the set of atoms
below $a$. It follows that the Boolean space $\mathsf{X}(B)$ is homeomorphic
with the the discrete space of atoms of $B$. Let $\theta\colon B\rightarrow C$
be a homomorphism of finite Boolean algebras. If $c\in C$ is an atom then
$c^{\uparrow}$ is an ultrafilter in $C$ and so $\theta^{-1}(c^{\uparrow})$ is
an ultrafilter in $B$. It follows that
$\theta^{-1}(c^{\uparrow})=b^{\uparrow}$, where $b$ is an atom in $B$. Thus
$x\geq b$ if and only if $\theta(x)\geq c$. We therefore have that
$\theta(b)\geq c$. But $b$ is the only atom of $B$ which has this property. In
this way, the classical theory of finite Boolean algebras can be derived from
Stone duality; that is, Theorem 2.11 is a special case of Theorem 2.25.

  2. (2)

Tarski proved that any two atomless, countably infinite Boolean algebras are
isomorphic [9, Chapter 16, Theorem 10]. It makes sense, therefore, to define
the Tarski algebra333Not an established term. to be an atomless, countably
infinite Boolean algebra. We describe the Stone space of the Tarski algebra.
An element $x$ of a topological space is said to be isolated if $\\{x\\}$ is
open. Suppose that $a$ is an atom of the Boolean algebra $B$. By definition
$\mathscr{U}_{a}$ is the set of all ultrafilters that contain $a$. But
$a^{\uparrow}$ is an ultrafilter containing $a$ by Lemma 2.16 and, evidently,
the only one. Thus $\mathscr{U}_{a}$ is an open set containing one point and
so the point $a^{\uparrow}$ is isolated. Suppose that $\mathscr{U}_{a}$
contains exactly one point $F$. Then $F$ is the only ultrafilter containing
$a$. Suppose that $a$ were not an atom. Then we could find $0\neq b<a$. Thus
$a=b\vee(a\wedge b^{\prime})$. By Corollary 2.18, there is an ultrafiler
$F_{1}$ containing $b$, and there is an ultrafilter $F_{2}$ containing
$a\wedge b^{\prime}$. Then $F_{1}\neq F_{2}$ but both contain $a$. This is a
contradiction. It follows that $a$ is an atom. We deduce that the atoms of the
Boolean algebra determine the isolated points of the associated Stone space.
It follows that a Boolean algebra has the property that every element is above
an atom (that is, it is atomic) if and only if the isolated points in its
Stone space form a dense subset. We deduce that the Stone space associated
with an atomless Boolean algebra has no isolated points. If $B$ is countable
then its Stone space is second-countable since $B$ is isomorphic to the set of
all clopen subsets of the Stone space of $B$. The Stone space of the Tarski
algebra is therefore a $0$-dimensional, second countable, compact, Hausdorff
space with no isolated points. Observe by [58, Theorem 9.5.10] that such a
space is metrizable. It follows by Brouwer’s theorem, [60, Theorem 30.3], that
the Stone space of the Tarski algebra is the Cantor space.

  3. (3)

The Cantor space described in Example 2 above often appears in disguise. Let
$A$ be any finite set with at least two elements. Denote by $X=A^{\omega}$ the
set of all right-infinite strings of elements over $A$. We can regard this set
as the product space $A^{\mathbb{N}}$ which is compact since $A$ is finite by
part (1) of Lemma 1.3. For each finite string $x\in A^{\ast}$ denote by $xX$
the subset of $X$ that consists of all elements of $X$ that begin with the
finite string $x$. This is an open set of $X$. If $a\in A$ denote by
$\hat{a}=A\setminus\\{a\\}$. Let $x=x_{1}\ldots x_{n}$ have length $n\geq 1$.
Then $\overline{xX}=\hat{x}_{1}X\cup x_{1}\hat{x}_{2}X\cup\ldots x_{1}\ldots
x_{n-1}\hat{x}_{n}X$. It follows that if $xX$ is open then $\overline{xX}$ is
open. Thus the sets $xX$ are clopen. The set $A^{\ast}$ is countably infinite
and so the number of clopen subsets is countably infinite. If $x,y\in
A^{\ast}$ then there are a few possibilities. If neither $x$ nor $y$ is the
prefix of the other then $xX\cap yX=\varnothing$. Now, suppose that $x=yu$.
Then $xX=yuX\subseteq yX$ from which it follows that $xX\cap yX=xX$. An open
subset $U$ of $X$ has the form $U=X_{1}\ldots X_{n}X$, where the $X_{i}$ are
subsets of $A$. This is a union of sets of the form $xX$ where $x\in
X_{1}\ldots X_{n}$. It follows that the sets $\varnothing$ and $xX$, where
$x\in A^{\ast}$, form a clopen base for the topology on $X$. If $w$ and
$w^{\prime}$ are distinct elements of $X$ then they will differ in the $n$th
position and so belong to disjoint sets of the form $xX$. It follows that $X$
is a second-countable Boolean space. This space cannot have any isolated
points: if $\\{w\\}$ is an open subset then it must be a union of sets of the
form $xX$ but this is impossible, It follows that $X$ is the Cantor space. We
refer the reader to [22, 23] and [32, Section 5] for more on this topological
space.

  4. (4)

We construct the Stone spaces of the powerset Boolean algebras
$\mathsf{P}(X)$. The isolated points of the Stone space of $\mathsf{P}(X)$
form a dense subset of the Stone space which is homeomorphic to the discrete
space $X$. Thus the Stone space of $\mathsf{P}(X)$ is a compact Hausdorff
space that contains a copy of the discrete space $X$. In fact, the Stone-Čech
compactification of the discrete space $X$ is precisely the Stone space of
$\mathsf{P}(X)$. See [53, Section 30, Theorem A] and [53, Section 75].

  5. (5)

Let $A$ be any finite alphabet. We shall use notation from the theory of
regular expressions so that $L+M$ means $L\cup M$ and $x$ can mean $\\{x\\}$
(but also the string $x$ in a different context). A language $L$ over $A$ is
said to be definite444Strictly speaking, ‘reverse definite’. if
$L=X+YA^{\ast}$ where $X,Y\subseteq A^{\ast}$ are finite languages. Denote the
set of definite languages by $\mathscr{D}$. This forms a Boolean algebra. The
Stone space of the Boolean algebra $\mathscr{D}$ can be described as follows.
Put $\mathscr{X}=A^{\ast}+A^{\omega}$. If $x$ and $y$ are distinct elements of
$\mathscr{X}$, define $x\wedge y$ to be the largest common prefix of $x$ and
$y$. Define $d(x,y)=0$ if $x=y$ else $d(x,y)=\exp(-|x\wedge y|)$. Then $d$ is
an ultrametric on $\mathscr{X}$ and $\mathscr{X}$ is a complete metric space
with respect to this ultrametric. The open balls are of the form $\\{x\\}$ or
$xA^{\ast}+xA^{\omega}$ where $x\in A^{\ast}$ and form a basis for the
topology. Thus $\mathscr{X}$ is a Boolean space. It can be proved that the
Stone space of $\mathscr{D}$ is this ultrametric space $\mathscr{X}$. See [44]
for more on this example.

###  2.4. Generalized Boolean algebras

There is a generalization of classical Stone duality, Theorem 2.25, that
relates what are termed generalized Boolean algebras to locally compact
Boolean spaces. At the level of objects, this was described in [56, Theorem 4]
and at the level of homomorphisms in [4].

In elementary work [12], Boolean algebras are usually defined with a top
element and globally defined complements. However, this is too restrictive for
the applications we have in mind; it corresponds in topological language to
only looking at compact spaces even though many mathematically interesting
spaces are locally compact. In this section, we shall study what are termed
‘generalized Boolean algebras’ or, to adapt terminology current in ring
theory, non-unital Boolean algebras. Similarly, in this section, a
distributive lattice will always have a bottom but not necessarily a top. Let
$D$ be a distributive lattice equipped with a binary operation $\setminus$
such that for all $x,y\in D$ we have that $0=y\wedge(x\setminus y)$ and
$x=(x\wedge y)\vee(x\setminus y)$. We say that such a distributive lattice is
a generalized Boolean algebra. Let $B$ be a generalized Boolean algebra. A
subset $C\subseteq B$ is said to be a subalgebra if it contains the bottom
element of $B$, is closed under meets and joins and is closed under the
operation $\setminus$. Such a $C$ is a generalized Boolean algebra in its own
right.

If $b\leq a$ in a lattice then $[b,a]$ denotes the set of all elements $x$ of
the lattice such that $b\leq x\leq a$. We call the set $[b,a]$ an interval. If
$c\in[b,a]$ then a complement of $c$ is an element $d\in[b,a]$ such that
$c\wedge d=b$ and $c\vee d=a$. We say that $[b,a]$ is complemented if every
element has a complement. Let $D$ be a distributive lattice. We say it is
relatively complemented if for every pair $b\leq a$, the interval $[b,a]$ is
complemented.

######  Lemma 2.27.

Let $D$ be a distributive lattice. Then the following are equivalent:

  1. (1)

$D$ is a generalized Boolean algebra.

  2. (2)

Each non-zero principal order-ideal of $D$ is a unital Boolean algebra.

  3. (3)

$D$ is relatively complemented.

###### Proof.

(1)$\Rightarrow$(2). Let $a\in B$ be non-zero. Then $a^{\downarrow}$ is a
distribitive lattice with bottom element $0$ and top element $a$. Let $b\leq
a$. Then $b\wedge(a\setminus b)=0$ and $b\vee(a\setminus b)=a$. It follows
that within $a^{\downarrow}$ we should define $b^{\prime}=a\setminus b$. Thus
each non-zero principal order ideal is a unital Boolean algebra.

(2)$\Rightarrow$(1). Let $x,y\in D$ where $x\neq 0$. Then $x\wedge y\leq x$.
Define $x\setminus y=(x\wedge y)^{\prime}$ where $(x\wedge y)^{\prime}$ is the
complement of $x\wedge y$ in the Boolean algebra $x^{\downarrow}$. By
definition $(x\wedge y)\vee(x\setminus y)=x$ and $y\wedge(x\setminus
y)=y\wedge((x\setminus y)\wedge x=(x\wedge y)\wedge(x\setminus y)=0$. If $x=0$
then define $0=(0\setminus 0)$.

(1)$\Rightarrow$(3). Let $b\leq a$. Let $x\in[b,a]$. Put $y=(a\setminus x)\vee
b$. Then $x\wedge y=b$ and $x\vee y=a$. We have proved that in each interval,
every element has a complement.

(3)$\Rightarrow$(1). Immediate. ∎

In the light of the above result, we shall regard generalized Boolean algebras
as distributive lattices with zero in which each non-zero principal order-
ideal is a Boolean algebra.

######  Example 2.28.

Let $B$ be the set of all finite subsets of $\mathbb{N}$. Then $B$ is a
generalized Boolean algebra but not a (unital) Boolean algebra.

We may define ultrafilters and prime filters in generalized Boolean algebras
just as we defined them in unital Boolean algebras. Let $B$ be a generalized
Boolean algebra. Define $\mathsf{X}(B)$ to be the set of ultrafilters on $B$.
If $a\in B$ denote by $\mathscr{U}_{a}$ the set of ultrafilters containing
$a$.

######  Lemma 2.29.

Let $B$ be a generalized Boolean algebra and let $a$ be any non-zero element.
Then there is an order-isomorphism between the filters in the Boolean algebra
$a^{\downarrow}$ and the filters in $B$ that contain $a$. Under this order-
isomorphism, proper filters correspond to proper filters, and ultrafilters to
ultrafilters.

###### Proof.

Let $F$ be a filter of $B$ that contains $a$. Put $F_{\downarrow}=F\cap
a^{\downarrow}$. Then $F_{\downarrow}$ is non-empty and it is straightforward
to show that it is a filter. Observe that if $F_{1}\subseteq F_{2}$ are
filters of $B$ that contain $a$ then
$(F_{1})_{\downarrow}\subseteq(F_{2})_{\downarrow}$.

Let $G$ be a filter of $a^{\downarrow}$. The proof that $G^{\uparrow}$, taken
in $B$, is a filter of $B$ that contains $a$ is straightforward. Observe that
if $G_{1}\subseteq G_{2}$ are both filters of $a^{\downarrow}$ then
$G_{1}^{\uparrow}\subseteq G_{2}^{\uparrow}$.

It is now routine to check that $F=(F_{\downarrow})^{\uparrow}$ and
$G=(G^{\uparrow})_{\downarrow}$. We have therefore established our order-
isomorphism. Since $B$ and $a^{\downarrow}$ have the same bottom element it is
routine to check that proper filters in $a^{\downarrow}$ are mapped to proper
filters in $B$, and that ultrafilters in $a^{\downarrow}$ are mapped to
ultrafilters in $B$. ∎

Part (1) below was proved as [57, Theorem 3] and part (2) below was proved as
[30, Proposition 1.6] and uses Lemma 2.29.

######  Lemma 2.30.

  1. (1)

In a distributive lattice every ultrafilter is a prime filter.

  2. (2)

A distributive lattice is a generalized Boolean algebra if and only if every
prime filter is an ultrafilter.

######  Remark 2.31.

It follows that in a generalized Boolean algebra, prime filters and
ultrafilters are the same.

Let $X$ be a Hausdorff space. Then $X$ is locally compact if each point of $X$
is contained in the interior of a compact subset [60, Theorem 18.2].

######  Lemma 2.32.

Let $X$ be a Hausdorff space. Then the following are equivalent.

  1. (1)

$X$ is locally compact and $0$-dimensional.

  2. (2)

$X$ has a base of compact-open sets.

###### Proof.

(1)$\Rightarrow$(2). Let $U$ be a clopen set (since the space is
$0$-dimensional). Let $x\in U$. Since $X$ is locally compact, there exists a
compact set $V$ such that $x\in V^{\circ}$. Now $x\in U\cap V^{\circ}$ is open
and $X$ has a basis of clopen sets. In particular, we can find a clopen set
$W$ such that $x\in W\subseteq U\cap V^{\circ}\subseteq V$. By part (3) of
Lemma 1.3, $W$ is a closed subset of the compact set $V$ and so $W$ is
compact. It follows that $x\in W\subseteq U$ where $W$ is compact-open. Thus
$X$ has a base of compact-open sets.

(2)$\Rightarrow$(1). By part (2) of Lemma 1.3, every compact subset of a
Hausdorff space is closed. It follows that $X$ has a basis of clopen subset.
It is immediate that the space is locally compact. ∎

We define a locally compact Boolean space to be a $0$-dimensional, locally
compact Hausdorff space. Let $X$ be a locally compact Boolean space. Denote by
$\mathsf{B}(X)$ the set of all compact-open subsets of $X$. The proof of the
following is straightforward using Lemma 1.3.

######  Lemma 2.33.

Let $X$ be a locally compact Boolean space. Then under the usual operations of
union and intersection, the poset $\mathsf{B}(X)$ is a generalized Boolean
algebra.

The proof of the following lemma is routine, once you recall that ultrafilters
and prime filters are the same thing in generalized Boolean algebras.

######  Lemma 2.34.

Let $B$ be a generalized Boolean algebra.

  1. (1)

$\mathscr{U}_{0}=\varnothing$.

  2. (2)

$\mathscr{U}_{a}\cap\mathscr{U}_{b}=\mathscr{U}_{a\wedge b}$.

  3. (3)

$\mathscr{U}_{a}\cup\mathscr{U}_{b}=\mathscr{U}_{a\vee b}$.

The above lemma tells us that the collection of all sets of the form
$\mathscr{U}_{a}$, where $a\in B$, is the base for a topology on
$\mathsf{X}(B)$. We shall first of all determine the salient properties of the
topological space $\mathsf{X}(B)$.

######  Lemma 2.35.

For each generalized Boolean algebra $B$, the topological space
$\mathsf{X}(B)$ is a locally compact Boolean space.

###### Proof.

Let $A$ and $B$ be distinct ultrafilters. Let $a\in A\setminus B$; such an
element exists since we cannot have that $A$ is a proper subset of $B$ since
both are ultrafilters. By Lemma 1.2, there exists $b\in B$ such that $a\wedge
b=0$. Observe that $\mathscr{U}_{a}\cap\mathscr{U}_{b}=\varnothing$ and
$A\in\mathscr{U}_{a}$ and $B\in\mathscr{U}_{b}$. We have proved that
$\mathsf{X}(B)$ is Hausdorff. It only remains to prove that each of the sets
$\mathscr{U}_{a}$ is compact. Suppose that
$\mathscr{U}_{a}\subseteq\bigcup_{i\in I}\mathscr{U}_{b_{i}}$. Observe that
$\mathscr{U}_{a}=\bigcup_{i\in I}\mathscr{U}_{a\wedge b_{i}}$. So, without
loss of generality, we can assume that $b_{i}\leq a$. Thus we are given that
$\mathscr{U}_{a}=\bigcup_{i\in I}\mathscr{U}_{b_{i}}$ where $b_{i}\leq a$. By
assumption, $a^{\downarrow}$ is a unital Boolean algebra. The result now
follows by Lemma 2.29 and Lemma 2.23. ∎

Just as before, the topological space $\mathsf{X}(B)$ is called the Stone
space of the generalized Boolean algebra $B$. We can now assemble Lemma 2.33
and Lemma 2.35 into the following result:

######  Proposition 2.36.

  1. (1)

Let $B$ be a generalized Boolean algebra. Then
$B\cong\mathsf{B}(\mathsf{X}(B))$, where here $\cong$ means an isomorphism of
generalized Boolean algebras.

  2. (2)

Let $X$ be a locally compact Boolean space. Then
$X\cong\mathsf{X}(\mathsf{B}(X))$, where here $\cong$ means a homeomorphism of
topological spaces.

###### Proof.

(1) Define $\alpha\colon B\rightarrow\mathsf{B}(\mathsf{X}(B))$ by
$a\mapsto\mathscr{U}_{a}$. Let $a$ and $b$ be elements of $B$. Suppose that
$\mathscr{U}_{a}=\mathscr{U}_{b}$. Then both $a$ and $b$ are in the order
ideal $(a\vee b)^{\downarrow}$. It follows by Lemma 2.29 and Proposition 2.24
that $a=b$. By Lemma 2.34, the bottom element is mapped to the bottom element
and binary meets and binary joins are preserved. It remains to show that it is
surjective. Let $U$ be a compact-open set of $\mathsf{B}(\mathsf{X}(B))$.
Since it is open it is a union of basic open sets and since it is compact it
is a union of a finite number of basic open sets. But this implies that $U$ is
a basic open set and so $U=\mathscr{U}_{a}$ for some $a\in B$.

(2) For each $x\in X$ denote by $O_{x}$ the set of all compact-open sets that
contain $x$. It is easy to check that $O_{x}$ is a prime filter in
$\mathsf{B}(X)$ and so we have defined a map from $X$ to
$\mathsf{X}(\mathsf{B}(X))$. This map is injective because locally compact
Boolean spaces are Hausdorff. Let $F$ be an arbitrary ultrafilter in
$\mathsf{B}(X)$. This is therefore an ultrafilter whose elements are compact-
open. Let $V\in F$. We now use part (2) of Lemma 1.3: compact subsets of
Hausdorff spaces are closed. Since $F$ is an ultrafilter, all intersections of
elements of $F$ with $V$ are non-empty and the set of sets so formed has the
finite intersection property. It follows that there is a point $x$ that
belongs to them all. Thus $x$ belongs to every element of $F$. Thus
$F\subseteq O_{x}$ from which we get equality since we are dealing with
ultrafilters. We have therefore established that we have a bijection. In
particular, every ultrafilter in $\mathsf{B}(X)$ is of the form $O_{x}$. We
prove that this function and its inverse are continuous. Let $V$ be a compact-
open set in $X$. Then the image of $V$ under our map is the set
$S=\\{O_{x}\colon x\in V\\}$. But $V$ is an element of $\mathsf{B}(X)$. The
set $S$ is just $\mathscr{U}_{V}$. We now show that the inverse function is
continuous. An element of a base for the topology on $\mathsf{B}(X)$ is of the
form $\mathscr{U}_{a}$ where $a\in\mathsf{B}(X)$. Let $a=V$ a compact-open
subset of $X$. A typical element of $\mathscr{U}_{V}$ is $O_{x}$ where $x\in
V$. It follows that the inverse image of $\mathscr{U}_{V}$ is $V$. ∎

Let $\theta\colon B\rightarrow C$ be a homomorphism of generalized Boolean
algebras. We say it is proper if $C=\mbox{\rm im}(\theta)^{\downarrow}$; in
other words, each element of $C$ is below an element of the image. A
continuous map between topological spaces is said to be proper if the inverse
images of compact sets are compact.

######  Theorem 2.37 (Commutative Stone duality II).

The category of generalized Boolean algebras and proper homomorphisms is
dually equivalent to the category of locally compact Boolean spaces and proper
continuous homomorphisms.

###### Proof.

Let $\theta\colon B_{1}\rightarrow B_{2}$ be a proper homomorphism between
generalized Boolean algebras and let $F$ be an ultrafilter in $B_{2}$. Then
$\theta^{-1}(F)$ is non-empty because the homomorphism is proper and it is an
ultrafilter in $B_{1}$. We therefore have a map
$\theta^{-1}\colon\mathsf{X}(B_{2})\rightarrow\mathsf{X}(B_{1})$. Let
$\phi\colon X_{1}\rightarrow X_{2}$ be a proper continuous map between locally
compact Boolean spaces and let $U$ be a compact-open subset of $X_{2}$. Then
$\phi^{-1}(U)$ is also compact-open. We therefore have a map
$\phi^{-1}\colon\mathsf{B}(X_{2})\rightarrow\mathsf{B}(X_{1})$. It is now
routine to check using Proposition 2.36, that we have a duality between
categories. ∎

The above theorem generalizes Theorem 2.25 since homomorphisms between unital
Boolean algebras are automatically proper, and in a Hausdorff space compact
sets are closed and closed subsets of compact spaces are themselves compact by
Lemma 1.3 and so continuous maps between Boolean spaces are automatically
proper. We shall now generealize the above theorem in the sections that
follow.

Every locally compact Hausdoff space admits a one-point compactification [53,
Section 37] in which the resulting space is compact Hausdorff. If the original
space if $0$-dimensional so, too, is its one-point compactification; see [9,
Exercise 43.19] and [56, page 387].

######  Lemma 2.38.

The one-point compactification of a locally compact Boolean space is a compact
Boolean space.

We shall return to this lemma later in Section 9.

##  3\. Boolean inverse semigroups

Let $S$ be an inverse semigroup. We denote its semilattice of idempotents by
$\mathsf{E}(S)$. If $X\subseteq S$, define $\mathsf{E}(X)=X\cap\mathsf{E}(S)$.
If $a\in S$, write $\mathbf{d}(a)=a^{-1}a$ and $\mathbf{r}(a)=aa^{-1}$. We say
that $a,b\in S$ are compatible, written $a\sim b$, if $a^{-1}b$ and $ab^{-1}$
are both idempotents. A pair of elements being compatible is a necessary
condition for them to have an upper bound. A subset is said to be compatible
if every pair of elements in that subset is compatible. The following was
proved in [20, Lemma 1.4.11, Lemma 1.4.12].

######  Lemma 3.1.

  1. (1)

$s\sim t$ if and only if $s\wedge t$ exists and $\mathbf{d}(s\wedge
t)=\mathbf{d}(s)\wedge\mathbf{d}(t)$ and $\mathbf{r}(s\wedge
t)=\mathbf{r}(s)\wedge\mathbf{r}(t)$.

  2. (2)

If $a\sim b$ then

$a\wedge b=ab^{-1}b=bb^{-1}a=aa^{-1}b=ba^{-1}a.$

We now suppose our inverse semigroup contains a zero. If $e$ and $f$ are
idempotents then we say they are orthogonal, written $e\perp f$, if $ef=0$. If
$a$ and $b$ are elements of an inverse semigroup with zero we say that they
are orthogonal, written $a\perp b$, if $\mathbf{d}(a)\perp\mathbf{d}(b)$ and
$\mathbf{r}(a)\perp\mathbf{r}(b)$. If $a\perp b$ then $a\sim b$; in this case,
if $a\vee b$ exists we often write $a\oplus b$ and talk about orthogonal
joins. This terminology can be extended to any finite set.

An inverse semigroup with zero is said to be distributive if it has all binary
compatible joins and multiplication distributes overs such joins. The
semilattice of idempotents of a distributive inverse semigroup is a
distributive lattice.

An inverse semigroup is a meet-semigroup if it has all binary meets. Let $S$
be an arbitrary inverse semigroup. A function $\phi\colon
S\rightarrow\mathsf{E}(S)$ is called a fixed-point operator if for each $a\in
S$ the element $\phi(a)$ is the largest idempotent less than or equal to $a$.
The proofs of the following can be found in [35] or follows from the
definition.

######  Lemma 3.2.

Let $S$ be an inverse semigroup.

  1. (1)

$S$ is a meet-semigroup if and only if it has a fixed-point operator.

  2. (2)

If $S$ is a meet-semigroup, then we may define $\phi$ by
$\phi(a)=a\wedge\mathbf{d}(a)(=a\wedge\mathbf{r}(a)$).

  3. (3)

If $\phi$ is a fixed-point operator then $\phi(ae)=\phi(a)e$ and
$\phi(ea)=e\phi(a)$ for all $e\in\mathsf{E}(S)$.

  4. (4)

If $\phi$ is a fixed-point operator then $a\wedge b=\phi(ab^{-1})b$.

It is important to be able to manipulate meets and joins in a distributive
inverse semigroup. The following result tells us how. Part (1) was proved as
[20, Proposition 1.4.17], part (2) was proved as [20, Proposition 1.4.9], and
parts (3), (4) and (5) were proved as [24, Lemma 2.5].

######  Lemma 3.3.

In a distributive inverse semigroup, the following hold.

  1. (1)

In a distributive inverse monoid, if $a\vee b$ exists then

$\mathbf{d}(a\vee b)=\mathbf{d}(a)\vee\mathbf{d}(b)\mbox{ and
}\mathbf{r}(a\vee b)=\mathbf{r}(a)\vee\mathbf{r}(b).$

  2. (2)

If $a\wedge b$ exists then $ac\wedge bc$ exists and $(a\wedge b)c=ac\wedge
bc$, and dually.

  3. (3)

Suppose that $\bigvee_{i=1}^{m}a_{i}$ and
$c\wedge\left(\bigvee_{i=1}^{m}a_{i}\right)$ exist. Then all the meets
$c\wedge a_{i}$ exist as does the join $\bigvee_{i=1}^{m}c\wedge a_{i}$ and we
have that

$c\wedge\left(\bigvee_{i=1}^{m}a_{i}\right)=\bigvee_{i=1}^{m}c\wedge a_{i}.$

  4. (4)

Suppose that $a$ and $b=\bigvee_{j=1}^{n}b_{j}$ are such that all the meets
$a\wedge b_{j}$ exist. Then $a\wedge b$ exists and is equal to
$\bigvee_{j}a\wedge b_{j}$.

  5. (5)

Let $a=\bigvee_{i=1}^{m}a_{i}$ and $b=\bigvee_{j=1}^{n}b_{j}$ and suppose that
all meets $a_{i}\wedge b_{j}$ exist. Then $\bigvee_{i,j}a_{i}\wedge b_{j}$
exists as does $a\wedge b$ and we have that $a\wedge
b=\bigvee_{i,j}a_{i}\wedge b_{j}$.

A distributive inverse semigroup is said to be Boolean if its semilattice of
idempotents is a generalized Boolean algebra.

######  Examples 3.4.

Here are some examples of Boolean inverse semigroups.

  1. (1)

Let $X$ be an infinite set. Denote by $\mathcal{I}^{\rm fin}(X)$ the set of
all partial bijections of the set $X$ with finite domains. Then this is a
generalized Boolean inverse semigroup that is not a Boolean inverse monoid.

  2. (2)

Symmetric inverse monoids $\mathcal{I}(X)$ are Boolean inverse monoids. Its
Boolean algebra of idempotents is isomorphic to the power set Boolean algebra
$\mathsf{P}(X)$. If $X$ is finite with $n$ elements then we denote the
symmetric inverse monoid on an $n$-element set by $\mathcal{I}_{n}$.

  3. (3)

Groups with zero adjoined, denoted by $G^{0}$. These may look like chimeras
but they are honest-to-goodness Boolean inverse monoids whose Boolean algebra
of idempotents is isomorphic to the 2-element Boolean algebra.

  4. (4)

$R_{n}$ the set of all $n\times n$ rook matrices [52]. These are all $n\times
n$ matrices over the numbers $0$ and $1$ such that each row and each column
contains at most one non-zero entry. In fact, $R_{n}$ is isomorphic to
$\mathcal{I}_{n}$.

  5. (5)

$R_{n}(G^{0})$ the set of all $n\times n$ rook matrices over a group with zero
[37]. These are all $n\times n$ matrices over the group with zero $G^{0}$ in
which each row and each column contains at most one non-zero entry.

  6. (6)

Let $S$ be a Boolean inverse semigroup. Then the set $M_{\omega}(S)$ of all
$\omega\times\omega$ generalized rook matrices over $S$ is a Boolean inverse
semigroup. See [16].

######  Lemma 3.5.

Let $S$ be a Boolean inverse semigroup. Let $b\leq a$. Then there is a unique
element, denoted by $(a\setminus b)$, such that the following properties hold:
$(a\setminus b)\leq a$, $b\perp(a\setminus b)$, and $a=b\vee(a\setminus b)$.

###### Proof.

Observe that there is an order-isomorphism from $a^{\downarrow}$ to
$\mathbf{d}(a)^{\downarrow}$ under the map $x\mapsto\mathbf{d}(x)$. If $b\leq
a$ then define $(a\setminus b)$ to be the unique element below $a$ that
corresponds to $\mathbf{d}(a)\setminus\mathbf{d}(b)$ under the above order-
isomorphism. ∎

Let $S$ be a Boolean inverse semigroup. If $X\subseteq S$ is any non-empty
subset, then $X^{\vee}$ denotes the set of all binary joins of compatible
pairs of elements of $X$. Clearly, $X\subseteq X^{\vee}$. A subset $I$ of $S$
is said to be an additive ideal if $I$ is a semigroup ideal of $S$ (that is,
$SI\subseteq I$ and $IS\subseteq I$) and $I$ is closed under binary compatible
joins. If $\theta\colon S\rightarrow T$ is a morphism of Boolean inverse
semigroups then the set $K=\\{s\in S\colon\theta(s)=0\\}$ is called the kernel
of $\theta$ and is clearly an additive ideal of $S$. We say that $S$ is
$0$-simplifying if $S\neq\\{0\\}$ and the only additive ideals are $\\{0\\}$
and $S$ itself. Let $e$ and $f$ be any idempotents. We say that a non-empty
finite set $X=\\{x_{1},\ldots,x_{n}\\}$ is a pencil from $e$ to $f$ if
$e=\bigvee_{i=1}^{n}\mathbf{d}(x_{i})$ and $\mathbf{r}(x_{i})\leq f$. Suppose
that $e=0$. Then all of the $x_{i}=0$. It follows that there is always a
pencil from $0$ to any idempotent $f$. On the other hand if $f=0$ then
$\mathbf{r}(x_{i})=0$ and so $x_{i}=0$ and it follows that $e=0$. It is easy
to check that if $I$ is an additive ideal and $f\in I$, where $f$ is an
idempotent, and there is a pencil from $e$ to $f$, where $e$ is an idempotent,
then $e\in I$.

######  Lemma 3.6.

Let $S$ be a Boolean inverse semigroup not equal to zero. Then $S$ is
$0$-simplifying if and only if whenever $e$ and $f$ are non-zero idempotents
there is a pencil from $e$ to $f$.

###### Proof.

Suppose that $S$ is $0$-simplifying, and let $e$ and $f$ be nonzero
idempotents. Then $(SfS)^{\vee}$ is a nonzero additive ideal of $S$. By
assumption $S=(SfS)^{\vee}$. It follows that $e\in(SfS)^{\vee}$. It is now
routine to check that there is a pencil from $e$ to $f$. We now prove the
converse. Let $I$ be a nonzero additive ideal of $S$ and let $s\in S$ be
arbitrary and non-zero. Let $f\in I$ be any non-zero idempotent. Then there is
a pencil from $s^{-1}s$ to $f$. It follows that $s^{-1}s\in I$. But $I$ is a
semigroup ideal and so $s\in I$. We have proved that $I=S$. ∎

######  Remark 3.7.

Let $X$ be a pencil from $e$ to $f$ in a Boolean inverse semigroup. We can
always assume, for any distinct $x,y\in X$, that
$\mathbf{d}(x)\perp\mathbf{d}(y)$.

Let $T$ be a Boolean inverse semigroup. An inverse subsemigroup $S$ of $T$ is
said to be a subalgebra if $S$ is closed under binary compatible joins taken
in $T$ and if $e,f\in\mathsf{E}(S)$ then $e\setminus f\in\mathsf{E}(S)$.
Observe that $S$ is a Boolean inverse semigroup for the induced operations;
observe that $\mathsf{E}(S)$ is a subalgebra of the generalized Boolean
algebra $\mathsf{E}(T)$.

######  Remark 3.8.

We should note that Wehrung [59, Definition 3.1.17] uses the term additive
inverse subsemigroup for what we have termed a subalgebra.

> Our perspective is that Boolean inverse semigroups are non-commutative
> generalizations of generalized Boolean algebras.

##  4\. Boolean groupoids

We shall assume that the reader is familiar with the basic ideas and
definitions of category theory as described in the first few chapters of Mac
Lane [41]. Our goal is just to present the perspective on categories needed in
this paper.

A category is usually regarded as a ‘category of structures’ of some kind,
such as the category of sets or the category of groups. A (small) category
can, however, also be regarded as an algebraic structure; that is, as a set
equipped with a partially defined binary operation satisfying certain axioms.
We shall need both perspectives in this paper, but the latter perspective will
be foremost. This algebraic approach to categories was an important ingredient
in Ehresmann’s work [5] and applied by Philip Higgins to prove some basic
results in group theory [10].

To define the algebraic notion of a category, we begin with a set $C$ equipped
with a partial binary operation which we denote by concatenation. We write
$\exists ab$ to mean that the product $ab$ is defined. An identity in such a
structure is an element $e$ such that if $\exists ae$ then $ae=a$ and if
$\exists ea$ then $ea=a$. A category is a set equipped with a partial binary
operation satisfying the following axioms:

(C1):

$\exists a(bc)$ if and only if $\exists(ab)c$ and when one is defined so is
the other and they are equal.

(C2):

$\exists abc$ if and only if $\exists ab$ and $\exists bc$.

(C3):

For each $a$ there is an identity $e$, perforce unique, such that $\exists
ae$, and there exists an identity $f$, perforce unique, such that $\exists
fa$. I shall write $\mathbf{d}(a)=e$ and $\mathbf{r}(a)=f$ and draw the
picture

$f\stackrel{{\scriptstyle a}}{{\longleftarrow}}e.$

The set of all elements from $e$ to $f$ is called a hom-set and denoted
$\mbox{hom}(e,f)$.

You can check that $\exists ab$ if and only if $\mathbf{d}(a)=\mathbf{r}(b)$.

######  Example 4.1.

A category with one identity is a monoid. Thus, viewed in this light,
categories are ‘monoids with many identities’.

The morphisms of categories are called functors; we shall think of functors as
generalizations of monoid homomorphisms.

Let $C$ be a category. If $A,B\subseteq C$ then we may define $AB$ to be that
subset of $C$ which consists of all products $ab$ where $a\in A$, $b\in B$ and
$ab$ is defined in the category. We call this subset multiplication.

We now define groupoids. An element $a$ of a category is said to be invertible
if there exists an element $b$ such that $ab$ and $ba$ are identities. If such
an element $b$ exists it is unique and is called the inverse of $a$; we denote
the inverse of $a$ when it exists by $a^{-1}$. A category in which every
element is invertible is called a groupoid.

######  Example 4.2.

  1. (1)

A groupoid with one identity is a group. Thus groupoids are ‘groups with many
identities’.

  2. (2)

A set can be regarded as a groupoid in which every element is an identity.

  3. (3)

Equivalence relations can be regarded as groupoids. They correspond to
principal groupoids; that is, those groupoids in which given any identities
$e$ and $f$ there is at most one element $g$ of the groupoid such that
$f\stackrel{{\scriptstyle g}}{{\longleftarrow}}e$. A special case of such
groupoids are the pair groupoids, $X\times X$, which correspond to equivalence
relations having exactly one equivalence class.

If $G$ is a groupoid and $A\subseteq G$ then $A^{-1}$ is the set of all
inverses of elements of $A$.

We shall need the following notation for the maps involved in defining a
groupoid (not entirely standard). Define $\mathbf{d}(g)=g^{-1}g$ and
$\mathbf{r}(g)=gg^{-1}$. If $g\in G$ define $\mathbf{i}(g)=g^{-1}$. Put

$G\ast G=\\{(g,h)\in G\times G\colon\mathbf{d}(g)=\mathbf{r}(h)\\},$

and define $\mathbf{m}\colon G\ast G\rightarrow G$ by $(g,h)\mapsto gh$. If
$U,V\subseteq G$, define $U\ast V=(U\times V)\cap(G\ast G)$. The set of
identities of $G$ is denoted by $G_{o}$. If $e$ is an identity in $G$ then
$G_{e}$ is the set of all elements $a$ such that $a^{-1}a=e=aa^{-1}$. We call
this the local group at $e$. Put $\mbox{Iso}(G)=\bigcup_{e\in G_{o}}G_{e}$.
This is called the isotropy groupoid of $G$.

We now show how to construct all groupoids. Let $G$ be a groupoid. We say that
elements $g,h\in G$ are connected, denoted $g\equiv h$, if there is an element
$x\in G$ such that $\mathbf{d}(x)=\mathbf{d}(h)$ and
$\mathbf{r}(x)=\mathbf{d}(g)$. The $\equiv$-equivalence classes are called the
connected components of the groupoid. If $\exists gh$ then necessarily
$g\equiv h$. It follows that $G=\coprod_{i\in I}G_{i}$ where the $G_{i}$ are
the connected components of $G$. Each $G_{i}$ is a connected groupoid. So, it
remains to describe the structure of all connected groupoids. Let $X$ be a
non-empty set and let $H$ be a group. The set of triples $X\times H\times X$
becomes a groupoid when we define
$(x,h,x^{\prime})(x^{\prime},h^{\prime},x^{\prime\prime})=(x,hh^{\prime},x^{\prime\prime})$
and $(x,h,y)^{-1}=(y,h^{-1},x)$. It is easy to check that $X\times H\times X$
is a connected groupoid. Now let $G$ be an arbitrary groupoid. Choose, and
fix, an identity $e$ in $G$. Denote the local group at $e$ by $H$. For each
identity $f$ in $G$ choose an element $x_{f}$ such that $\mathbf{d}(x_{f})=e$
and $\mathbf{r}(x_{f})=f$. Put $X=\\{x_{f}\colon f\in G_{o}\\}$. We prove that
$G$ is isomorphic to $X\times H\times X$. Let $g\in G$. Then
$x_{\mathbf{r}(g)}^{-1}gx_{\mathbf{d}(g)}\in H$. Define a map from $G$ to
$X\times H\times X$ by
$g\mapsto(x_{\mathbf{r}(g)},x_{\mathbf{r}(g)}^{-1}gx_{\mathbf{d}(g)},x_{\mathbf{d}(g)})$.
It is easy to show that this is a bijective functor.

We shall need some special kinds of functors in our duality theory called
covering functors. These we define now. Let $G$ be any groupoid and $e$ any
identity. The star of $e$, denoted by $\mbox{St}_{e}$, consists of all
elements $g\in G$ such that $\mathbf{d}(g)=e$. Let $\theta\colon G\rightarrow
H$ be a functor between groupoids. Then for each identity $e\in G$, the
functor $\theta$ induces a function $\theta_{e}$ mapping $\mbox{St}_{e}$ to
$\mbox{St}_{\theta(e)}$. If all these functions are injective (respectively,
surjective) then we say that $\theta$ is star-injective (respectively, star-
surjective). A covering functor is a functor which is star-bijective. The
following was proved as [24, Lemma 2.26].

######  Lemma 4.3.

Let $\theta\colon G\rightarrow H$ be a covering functor between groupoids.
Suppose that the product $ab$ is defined in $H$ and that $\theta(x)=ab$. Then
there the exist $u,v\in G$ such that $x=uv$ and $\theta(u)=a$ and
$\theta(v)=b$.

The key definition needed to relate groupoids and inverse semigroups in our
non-commutative generalization of Stone duality is the following. A subset
$A\subseteq G$ is called a local bisection if $A^{-1}A,AA^{-1}\subseteq
G_{o}$.

######  Lemma 4.4.

A subset $A\subseteq G$ is a local bisection if and only if $a,b\in A$ and
$\mathbf{d}(a)=\mathbf{d}(b)$ implies that $a=b$, and
$\mathbf{r}(a)=\mathbf{r}(b)$ implies that $a=b$.

###### Proof.

Suppose that $A$ is a local bisection. Let $a,b\in A$ such that
$\mathbf{d}(a)=\mathbf{d}(b)$. Then the product $ab^{-1}$ exists and, by
assumption, is an identity. It follows that $a=b$. A similar argument shows
that if $a,b\in A$ are such that $\mathbf{r}(a)=\mathbf{r}(b)$ then $a=b$. We
now prove the converse. We prove that $A^{-1}A\subseteq G_{o}$. Let $a,b\in A$
and suppose that $a^{-1}b$ is exists. Then $\mathbf{r}(a)=\mathbf{r}(b)$. By
assumption, $a=b$ and so $a^{-1}b$ is an identity. The proof that
$AA^{-1}\subseteq G_{o}$ is similar. ∎

What follows is based on [47, 54].

Just as we can study topological groups, so we can study topological
groupoids. A topological groupoid is a groupoid $G$ equipped with a topology,
and $G_{o}$ is equipped with the subspace topology, such that the maps
$\mathbf{d},\mathbf{r},\mathbf{m},\mathbf{i}$ are all continuous functions
where $\mathbf{d},\mathbf{r}\colon G\rightarrow G$ and $\mathbf{m}\colon G\ast
G\rightarrow G$. Clearly, it is just enough to require that $\mathbf{m}$ and
$\mathbf{i}$ are continuous.

A topological groupoid is said to be open if the map $\mathbf{d}$ is an open
map; it is said to be étale if the map $\mathbf{d}$ is a local homeomorphism.

######  Remark 4.5.

It is worth observing (see [54, page 12]) that the definition of étale is
based on the function $\mathbf{d}\colon G\rightarrow G$.

In this paper, we shall focus on étale topological groupoids. The obvious
question is why should étale groupoids be regarded as a nice class of
topological groupoids? The following result due to Pedro Resende provides us
with one reason. For the following see [47, Exercises I.1.8].

######  Lemma 4.6.

Let $G$ be a topological groupoid. Then $G$ is étale if and only if
$\Omega(G)$, the set of all open subsets of $G$, is a monoid under subset
multiplication with $G_{o}$ as the identity.

We may paraphrase the above theorem by saying that étale groupoids are those
topological groupoids that have an algebraic alter ego.

We say that an étale topological groupoid is Boolean if its space of
identities is a locally compact Boolean space.

> Our perspective is that Boolean groupoids are ‘non-commutative’
> generalizations of locally compact Boolean spaces.

Thinking of topological groupoids as non-commutative spaces in this way goes
back to [18, 46].

##  5\. From Boolean groupoids to Boolean inverse semigroups

The goal of this section is to show how to construct Boolean inverse
semigroups from Boolean groupoids.

######  Lemma 5.1.

The set of all local bisections of a discrete groupoid forms a Boolean inverse
monoid under subset multiplication.

###### Proof.

Let $A$ and $B$ be local bisections. We prove that $AB$ is a local bisection.
We calculate $(AB)^{-1}AB$. This is equal to $B^{-1}A^{-1}AB$. Now $A^{-1}A$
is a set of identities. Thus $B^{-1}A^{-1}AB\subseteq B^{-1}B$. But $B^{-1}B$
is a set of identities. It follows that $(AB)^{-1}AB$ is a set of identities.
By a similar argument we deduce that $AB(AB)^{-1}$ is a set of identities. We
have therefore proved that the product of two local bisections is a local
bisection. The proof of associativity is straightforward. Since $G_{o}$ is a
local bisection, we have proved that the set of local bisections is a monoid.
Observe that if $A$ is a local bisection, then $A=AA^{-1}A$ and
$A=A^{-1}AA^{-1}$. Thus the semigroup is regular. Suppose that $A^{2}=A$,
where $A$ is a local bisection. Then $a=bc$ where $b,c\in A$. But
$\mathbf{d}(a)=\mathbf{d}(c)$, and so $a=c$, and
$\mathbf{r}(a)=\mathbf{r}(b)$, and so $a=b$. It follows that $a=a^{2}$. But
the only idempotents in groupoids are identities and so $a$ is an identity. We
have shown that if $A^{2}=A$ then $A\subseteq G_{o}$. It is clear that if
$A\subseteq G_{o}$ then $A^{2}=A$. We have therefore proved that the
idempotent local bisections are precisely the subsets of the set of
identities. The product of any two such idempotents is simply their
intersection and so idempotents commute with each other. It follows that our
monoid is inverse. It is easy to check that $A\leq B$ in this inverse
semigroup precisely when $A\subseteq B$. Now, the idempotents are the subsets
of the set of identities and the natural partial order is subset inclusion. It
follows that the set of identities is a Boolean algebra, since it is
isomorphic to the Boolean algebra of all subsets of $G_{o}$. Suppose that $A$
and $B$ are local bisections such that $A\sim B$. Then it is easy to check
that $A\cup B$ is a local bisection. Clearly, subset multiplication
distributes over such unions. We have therefore proved that the monoid is a
Boolean inverse monoid. ∎

A subset $A\subseteq G$ of a groupoid is called a bisection if

$A^{-1}A,AA^{-1}=G_{o}.$

The following is immediate by Proposition 5.1 and tells us that we may also
construct groups from groupoids.

######  Corollary 5.2.

The set of bisections of a discrete groupoid is just the group of units of the
inverse monoid of all local bisections of the discrete groupoid.

Definition.  Let $G$ be a Boolean groupoid. Denote by $\mathsf{KB}(G)$ the set
of all compact-open local bisections of $G$.

######  Proposition 5.3.

Let $G$ be a Boolean groupoid. Then $\mathsf{KB}(G)$ is a Boolean inverse
semigroup.

###### Proof.

Let $U$ and $V$ be two compact-open local bisections. Since the groupoid $G$
is étale, the product $UV$ is open by Lemma 4.6. The product of local
bisections is a local bisection by the proof of Lemma 5.1. It remains to show
that $UV$ is compact. Let $UV\subseteq\bigcup_{i\in I}A_{i}$ where the $A_{i}$
are open local bisections — since the open local bisections of an étale
groupoids form a base for the topology. Then $U^{-1}U\cap
VV^{-1}=U^{-1}UVV^{-1}\subseteq\bigcup_{i\in I}U^{-1}A_{i}V^{-1}$. The sets
$U^{-1}A_{i}V^{-1}$ are open local bisections. By assumption $U^{-1}U\cap
VV^{-1}$ is compact-open; here we use the fact that the identity space of a
Boolean groupoid is a locally compact Boolean space. Thus $U^{-1}U\cap
VV^{-1}=U^{-1}UVV^{-1}\subseteq\bigcup_{i=1}^{n}U^{-1}A_{i}V^{-1}$,
relabelling if necessary. It follows that
$UV\subseteq\bigcup_{i=1}^{n}UU^{-1}A_{i}V^{-1}V\subseteq\bigcup_{i=1}^{n}A_{i}$
and so $UV$ is compact. It follows that $\mathsf{KB}(G)$ is a semigroup. The
proof that it is a Boolean inverse semigroup now follows easily from what we
have done above and Lemma 5.1. ∎

##  6\. From Boolean inverse semigroups to Boolean groupoids

The goal of this section is to show how to construct Boolean groupoids from
Boolean inverse semigroups.

Let $S$ be an inverse semigroup. A non-empty subset $A\subseteq S$ is said to
be a (respectively, proper) filter (respectively, if $0\notin A$), if the set
$A$ is downwardly directed and upwardly closed. A maximal proper filter is
called an ultrafilter. Let $A$ be a proper filter in a distributive inverse
semigroup $S$. We say it is prime if $a\vee b\in A$ implies that $a\in A$ or
$b\in A$. The proof of the following is straightforward.

######  Lemma 6.1.

Let $S$ be a Boolean inverse semigroup. Then $A$ is an ultrafilter if and only
if $A^{-1}$ is an ultrafilter.

A subset $A$ of an inverse semigroup $S$ is said to be a coset if $a,b,c\in A$
implies that $ab^{-1}c\in A$. The following extends [24, Lemma 2.6].

######  Lemma 6.2.

Every filter is a coset.

###### Proof.

Let $A$ be a filter and let $a,b,c\in A$. Let $d\in A$ where $d\leq a,b,c$.
Then $d=dd^{-1}d\leq ab^{-1}c$. It follows that $ab^{-1}c\in A$. ∎

We define an idempotent filter to be a filter that contains an idempotent. The
following was proved for cosets as [19, Proposition 1.5] and so follows by
Lemma 6.2.

######  Lemma 6.3.

A filter is idempotent if and only if it is an inverse subsemigroup.

We now relate idempotent filters in $S$ to filters in $\mathsf{E}(S)$.

######  Lemma 6.4.

Let $S$ be an inverse semigroup. There is an order-isomorphism between the
idempotent filters in $S$ and the filters in $\mathsf{E}(S)$, in which proper
filters correspond to proper filters.

###### Proof.

Let $A$ be an idempotent filter in $S$. We prove first that $\mathsf{E}(A)$ is
a filter in $\mathsf{E}(S)$. Let $e,f\in\mathsf{E}(A)$. By assumption, $e,f\in
A$. Thus there is an element $i\in A$ such that $i\leq e,f$. But the set of
idempotents of an inverse semigroup is an order-ideal. It follows that $i$ is
an idempotent. But $i\leq e,f$ and so $i\leq ef$. However, $A$ is a filter and
so since $i\in A$ we have that $ef\in A$ and so $ef\in\mathsf{E}(A)$. Let
$e\in\mathsf{E}(A)$ and let $e\leq f$ where $f$ is an idempotent. Then $f\in
A$ and so $f\in\mathsf{E}(A)$. We have therefore proved that $\mathsf{E}(A)$
is a filter in $\mathsf{E}(S)$. It is clear that if $A\subseteq B$ then
$\mathsf{E}(A)\subseteq\mathsf{E}(B)$. Now, let $F$ be a filter in
$\mathsf{E}(S)$ is is easy to check that $F^{\uparrow}$ is an idempotent
filter in $S$. It is clear that if $F\subseteq G$ then $F^{\uparrow}\subseteq
G^{\uparrow}$. Let $A$ be an idempotent filter in $S$. We prove that
$A=\mathsf{E}(A)^{\uparrow}$. It is clear that
$\mathsf{E}(A)^{\uparrow}\subseteq A$. Let $a\in A$. By assumption $e\in A$
for some idempotent $e$. But $a,e\in A$. There exists $f\leq a,e$ which has to
be an idempotent since the set of idempotents of an inverse semigroup is an
order-ideal. It follows, in particular, that $f\leq a$, which proves the
claim. Let $F$ be a filter in $\mathsf{E}(S)$. It is now routine to check that
$F=\mathsf{E}(F^{\uparrow})$. We have therefore proved our order-isomorphism
and it is clear that proper filters map to proper filters. ∎

######  Lemma 6.5.

Let $A$ be an idempotent filter in a distributive inverse semigroup $S$.

  1. (1)

$A$ is a prime filter in $S$ if and only if $\mathsf{E}(A)$ is a prime filter
in $\mathsf{E}(S)$.

  2. (2)

$A$ is an ultrafilter in $S$ if and only if $\mathsf{E}(A)$ is an ultrafilter
in $\mathsf{E}(S)$.

###### Proof.

(1) Suppose first that $A$ is a prime filter in $S$. We saw in Lemma 6.4, that
$\mathsf{E}(A)$ is a proper filter in $\mathsf{E}(S)$. Suppose that $e\vee
f\in\mathsf{E}(A)$. Then $e\vee f\in A$. It follows that $e\in A$ or $f\in A$.
It is now immediate that $\mathsf{E}(A)$ is a prime filter in $\mathsf{E}(S)$.
We now prove the converse. Suppose that $\mathsf{E}(A)$ is a prime filter in
$\mathsf{E}(S)$. We prove that $A$ is a prime filter in $S$. Let $a\vee b\in
A$. Then $\mathbf{d}(a\vee b)=\mathbf{d}(a)\vee\mathbf{d}(b)\in\mathsf{E}(A)$.
It follows that $\mathbf{d}(a)\in\mathsf{E}(A)$ or
$\mathbf{d}(b)\in\mathsf{E}(A)$. Without loss of generality, we suppose the
former. So, $a\vee b\in A$ and $\mathbf{d}(a)\in A$. But $A$ is an inverse
subsemigroup and so $a=(a\vee b)\mathbf{d}(a)\in A$. This proves that $A$ is a
prime filter in $S$.

(2) Suppose that $A$ is an ultrafilter in $S$. We prove that $\mathsf{E}(A)$
is an ultrafilter in $\mathsf{E}(S)$. Suppose that $F$ is a proper filter in
$\mathsf{E}(S)$ such that $\mathsf{E}(A)\subseteq F$. We claim that
$F^{\uparrow}$ is a proper filter in $A$. Let $a,b\in F^{\uparrow}$. The
$e\leq a$ and $f\leq b$ where $e,f\in F$. But then $ef\leq a,b$ and $ef\in F$.
The set $F^{\uparrow}$ is clearly upwardly closed and evidently proper. But
$A\subseteq F^{\uparrow}$ and $A$ is a maximal proper filter. We deduce that
$A=F^{\uparrow}$. It is now immediate that $\mathsf{E}(A)=F$. We have
therefore proved that $\mathsf{E}(A)$ is an ultrafilter in $\mathsf{E}(S)$. We
now prove the converse. Suppose that $\mathsf{E}(A)$ is an ultrafilter in
$\mathsf{E}(S)$. We prove that $A$ is an ultrafilter in $S$. Suppose that
$A\subseteq B$ where $B$ is a proper filter in $S$. Since $A$ is an idempotent
filter so too is $B$. We therefore have that
$A=\mathsf{E}(A)^{\uparrow}\subseteq\mathsf{E}(B)^{\uparrow}=B$. But
$\mathsf{E}(A)\subseteq\mathsf{E}(B)$ and, by assumption, $\mathsf{E}(A)$ is
an ultrafilter in $\mathsf{E}(S)$. It follows that
$\mathsf{E}(A)=\mathsf{E}(B)$ from which we deduce that $A=B$. ∎

######  Lemma 6.6.

Let $A$ be a (respectively, proper) filter. Then $(A^{-1}A)^{\uparrow}$ is a
(respectively, proper) idempotent filter. Likewise, $(AA^{-1})^{\uparrow}$ is
a filter.

###### Proof.

We prove that $(A^{-1}A)^{\uparrow}$ is a filter. Let
$x,y\in(A^{-1}A)^{\uparrow}$. Then $a^{-1}b\leq x$ and $c^{-1}d\leq y$ where
$a,b,c,d\in A$. Let $z\leq a,b,c,d$ where $z\in A$. Then $z^{-1}z\leq x,y$. It
is clear that $(A^{-1}A)^{\uparrow}$ is upwardly closed. It is also clear that
if $A$ is a proper filter so too is $(A^{-1}A)^{\uparrow}$. The fact that
$(A^{-1}A)^{\uparrow}$ is an idempotent filter is immediate. ∎

Let $A$ be a (respectively, proper) filter. Then $A^{-1}$ is a (respectively,
proper) filter. If $A$ is a (proper) filter define
$\mathbf{d}(A)=(A^{-1}A)^{\uparrow}$ and $\mathbf{r}(A)=(AA^{-1})^{\uparrow}$.
By Lemma 6.6, these are both (proper) idempotent filters. In fact, we have the
following which is easy to prove using Lemma 6.3.

######  Lemma 6.7.

Let $S$ be a Boolean inverse semigroup. If $A$ is an idempotent ultrafilter
then $\mathbf{d}(A)=A$ and $\mathbf{r}(A)=A$.

We can now explain why we have used the term ‘coset’.

######  Lemma 6.8.

Let $A$ be a filter in an inverse semigroup. Then
$A=(a\mathbf{d}(A))^{\uparrow}$ and $A=(\mathbf{r}(A)a)^{\uparrow}$, where
$a\in A$.

###### Proof.

We prove that $A=(a\mathbf{d}(A))^{\uparrow}$. Let $x\in A$. Then, since $a\in
A$, there exists $b\in A$ such that $b\leq x,a$. Observe that $ab^{-1}b\leq
xx^{-1}x=x$ and $ab^{-1}b\in a\mathbf{d}(A)$. Thus
$A\subseteq(a\mathbf{d}(A))^{\uparrow}$. To prove the reverse inclusion let
$x\in(a\mathbf{d}(A))^{\uparrow}$. Then $ab^{-1}c\leq x$ where $a,b,c\in A$.
But we proved above that $A$ was a coset and so $ab^{-1}c\in A$. It follows
that $x\in A$. ∎

######  Lemma 6.9.

Let $A$ be a filter in a distributive inverse semigroup.

  1. (1)

$A$ is a prime filter if and only if $\mathbf{d}(A)$ is a prime filter, and
dually.

  2. (2)

$A$ is an ultrafilter if and only if $\mathbf{d}(A)$ is an ultrafilter, and
dually.

###### Proof.

(1) Suppose that $A$ is a prime filter. We prove that $\mathbf{d}(A)$ is a
prime filter. Let $x\vee y\in\mathbf{d}(A)$. Then $a^{-1}b\leq x\vee y$. It
follows that $aa^{-1}b\leq a(x\vee y)$. But $aa^{-1}b\in A$ since $A$ is a
coset. Thus $ax\vee ay\in A$. Without loss of generality, suppose that $ax\in
A$. Then $a^{-1}(ax)\in A^{-1}A$ and so $x\in\mathbf{d}(A)$. We have therefore
proved that $\mathbf{d}(A)$ is a prime filter. Suppose now that
$\mathbf{d}(A)$ is a prime filter. We prove that $A$ is a prime filter. Let
$x\vee y\in A$. Then $\mathbf{d}(x\vee y)\in\mathbf{d}(A)$. Without loss of
generality, suppose that $\mathbf{d}(x)\in\mathbf{d}(A)$. It follows that
$a^{-1}b\leq\mathbf{d}(x)$ where $a,b\in A$. Thus $(x\vee y)a^{-1}b\leq(x\vee
y)\mathbf{d}(x)=x$ where $(x\vee y)a^{-1}b\in A$ since $A$ is a coset. We have
proved that $x\in A$.

(2) Suppose that $A$ is an ultrafilter. We prove that $\mathbf{d}(A)$ is an
ultrafilter. Suppose that $\mathbf{d}(A)\subseteq B$ where $B$ is a proper
filter in $S$. Observe that $B$ is an idempotent filter. Let $a\in A$. Then
$A=(a\mathbf{d}(A))^{\uparrow}$. It follows that $A\subseteq(aB)^{\uparrow}$.
By assumption $A=(aB)^{\uparrow}$. Thus $\mathbf{d}(A)=B$, as required.
Suppose now that $\mathbf{d}(A)$ is an ultrafilter. We prove that $A$ is an
ultrafilter. Suppose that $A\subseteq B$ where $B$ is a proper filter. Then
$\mathbf{d}(A)\subseteq\mathbf{d}(B)$. By assumption
$\mathbf{d}(A)=\mathbf{d}(B)$ from which it follows that $A=B$. ∎

The following result is important. In particular, prime filters are easier to
work with than ultrafilters.

######  Lemma 6.10.

In a Boolean inverse semigroup, prime filters are the same as ultrafilters.

###### Proof.

The proper filter $A$ is prime if and only if $\mathbf{d}(A)$ is prime by
Lemma 6.9. By Lemma 6.5, the proper filter $\mathbf{d}(A)$ is prime if and
only if $\mathsf{E}(\mathbf{d}(A))$ is a prime filter in the generalized
Boolean algebra $\mathsf{E}(S)$. But in a generalized Boolean algebra, byLemma
2.30, prime filters are the same as ultrafilters. We can now work our way
backwards to etablish the claim. ∎

Definition. Let $A$ and $B$ be ultrafilters. Define $A\cdot B$ precisely when
$\mathbf{d}(A)=\mathbf{r}(B)$. In which case, $A\cdot B=(AB)^{\uparrow}$.
Observe that $\mathbf{d}(A^{-1})=\mathbf{r}(A)$. It follows that $A^{-1}\cdot
A=\mathbf{d}(A)$ and $A\cdot A^{-1}=\mathbf{r}(A)$.

######  Lemma 6.11.

Let $S$ be a Boolean inverse semigroup.

  1. (1)

If $A$ is an ultrafilter then both $\mathbf{d}(A)$ and $\mathbf{r}(A)$ are
ultrafilters.

  2. (2)

If $A$ and $B$ are ultrafilters of $S$ such that $A\cdot B$ is defined, then
$A\cdot B$ is an ultrafilter of $S$ such that $\mathbf{d}(A\cdot
B)=\mathbf{d}(B)$ and $\mathbf{r}(A\cdot B)=\mathbf{r}(A)$.

  3. (3)

If $A$ is an ultrafilter in $S$ then $A\cdot\mathbf{d}(A)$ is defined and
$A\cdot\mathbf{d}(A)=A$. Similarly, $\mathbf{r}(A)\cdot A=A$.

  4. (4)

$(A\cdot B)\cdot C=A\cdot(B\cdot C)$ when the product are defined.

###### Proof.

(1) This follows by Lemma 6.9.

(2) We prove first that if $A$ and $B$ are proper filters then $A\cdot B$ is a
proper filter. Let $x,y\in A\cdot B$. Then $ab\leq x$ and $cd\leq y$ where
$a\in A$, $b\in B$, $c\in A$ and $d\in B$. Let $u\leq a,c$ and $u\in A$, and
let $v\leq b,d$ and $v\in B$. Then $uv\leq ab\leq x$ and $uv\leq cd\leq y$.
Observe that $uv\in AB$ and so we have proved that $(AB)^{\uparrow}$ is
downwardly directed. It is clearly upwardly directed. Suppose that
$0\in(AB)^{\uparrow}$. Then $ab=0$ for some $a\in A$ and $b\in B$. But
$(A^{-1}A)^{\uparrow}=(BB^{-1})^{\uparrow}$. It follows that
$a^{-1}a\in(BB^{-1})^{\uparrow}$. Thus $bb^{-1}\leq a^{-1}a$ where $b\in B$.
It follows that $0=ab=a(aa^{-1})b\geq bb^{-1}b=b$. It follows that $b=0$ which
contradicts the assumption that $B$ is a proper filter. It is routine to prove
that $\mathbf{d}(A\cdot B)=\mathbf{d}(B)$. The result now follows by Lemma
6.9.

(3) This is straightforward on the basis of Lemma 6.2.

(4) This follows by (2) above and Lemma 6.8. ∎

Definition.  Let $S$ be a Boolean inverse semigroup. Denote by $\mathsf{G}(S)$
the set of prime filters on $S$.

Using the identification between prime filters and ultrafilters proved in
Lemma 6.10 together with Lemma 6.11, Lemma 6.1 and Lemma 6.7, we have proved
the following.

######  Lemma 6.12.

Let $S$ be a Boolean inverse semigroup. Then $\mathsf{G}(S)$ is a groupoid
with respect to the partially defined operation $\cdot$, where the identities
of this groupoid are precisely the idempotent ultrafilters.

Let $S$ be a Boolean inverse semigroup. Let $a\in S$. Denote by
$\mathscr{U}_{a}$ the set of all primefilters in $S$ that contain $a$. Observe
that if $a=0$ then $\mathscr{U}_{0}=\varnothing$.

######  Lemma 6.13.

Let $S$ be a Boolean inverse semigroup.

  1. (1)

Each non-zero element of $S$ is contained in a prime filter. Thus
$\mathscr{U}_{a}\neq\varnothing$ if and only if $a\neq 0$.

  2. (2)

$\left(\mathscr{U}_{a}\right)^{-1}=\mathscr{U}_{a^{-1}}$.

  3. (3)

Let $A\in\mathscr{U}_{a}\cap\mathscr{U}_{b}$. Then there exists $c\leq a,b$
such that $A\in\mathscr{U}_{c}\subseteq\mathscr{U}_{a}\cap\mathscr{U}_{b}$.

  4. (4)

If $a\sim b$ then $\mathscr{U}_{a\vee b}=\mathscr{U}_{a}\cup\mathscr{U}_{b}$.

###### Proof.

(1) Let $a\neq 0$. Then $\mathbf{d}(a)\neq 0$. Let $F$ be any prime filter in
$\mathsf{E}(S)$ that contains $\mathbf{d}(a)$. Then $F^{\uparrow}$ is an
idempotent prime filter in $S$. Thus $A=(aF^{\uparrow})^{\uparrow}$ is a prime
filter in $S$ that contains $a$.

(2) Straightforward.

(3) Let $A\in\mathscr{U}_{a}\cap\mathscr{U}_{b}$. It follows that $a,b\in A$.
But $A$ is a filter and so there is $c\in A$ such that $c\leq a,b$. It follows
that $A\in\mathscr{U}_{c}$ and that
$\mathscr{U}_{c}\subseteq\mathscr{U}_{a}\cap\mathscr{U}_{b}$.

(4) We suppose that $a\sim b$ and so $a\vee b$ exists. The inclusion
$\mathscr{U}_{a}\cup\mathscr{U}_{b}\subseteq\mathscr{U}_{a\vee b}$ is
immediate. The reverse inclusion follows from the fact that ultrafilters are
the same as prime filters in a Boolean inverse semigroup by Lemma 6.10 ∎

By Lemma 6.13, the collection of all sets $\mathscr{U}_{a}$, where $a\in S$,
forms a base for a topology on $\mathsf{G}(S)$. Denote by $\mathsf{F}(S)$ the
set of all idempotent prime filters. This can be topologized by giving it the
subspace topology. Thus the sets of the form
$\mathscr{U}_{a}\cap\mathsf{F}(S)$ form a base for the topology on
$\mathsf{F}(S)$. Observe that

$\mathscr{U}_{a}\cap\mathsf{F}(S)=\bigcup_{f\leq a,f^{2}=f}\mathscr{U}_{f}.$

It follows by the above observation and Lemma 6.3 that the collection of sets
$\mathscr{U}_{e}$, where $e$ is an idempotent, forms a base for the subspace
topology on $\mathsf{F}(S)$. By Lemma 6.4 and Lemma 6.10, we see that there is
a bijection between the set $\mathsf{F}(S)$ and the set of prime filters of
$\mathsf{E}(S)$. If $e\in\mathsf{E}(S)$ then we denote the set of ultrafilters
of $\mathsf{E}(S)$ that contain $e$ by $\mathscr{V}_{e}$. The bijection above
restricts to a bijection between $\mathscr{U}_{e}$ and $\mathscr{V}_{e}$ for
each idempotent $e$. We have therefore proved the following.

######  Lemma 6.14.

Let $S$ be a Boolean inverse semigroup. Then the topological space of
idempotent prime filters is homeomorphic to the Stone space of
$\mathsf{E}(S)$.

######  Lemma 6.15.

Let $A$ and $B$ be filters such that $A\cap B\neq\varnothing$ and
$\mathbf{d}(A)=\mathbf{d}(B)$ (respectively, $\mathbf{r}(A)=\mathbf{r}(B)$).
Then $A=B$.

###### Proof.

Let $a\in A\cap B$. Put $\mathbf{d}(A)=C$. Then $A=(aC)^{\uparrow}=B$. The
proof of the other case is similar. ∎

######  Proposition 6.16.

Let $S$ be a Boolean inverse semigroup. Then $\mathsf{G}(S)$ is a Boolean
groupoid.

###### Proof.

First we show that $\mathsf{G}(S)$ is a topological groupoid. By part (2) of
Lemma 6.13, the inversion map is continuous. We observe that

$\mathbf{m}^{-1}(\mathscr{U}_{s})=\left(\bigcup_{0\neq ab\leq
s}\mathscr{U}_{a}\times\mathscr{U}_{b}\right)\cap(\mathsf{G}(S)\ast\mathsf{G}(S))$

for all $s\in S$. The proof is straightforward and the same as Step 3 of the
proof of [24, Proposition 2.22] and shows that $\mathbf{m}$ is a continuous
function.

We show that $\mathsf{G}(S)$ is étale. It is enough to show that the map from
$\mathscr{U}_{a}$ to $\mathscr{U}_{\mathbf{d}(a)}$ given by
$A\mapsto\mathbf{d}(A)$ is a homeomorphism. The proof of this is the same as
the proof of Step 4 of the proof of [24, Proposition 2.22].

The fact that the identity space of $\mathsf{G}(S)$ is homeomorphic to the
Stone space of $\mathsf{E}(S)$ follows by Lemma 6.14. This tells us that our
étale topological groupoid is a Boolean groupoid. ∎

Definition. If $S$ is a Boolean inverse semigroup, then we refer to
$\mathsf{G}(S)$ as the Stone groupoid of $S$.

##  7\. Non-commutative Stone duality

In this section, we shall generalize Theorem 2.37 by replacing generalized
Boolean algebras by Boolean inverse semigroups, and locally compact Boolean
spaces by Boolean groupoids.

###  7.1. Properties of prime filters

Our first goal now is to prove that we have enough ultrafilters in a Boolean
inverse semigroup. We adapt to our setting the proofs to be found in [12,
Chapter I, Section 2]. Any proofs that are omitted can be found in [30]. Let
$S$ be a distibutive inverse semigroup. An order-ideal of $S$ closed under
binary joins is called an additive order-ideal. If $A$ is an order-ideal then
$A^{\vee}$ denotes the set of all binary joins of compatible pairs of elements
of $A$. If $A$ is an order-ideal then $A^{\vee}$ is an additive order-ideal
containing $A$. We say that an additive order-ideal $A$ is prime if
$a^{\downarrow}\cap b^{\downarrow}\subseteq A$ implies that $a\in A$ or $b\in
A$. The proof of the following is straightforward or can be found as [30,
Lemma 3.10].

######  Lemma 7.1.

Let $S$ be a distributive inverse semigroup. Then $A$ is a prime filter if and
only if $S\setminus A$ is a prime additive order-ideal.

The following can be proved using Zorn’s Lemma.

######  Lemma 7.2.

Let $S$ be a distributive inverse semigroup. Let $I$ be an additive order-
ideal of $S$ and let $F$ be a filter disjoint from $I$. Then there is an
additive order-ideal $J$ maximal with respect to two properties: $I\subseteq
J$ and $J\cap F=\varnothing$.

The proof of the following is straightforward.

######  Lemma 7.3.

Let $S$ be a distributive inverse semigroup. Let $I$ be an additive order-
ideal and let $a$ be an arbitrary element of $S$. Then $I\cup a^{\downarrow}$
is an order-ideal and

$\left(I\cup a^{\downarrow}\right)^{\vee}=\\{x\vee b\colon x\in I,b\leq
a,x\sim b\\}.$

######  Lemma 7.4.

Let $S$ be a distributive inverse semigroup. Let $F$ be a filter in $S$ and
let $J$ be an additive order-ideal maximal amongst all additive order-ideals
disjoint from $F$. Then $J$ is a prime additive order-deal.

The following lemma is crucial to our program.

######  Lemma 7.5.

Let $S$ be a Boolean inverse semigroup. Let $a,b\in S$ such that $b\nleq a$.
Then there is an ultrafilter that contains $b$ but omits $a$.

###### Proof.

By assumption, $b^{\uparrow}\cap a^{\downarrow}=\varnothing$. Now
$b^{\uparrow}$ is a filter and $a^{\downarrow}$ is an additive order-ideal. By
Lemma 7.2, there is an additive order-ideal $J$ maximal with respect to two
properties: $a^{\downarrow}\subseteq J$ and $J\cap b^{\uparrow}=\varnothing$.
By Lemma 7.4, we have that $J$ is prime. By Lemma 7.1, the complement of $J$
in $S$ is a prime filter containing $b$ but omitting $a$. ∎

######  Lemma 7.6.

Let $S$ be a Boolean inverse semigroup.

  1. (1)

$\mathscr{U}_{a}\subseteq\mathscr{U}_{b}$ if and only if $a\leq b$.

  2. (2)

$\mathscr{U}_{a}=\mathscr{U}_{b}$ if and only if $a=b$.

  3. (3)

$\mathscr{U}_{a}\cdot\mathscr{U}_{b}=\mathscr{U}_{ab}$.

  4. (4)

$\mathscr{U}_{a}$ contains only idempotent ultrafiters if and only if $a$ is
an idempotent.

  5. (5)

$\mathscr{U}_{a}$ is compact.

  6. (6)

$\mathscr{U}_{a}$ is an idempotent if and only if $a$ is an idempotent.

  7. (7)

$\mathscr{U}_{a}\sim\mathscr{U}_{b}$ if and only if $a\sim b$.

###### Proof.

(1) Suppose that $\mathscr{U}_{a}\subseteq\mathscr{U}_{b}$. If $a\nleq b$ then
by Lemma 7.5 there exists an ultrafilter that contains $a$ and omits $b$,
which contradicts our assumption. Thus $a\leq b$ as required.

(2) Immediate by (1) above.

(3) Let $A\in\mathscr{U}_{a}$ and $B\in\mathscr{U}_{b}$ such that $A\cdot B$
is defined. Then $A\cdot B\in\mathscr{U}_{ab}$. The proof of the reverse
inclusion is the same as the proof of part (4) of [24].

(4) Only one direction needs proving. Observe that
$\mathscr{U}_{a}\subseteq\mathscr{U}_{a^{2}}$ since any ultrafilter that
contains $a$ is an inverse subsemigroup and so must also contain $a^{2}$. We
therefore have that $a\leq a^{2}$ by (1) above from which it follows that
$a=a^{2}$.

(5) By Lemma 6.8, there is a bijection between the set $\mathscr{U}_{a}$ and
the set $\mathscr{U}_{\mathbf{d}(a)}$ given by $A\mapsto\mathbf{d}(A)$. A base
for the open sets of $\mathscr{U}_{a}$ is the collection $\mathscr{U}_{c}$
where $c\leq b$. By the above bijection, the set $\mathscr{U}_{c}$ is mapped
to the set $\mathscr{U}_{\mathbf{d}(c)}$. It follows that there is a
homeomorphism between $\mathscr{U}_{a}$ and the set
$\mathscr{U}_{\mathbf{d}(a)}$. But by Lemma 6.14, the space
$\mathscr{U}_{\mathbf{d}(a)}$ is homeomorphic with the space
$\mathscr{V}_{\mathbf{d}(a)}$ of the prime filters in $\mathsf{E}(S)$ which
contain $\mathbf{d}(a)$. But the sets $\mathscr{V}_{\mathbf{d}(a)}$ are
compact by the proof of Lemma 2.35. It follows that $\mathscr{U}_{a}$ is
compact.

(6) This follows by part (3) above and part (2).

(7) Only one direction needs proving. Suppose that
$\mathscr{U}_{a}\sim\mathscr{U}_{b}$. Then by part (2) of Lemma 6.13 and part
(3) above, both $\mathscr{U}_{a^{-1}b}$ and $\mathscr{U}_{ab^{-1}}$ are
idempotents. It follows by part (6) above, that $a\sim b$. ∎

###  7.2. Properties of compact-open local bisections

Let $G$ be a Boolean groupoid and let $g\in G$. Define $\mathscr{F}_{g}$ to be
the set of compact-open local bisections of $G$ that contain $g$.

######  Lemma 7.7.

Let $G$ be a Boolean groupoid.

  1. (1)

$\mathscr{F}_{g}$ is a prime filter in $\mathsf{KB}(G)$.

  2. (2)

Every prime filter in $\mathsf{KB}(G)$ is of the form $\mathscr{F}_{g}$ for
some $g\in G$.

  3. (3)

$\mathscr{F}_{g}\cdot\mathscr{F}_{g^{-1}}=\mathscr{F}_{\mathbf{r}(g)}$ and
$\mathscr{F}_{g^{-1}}\cdot\mathscr{F}_{g}=\mathscr{F}_{\mathbf{d}(g)}$.

  4. (4)

Suppose that $gh$ is defined in $G$. Then
$\mathbf{d}(\mathscr{F}_{g})=\mathbf{r}(\mathscr{F}_{h})$ and
$\mathscr{F}_{g}\cdot\mathscr{F}_{h}=\mathscr{F}_{gh}$.

  5. (5)

If $\mathscr{F}_{g}=\mathscr{F}_{h}$. Then $g=h$.

###### Proof.

(1) Let $U,V\in\mathscr{F}_{g}$. Then $U\cap V$ is an open set containing $g$.
We now use the fact that the compact-open local bisections of $G$ form a base.
There is therefore a compact-open local bisection $W$ containing $g$ such that
$W\subseteq U\cap V$. It follows that $\mathscr{F}_{g}$ is downwardly
directed. It is clearly closed upwards and doesn’t contain the empty set. It
is clearly a prime filter.

(2) The proof is the same as the proof of [24, part (5) of Lemma 2.19].

(3) We shall prove
$\mathscr{F}_{g}\cdot\mathscr{F}_{g^{-1}}=\mathscr{F}_{\mathbf{r}(g)}$ since
the proof of the other case is similar. We use the fact that in a Boolean
groupoid the product of compact-open local bisections is a compact-open local
bisection by Proposition 5.3. Thus
$\mathscr{F}_{g}\cdot\mathscr{F}_{g^{-1}}\subseteq\mathscr{F}_{\mathbf{r}(g)}$.
But both left-hand side and right-hand side are prime filters in a Boolean
inverse semigroup. It follows that both are ultrafilters and so must be equal.

(4) The proof is similar to the proof of (3).

(5) Let $g,h\in U\in\mathscr{F}_{g}=\mathscr{F}_{h}$. We have that
$\mathbf{d}(\mathscr{F}_{g})=\mathbf{d}(\mathscr{F}_{h})$. By definition,
$\mathbf{d}(\mathscr{F}_{g})=\mathscr{F}_{g}^{-1}\cdot\mathscr{F}_{g}$ and
this is equal to $\mathscr{F}_{\mathbf{d}(g)}$ by (3) above. Thus
$\mathscr{F}_{\mathbf{d}(g)}=\mathscr{F}_{\mathbf{d}(h)}$. It follows that the
compact-open subsets of $G_{o}$ that contain $\mathbf{d}(g)$ are the same as
the compact-open subsets of $G_{o}$ that contain $\mathbf{d}(h)$. But the
groupoid $G$ is Boolean and so $G_{o}$ is a locally compact Boolean space. It
follows that $\mathbf{d}(g)=\mathbf{d}(h)$. But $g$ and $h$ certainly both
belong to the same compact-open local bisection and so $g=h$. ∎

###  7.3. Proof of the first part of the main theorem

Given a Boolean inverse semigroup $S$, then by Proposition 6.16 we have shown
how to construct a Boolean groupoid $\mathsf{G}(S)$, and given a Boolean
groupoid $G$, then by Propsition 5.3 we have shown how to construct a Boolean
inverse semigoup $\mathsf{KB}(G)$. The following result tells us what happens
when we iterate these two constructions.

######  Proposition 7.8.

  1. (1)

Let $S$ be a Boolean inverse semigroup. Define a function $\alpha\colon
S\rightarrow\mathsf{KB}(\mathsf{G}(S))$ by $\alpha(a)=\mathscr{U}_{a}$. Then
this is an isomorphism of semigroups.

  2. (2)

Let $G$ be a Boolean groupoid. Define a function $\beta\colon
G\rightarrow\mathsf{G}(\mathsf{KB}(G))$ by $\beta(g)=\mathscr{F}_{g}$. Then
$\beta$ is an isomorphism of groupioids and a homeomorphism.

###### Proof.

(1) By Lemma 6.15, the set $\mathscr{U}_{a}$ is a local bisection. It is open
by definition of the topology. It is compact by part (5) of Lemma 7.6. Thus
$\mathscr{U}_{a}$ is a compact-open local bisection. It follows that the
function is well-defined. It is a semigroup homomorphism by part (3) of Lemma
7.6, since $\mathscr{U}_{a}\mathscr{U}_{b}=\mathscr{U}_{ab}$. It is injective
by part (2) of Lemma 7.6. It remains to show that it is surjective. Let $U$ be
any compact-open local bisection of $\mathsf{G}(S)$. Since it is open it is a
union of sets of the form $\mathscr{U}_{a}$ and since it is compact it is a
union of a finite numbers of sets of this form. It follows that
$U=\bigcup_{i=1}^{n}\mathscr{U}_{a_{i}}$. But $\mathscr{U}_{a_{i}}\subseteq U$
which is the natural partial order in the inverse semigroup
$\mathsf{KB}(\mathsf{G}(S))$. It follows that the set of elements of the form
$\mathscr{U}_{a_{i}}$ is compatible. By part (7) of Lemma 7.6, it follows that
set $\\{a_{1},\ldots,a_{m}\\}$ is compatible. Put $a=\bigvee_{i=1}^{n}a_{i}$.
Then $\alpha(a)=U$. We have therefore proved that $\alpha$ is an isomorphism
of semigroups.

(2) By Lemma 7.7, $\beta$ is a bijective functor. It remains to show that it
is a homeomorphism. Since $G$ is a Boolean groupoid, a base for the topology
on $G$ is provided by the compact-open local bisections. Let $U$ be a comapct-
open local bisection of $G$. Then $U\in\mathsf{KB}(G)$. We may therefore form
the set $\mathscr{U}_{U}$ which is a typical element of the base for the
topology on $\mathsf{G}(\mathsf{KB}(G))$. It is now easy to check (or see the
proof of [24, Part (1) of Proposition 2.23]), that the bijection $\beta$
restricts to a bijection between $U$ and $\mathscr{U}_{U}$. ∎

###  7.4. Proof of the main theorem

Our goal now is to take account of appropriate morphisms in our constructions.
We refer the reader to [17] for information about more general kinds of
morphisms.

The statement and proof of [30, part (3), Lemma 3.11] is incorrect. We now
give the correct statement and proof.

######  Lemma 7.9.

Let $\theta\colon S\rightarrow T$ be a morphism of distributive inverse
semigroups. Then for each prime filter $P$ we have that $\theta^{-1}(P)$ is
non-empty if and only if each $t\in T$ can be written
$t=\bigvee_{i=1}^{n}t_{i}$ where each $t_{i}\leq\theta(s_{i})$ for some
$s_{i}\in S$.

###### Proof.

We prove the easy direction first. Suppose that for each $t\in T$ we can write
$t=\bigvee_{i=1}^{n}t_{i}$ where each $t_{i}\leq\theta(s_{i})$ for some
$s_{i}\in S$. Let $P$ be any prime filter. By assumption it is non-empty. Let
$t\in P$. By assumption, we can write $t=\bigvee_{i=1}^{n}t_{i}$ where each
$t_{i}\leq\theta(s_{i})$ for some $s_{i}\in S$. But $P$ is a prime filter.
Thus $t_{i}\in P$ for some $i$. It follows that $\theta(s_{i})\in P$ for some
$s_{i}\in S$. It follows that $\theta^{-1}(P)$ is non-empty. We now prove the
converse. Suppose that $t\in S$ which cannot be written in the stated form.
Then $t\notin\left(\mbox{im}(\theta)^{\downarrow}\right)^{\vee}$. Put
$I=\left(\mbox{im}(\theta)^{\downarrow}\right)^{\vee}$. Then $t^{\uparrow}\cap
I=\emptyset$. We now use Section 7.1 to deduce that there is a prime filter
$P$ that contains $t$ and is disjoint from $J$. But this implies that
$\theta^{-1}(P)$ is empty which is a contradiction. It follows that no such
element $t$ exists. ∎

A morphism $\theta\colon S\rightarrow T$ of Boolean inverse semigroups is said
to be weakly-meet-preserving if $t\leq\theta(a),\theta(b)$ there exists $c\leq
a,b$ such that $t\leq\theta(c)$. The following is [59, Proposition 3-4.6].

######  Lemma 7.10.

Let $S$ be a Boolean inverse semigroup. Let $I$ be an additive ideal of $S$.

  1. (1)

Define $(a,b)\in\varepsilon_{I}$ if and only if there exists $c\leq a,b$ such
that $a\setminus c,b\setminus c\in I$. Then $\varepsilon_{I}$ is an additive
congruence with kernel $I$.

  2. (2)

If $\sigma$ is any additive congruence with kernel $I$ then
$\varepsilon_{I}\subseteq\sigma$.

An additive congruence is ideal-induced if it equals $\varepsilon_{I}$ for
some additive ideal $I$. The following result is due to Ganna Kudryavtseva
(private communication).

######  Proposition 7.11.

A morphism of Boolean inverse semigroups is weakly-meet-preserving if and only
if its associated congruence is ideal-induced.

###### Proof.

Let $I$ be an additive ideal of $S$ and let $\varepsilon_{I}$ be its
associated additive congruence on $S$. Denote by $\nu\colon S\rightarrow
S/\varepsilon_{I}$ its associated natural morphism. We prove that $\nu$ is
weakly-meet-preserving. Denote the $\varepsilon_{I}$-class containing $s$ by
$[s]$. Let $[t]\leq[a],[b]$. Then $[t]=[at^{-1}t]$ and $[t]=[bt^{-1}t]$. By
definition there exist $u,v\in S$ such that $u\leq t,at^{-1}t$ and $v\leq
t,bt^{-1}t$ such that $t\setminus u,at^{-1}t\setminus u,t\setminus
v,bt^{-1}t\setminus v\in I$. Now $[t]=[u]=[at^{-1}t]$ and
$[t]=[v]=[bt^{-1}t]$. Since $u,v\leq t$ it follows that $u\sim v$ and so
$u\wedge v$ exists. Clearly, $u\wedge v\leq a,b$. In addition $[t]=[u\wedge
v]$. We have proved that $\nu$ is weakly-meet-preserving.

Conversely, let $\theta\colon S\rightarrow T$ be weakly-meet-preserving. We
prove that it is determined by its kernel $I$. By part (2) of Lemma 7.10, it
is enough to prove that if $\theta(a)=\theta(b)$ then we can find $c\leq a,b$
such that $a\setminus c,b\setminus c\in I$. Put $t=\theta(a)=\theta(b)$. Then
there exists $c\leq a,b$ such that $t\leq\theta(c)$. It is easy to check that
$\theta(a\setminus c)=0=\theta(b\setminus c)$. We have therefore proved that
$a\setminus c,b\setminus c\in I$ and so $(a,b)\in\varepsilon_{I}$. ∎

We now combine the above two properties. A morphism $\theta\colon S\rightarrow
T$ of Boolean inverse semigroups is said to be callitic555I made this word up.
It comes from the Greek word ‘kallos’ meaning beauty. I simply wanted to
indicate that these maps were sufficiently ‘nice’. if it satisfies two
conditions:

  1. (1)

We require that $\theta$ be proper. This means that for each $t\in T$ we can
write $t=\bigvee_{i=1}^{n}t_{i}$ where each $t_{i}\leq\theta(s_{i})$ for some
$s_{i}\in S$.

  2. (2)

We require that $\theta$ be weakily-meet-preserving.

A continuous function between topological spaces is said to be coherent if the
inverse images of compact-open subsets are compact-open. You can easily check
that the collection of Boolean inverse semigroups and callitic morphisms forms
a category, as does the collection of Boolean groupoids and coherent,
continuous covering functors. We can now state and prove the main theorem of
this paper.

######  Theorem 7.12 (Non-commutative Stone duality).

The category of Boolean inverse semigroups and callitic morphisms is dually
equivalent to the category of Boolean groupoids and coherent, continuous
covering functors.

###### Proof.

Let $\theta\colon S\rightarrow T$ be a callitic morphism between Boolean
inverse semigroups. Let $B$ be a prime filter in $T$. We prove that
$\theta^{-1}(B)$ is a prime filter in $S$. By Lemma 7.9 this set is non-empty.
Let $x,y\in\theta^{-1}(B)$. Then $\theta(x),\theta(y)\in B$. But $B$ is a
filter and so there is an element $b\in B$ such that
$b\leq\theta(x),\theta(y)$. Since $\theta$ is weakly-meet-perserving, there is
$s\in S$ such that $s\leq x,y$ and $b\leq\theta(s)$. But $b\in B$ and so
$\theta(s)\in B$ and so $s\in\theta^{-1}(B)$. Let $x\in\theta^{-1}(B)$ and
$x\leq y$. Then $\theta(x)\leq\theta(y)$ and $\theta(x)\in B$. It follows that
$\theta(y)\in B$ and so $y\in\theta^{-1}(B)$. Let $x\vee y\in\theta^{-1}(B)$.
Then $\theta(x\vee y)\in B$. Now we use the fact that $\theta$ is also a
morphism to get that $\theta(x)\vee\theta(y)\in B$. But $B$ is a prime filter.
Without loss of generality, suppose that $\theta(x)\in B$ and so
$x\in\theta^{-1}(B)$. Put $\theta^{\star}=\theta^{-1}$. We have therefore
defined a function
$\theta^{\star}\colon\mathsf{G}(T)\rightarrow\mathsf{G}(S)$. It remains to
show that $\theta^{\star}$ is a coherent, continuous covering functor.

The bulk of the proof is taken up with showing that $\theta^{-1}$ is a
functor. Let $F$ be an idempotent prime filter in $T$. Thus by Lemma 6.3, this
is an inverse subsemigroup of $T$. Then $\theta^{-1}(F)$ is a prime filter in
$S$ and the inverse image of an inverse subsemigroup is an inverse
subsemigroup. It follows that $\theta^{-1}(F)$ is an idempotent prime filter.
We have therefore shown that $\theta^{-1}$ maps identities to identities.

We next prove that if $F$ and $G$ are prime filters such that $F^{-1}\cdot
F=G\cdot G^{-1}$ then

$(\theta^{-1}(F)\theta^{-1}(G))^{\uparrow}=\theta^{-1}((FG)^{\uparrow}).$

We prove first that

$\theta^{-1}(F)\theta^{-1}(G)\subseteq\theta^{-1}(FG).$

Let $s\in\theta^{-1}(F)\theta^{-1}(G)$. Then $s=ab$ where $a\in\theta^{-1}(F)$
and $b\in\theta^{-1}(G)$. Thus $\theta(s)=\theta(a)\theta(b)\in FG$. It
follows that $s\in\theta^{-1}(FG)$. Observe that
$\theta^{-1}(X)^{\uparrow}\subseteq\theta^{-1}(X^{\uparrow})$. It follows that

$(\theta^{-1}(F)\theta^{-1}(G))^{\uparrow}\subseteq\theta^{-1}((FG)^{\uparrow}).$

We now prove the reverse inclusion. Let $s\in\theta^{-1}((FG)^{\uparrow}).$
Then $\theta(s)\in F\cdot G$ and so $fg\leq\theta(s)$ for some $f\in F$ and
$g\in G$. The map $\theta$ is assumed proper and so we may quickly deduce that
there exists $v\in S$ such that $\theta(v)\in G$. Consider the product
$\theta(s)\theta(v)^{-1}$. Since $\theta(s)\in F\cdot G$ and
$\theta(v)^{-1}\in G^{-1}$ we have that $\theta(s)\theta(v)^{-1}\in F\cdot
G\cdot G^{-1}=F\cdot F^{-1}\cdot F=F$. Thus $\theta(sv^{-1})\in F$, and we
were given $\theta(v)\in G$, and clearly $(sv^{-1})v\leq s$. Put $a=sv^{-1}$
and $b=v$. Then $ab\leq s$ where $\theta(a)\in F$ and $\theta(b)\in G$. It
follows that $s\in(\theta^{-1}(F)\theta^{-1}(G))^{\uparrow}$.

We may now show that $\theta^{-1}$ is a functor. Let $F$ be a prime filter.
Observe that $\theta^{-1}(F)^{-1}=\theta^{-1}(F^{-1})$. We have that

$(\theta^{-1}(F^{-1})\theta^{-1}(F))^{\uparrow}=(\theta^{-1}(F)^{-1}\theta^{-1}(F))^{\uparrow}=\mathbf{d}(\theta^{-1}(F))$

and

$\theta^{-1}((F^{-1}F)^{\uparrow})=\theta^{-1}(\mathbf{d}(F)).$

Hence by our result above

$\theta^{-1}(\mathbf{d}(F))=\mathbf{d}(\theta^{-1}(F)).$

A dual result also holds and so $\theta^{-1}$ preserves the domain and
codomain operations. Suppose that $\mathbf{d}(F)=\mathbf{r}(G)$ so that
$F\cdot G$ is defined. By our calculation above
$\mathbf{d}(\theta^{-1}(F))=\mathbf{r}(\theta^{-1}(G))$ and so the product
$\theta^{-1}(F)\cdot\theta^{-1}(G)$ is defined. By our main result above we
have that

$\theta^{-1}(F\cdot G)=\theta^{-1}(F)\cdot\theta^{-1}(G),$

as required. We have therefore shown that $\theta^{-1}$ is a functor. The
proof that $\theta^{-1}$ is a covering functor follows the same lines as the
proof of [24, Proposition 2.15]: the proof of star injectivity uses Lemma
6.15, and the proof of star surjectivity uses this same lemma and Lemma 6.9.
To show that $\theta^{-1}$ is continuous, observe that a basic open set of
$\mathsf{G}(S)$ has the form $\mathscr{U}_{s}$ for some $s\in S$. It is simple
to check that this is pulled back under the map $(\phi^{-1})^{-1}$ to the set
$\mathscr{U}_{\theta(s)}$. We now prove coherence. Let $X$ be a compact-open
subset of $\mathsf{G}(S)$. Then $X$ may be written as a union of compact-open
local bisections. Thus, by compactness, we can write $X$ as a union of a
finite number of compact-open local bisections. We now use the previous result
to deduce that the inverse image of $X$ is a finite union of compact-open
local bisections and so is itself a compact-open set.

Let $\phi\colon G\rightarrow H$ be a coherent, continuous covering functor.
Let $U$ be a compact-open local bisection of $H$. Then $\phi^{-1}(U)$ is
compact-open since $\phi$ is coherent. Without loss of generality, we can
assume that it is non-empty. We prove that it is a local bisection. Let
$g,h\in\phi^{-1}(U)$ be such that $\mathbf{d}(g)=\mathbf{d}(h)$. Then
$\theta(g),\theta(h)\in U$ and $\mathbf{d}(\theta(g))=\mathbf{d}(\theta(h))$.
By assumption, $U$ is a local bisection and so $\theta(g)=\theta(h)$. We now
use the fact that $\phi$ is star-injective to deduce that $g=h$. A dial result
proves that $U$ is a local bisection. Put $\phi^{-1}=\phi_{\star}$. We have
therefore defined a function
$\phi_{\star}\colon\mathsf{KB}(H)\rightarrow\mathsf{KB}(G)$. It remains to
show that $\phi_{\star}$ is a callitic morphism.

The proof that it is a semigroup homomorphism follows the same lines as the
proof in [24, Proposition 2.17] where we use Lemma 4.3. We show that this map
is proper and weakly-meet-preserving. We prove first that $\phi^{-1}$ is
proper. Let $B\in\mathsf{KB}(G)$. Then $B$ is a non-empty compact-open local
bisection in $G$. Let $g\in B$. Then $\phi(g)\in H$. Clearly, $H$ is an open
set containing $\phi(g)$. Since $H$ is étale, it follows that $H$ is a union
of compact-open local bisections and so $\phi(g)\in C_{g}$ for some an
compact-open local bisection $C_{g}$ in $H$. Since $\phi$ is continuous and
coherent $g\in\phi^{-1}(C_{g})$ is compact-open and because $\phi$ is a
covering functor $\phi^{-1}(C_{g})$ is a local bisection. It follows that
$B\subseteq\bigcup_{g\in B}\phi^{-1}(C_{g})$. Since $B$ is compact, we may in
fact write $B\subseteq\bigcup_{i=1}^{m}\phi^{-1}(C_{g_{i}})$ for some finite
set of elements $g_{1},\ldots,g_{m}\in B$. Put
$B_{i}=B\cap\phi^{-1}(C_{g_{i}})$. This is clearly an open local bisection and
$B=\bigcup_{i=1}^{m}B_{i}$ and each $B_{i}\subseteq\phi^{-1}(C_{i})$ where
$C_{i}=C_{g_{i}}$. We prove that we may find compact-open local bisections
$D_{i}$ such that $B$ is the union of the $D_{i}$ and
$D_{i}\subseteq\phi^{-1}(C_{i})$. Since $B_{i}$ is an open local bisection it
is a union of compact-open local bisections. Amalgamating these unions we have
that $B$ is a union of compact-open local bisections each of which is a subset
of one of the $\phi^{-1}(C_{i})$. It follows that $B$ is a union of a finite
number of such compact-open local bisections. Define $D_{i}$ to be the union
of those which are contained in $\phi^{-1}(C_{i})$ and the result follows. We
now prove that $\phi^{-1}$ is weakly-meet-preserving. Let
$A,B\in\mathsf{KB}(H)$. Then $A$ and $B$ are compact-open local bisections of
$H$. Let $Y$ be any compact-open local bisection of $G$ such that
$Y\subseteq\phi^{-1}(A),\phi^{-1}(B)$. Clearly, $Y\subseteq\phi^{-1}(A\cap
B)$. We can at least say that $A\cap B$ is an open local bisection and
$\theta(Y)\subseteq A\cap B$. Since $\phi$ is continuous, we know that
$\phi(Y)$ is compact. Now $H$ has a base of compact-open local bisections. It
follows that $A\cap B$ is a union of compact-open local bisections. But
$\theta(Y)$ is compact and so $\theta(Y)$ is contained in a finite union of
compact-open local bisections that is also contained in $A\cap B$. Thus
$\theta(Y)\subseteq V=\bigcup_{i=1}^{m}V_{i}\subseteq A\cap B$. Now $A\cap B$
a local bisection implies that $V$ is a local bisection. It is evident that
$V$ is a compact-open local bisection itself. We therefore have
$\theta(Y)\subseteq V\subseteq A\cap B$. Hence
$Y\subseteq\theta^{-1}(\theta(Y))\subseteq\theta^{-1}(V)\subseteq\theta^{-1}(A\cap
B)$.

Let $\theta\colon S\rightarrow T$ be a callitic morphism between two Boolean
inverse semigroups. Define $\mathsf{G}(\theta)=\theta^{\star}$. Let
$\phi\colon G\rightarrow H$ be a coherent continuous covering functor between
Boolean groupoids. Define $\mathsf{KB}(\mathsf{\phi})=\phi_{\star}$. It is now
routine to check that we have defined functors: where $\mathsf{G}$ takes us
from Boolean inverse semigroups and callitic morphisms to the dual of the
category of Boolean groupoids and coherent continuous covering functors; and
where $\mathsf{KB}$ takes us from the category of Boolean groupoids and
coherent continuous covering functors to the dual of the category of Boolean
inverse semigroups and callictic morphisms.

Let $\theta\colon S\rightarrow T$ be a callitic morphism between two Boolean
inverse semigroups. We shall compare this to the callitic morphism
$(\theta^{\star})_{\star}\colon\mathsf{KB}(\mathsf{G}(S))\rightarrow\mathsf{KB}(\mathsf{G}(T))$
using Proposition 7.8. We have to compute
$(\theta^{\star})_{\star}(\mathscr{U}_{s})$. It is routine to check that this
is $\mathscr{U}_{\theta(s)}$. Let $\phi\colon G\rightarrow H$ be a coherent
continuous covering functor between Boolean groupoids. We shall compare this
to the coherent continuous covering functor
$(\phi_{\star})^{\star}\colon\mathsf{G}(\mathsf{KB}(G))\rightarrow\mathsf{G}(\mathsf{KB}(H))$
using Proposition 7.8. We have to compute
$(\phi_{\star})^{\star}(\mathscr{F}_{g})$. It is routine to check that this is
$\mathscr{F}_{\phi(g)}$.

The functor $\mathsf{KB}\circ\mathsf{G}$ is now clearly naturally isomorphic
with the identity functor on the category of Boolean inverse semigroups and
callitic morphisms, whereas the functor $\mathsf{G}\circ\mathsf{KB}$ is now
clearly naturally isomorphic with the identity functor on the category of
Boolean groupoids and coherent continuous covering functors. ∎

##  8\. Special cases

In this section, we shall describe some special cases of Theorem 7.12. None of
these is new, but they have not appeared altogether in this way before.

By the results of Section 2.3 and 2.4, a generalized Boolean algebra with an
identity is nothing other than a Boolean algebra. It follows that the Boolean
inverse semigroups with the property that the space of identities of their
Stone groupoids are compact are precisely the Boolean inverse monoids.

The group of units of a Boolean inverse monoid is just the set of all compact-
open bisections of the associated Stone groupoid; the compact-open bisections
form what is known as the topological full group of the Boolean groupoid.

Suppose that $S$ is a countable Boolean inverse semigroup. Then the Stone
groupoid of $S$ must have a countable base of compact-open local bisections.
It follows that its Stone groupoid is second-countable.

The Stone space of the Tarski algebra is the Cantor space by Example (2) of
Example 2.26. Define a Tarski monoid to be a countably infinite Boolean
inverse monoid whose semilattice of idempotents forms a Tarski algebra. Define
a Tarski groupoid to be a second-countable Boolean groupoid whose space of
identities is the Cantor space.

Following Wehrung [59], define a semisimple Boolean inverse semigroup to be
one in which for each element $a$ the principal order ideal $a^{\downarrow}$
is finite.

######  Proposition 8.1.

Let $S$ be a Boolean inverse semigroup. Then $S$ is semisimple if and only if
$\mathsf{G}(S)$ carries the discrete topology.

###### Proof.

Let $a$ be any element of such a semigroup. Then
$\mathscr{U}_{a}=\bigcup_{i=1}^{n}\mathscr{U}_{a_{i}}$, where
$a_{1},\ldots,a_{n}$ are all the atoms below $a$; this is proved by observinng
that in a semisimple Boolean inverse semigroup, every non-zero element lies
above an atom and so every non-zero element is a join of atoms. If $a$ is an
atom then $\mathscr{U}_{a}$ contains exactly one element. It follows that the
sets $\mathscr{U}_{a}$, where $a$ is an atom, form a base for the topology on
$\mathsf{G}(S)$, and so this set is equipped with the discrete topology.
Conversely, suppose that $\mathsf{G}(S)$ is equipped with the discrete
topology. Then, for each $a\in S$, we have that $\mathscr{U}_{a}$ is finite
since it is compact-open. But $b\leq a$ if and only if
$\mathscr{U}_{b}\subseteq\mathscr{U}_{a}$. This proves that the Boolean
inverse semigroup $S$ is semisimple. ∎

If $S$ is a semisimple Boolean inverse semigroup, then the Stone groupoid
$\mathsf{G}(S)$ is isomorphic to the set of atoms of $S$ equipped with the
restricted product; see [29] for the structure of semisimple Boolean inverse
semigroups. The significance of semisimple Boolean inverse semigroups is
explained by the following result. We say that a Boolean inverse semigroup is
atomless if it has no atoms.

######  Proposition 8.2 (Dichotomy theorem).

Let $S$ be a $0$-simplifying Boolean inverse semigroup. Then either $S$ is
semisimple or $S$ is atomless.

###### Proof.

Suppose that $a$ is an atom. Then $\mathbf{d}(a)$ is an atom. We can therefore
assume, without loss of generality, that there is at least one idempotent atom
$e$. Let $x$ be any non-zero element of $S$. We prove that $x^{\downarrow}$ is
finite. It is enough to prove that $\mathbf{d}(x)^{\downarrow}$ is finite. Let
$f\leq\mathbf{d}(x)$. Since the semigroup is assumed to be $0$-simplifying,
there is a pencil from $f$ to $e$. There is therefore a finite set of elements
$\\{x_{1},\ldots,x_{n}\\}$ such that $f=\bigvee_{i=1}^{n}\mathbf{d}(x_{i})$
and $\mathbf{r}(x_{i})\leq e$. But $e$ is an atom. Without loss of generality,
we assume that all the $x_{i}$ are non-zero. Thus $\mathbf{r}(x_{i})=e$. It
follows that each $\mathbf{d}(x_{i})$ is an atom and we have proved that $f$
is a join of a finite number of atoms. It follows that
$\mathbf{d}(x)^{\downarrow}$ is a finite Boolean algebra. ∎

Our next result tells us when the Stone groupoid is Hausdorff.

######  Proposition 8.3.

Let $S$ be a Boolean inverse semigroup. Then $S$ is a meet-semigroup if and
only if $\mathsf{G}(S)$ is Hausdorff.

###### Proof.

Let $S$ be a Boolean inverse semigroup and suppose that it is a meet-
semigroup. Let $F$ and $G$ be distinct ultrafilters in $\mathsf{G}(S)$, since
neither ultrafilter can be a subset of the other, we can find elements $a\in
F\setminus G$ and $b\in G\setminus F$. The element $a\wedge b$ exists by
assumption and also by assumption $a\neq a\wedge b$ and $b\neq a\wedge b$.
Observe that $F\in\mathscr{U}_{a\setminus(a\wedge b)}$ and
$G\in\mathscr{U}_{b\setminus(a\wedge b)}$. Let $x\leq a\setminus(a\wedge
b),b\setminus(a\wedge b)$. Then $x\leq a\wedge b$. It follows that $x=0$. We
have proved that $\mathscr{U}_{a\setminus(a\wedge
b)}\cap\mathscr{U}_{b\setminus(a\wedge b)}=\varnothing$ and so we have proved
that $\mathscr{G}(S)$ is Hausdorff. Conversely, suppose that $\mathscr{G}(S)$
is Hausdorff. Let $A$ and $B$ be two compact-open local bisections. By part
(2) of Lemma 1.3, both $A$ and $B$ are clopen. Thus $A\cap B$ is clopen. But
by part (3) of Lemma 1.3, $A\cap B$ is also compact, and so it is a compact-
open local bisection. The result, that $S$ is a meet-semigroup, follows by
Proposition 7.8. ∎

######  Remark 8.4.

The above result was essentially the basis of my paper [24]. This paper won
the Mahony-Neumann-Room Prize of the Australian Mathematical Society in 2017.

Recall that an inverse semigroup is fundamental if the only elements that
commute with all idempotents are themselves idempotents. If we denote the
centralizer of the idempotents by $\mathsf{Z}(\mathsf{E}(S))$ then an inverse
semigroup is fundmanetal when $\mathsf{E}(S)=\mathsf{Z}(\mathsf{E}(S))$ An
étale topological groupoid $G$ is said to be effective if the interior of the