# The Stone’s Representation Theorems and compactifications of a discrete
space

Hussain Rashed  University of Tennessee, Knoxville, TN 37996, USA
<EMAIL_ADDRESS>
###### Abstract.

This paper is meant to give a short exposition of the Stone’s Representation
Theorems. We provide three equivalent approaches to construct a Stone’s space
from a given Boolean algebra. Finally, we utilize the Stone’s Representation
Theorems to study the compactifications of a discrete space.

###### Key words and phrases:

Boolean Algebras, Boolean rings, Stone’s Spaces, Stone’s Representation
Theorems, Spectrum of Rings, Compactifications of a discrete space,
Stone-$\breve{C}$ech compactification of a discrete space

###### 2000 Mathematics Subject Classification:

Primary 54D35; Secondary 20F69

###### Contents

  1. 1 Introduction
  2. 2 Boolean algebras and their homomorphisms
  3. 3 Stone’s space of a Boolean algebra
    1. 3.1 Characterization of locally compact totally disconnected spaces
    2. 3.2 Stone’s space of a Boolean algebra as the family of its ultrafilters
    3. 3.3 Stone’s space of a Boolean algebra $B$ as the family of all homomorphisms from $B$ to $\mathbb{Z}_{2}$
  4. 4 The Stone’s representation theorems
  5. 5 Stone’s space of a Boolean algebra as the spectrum of its induced Boolean ring
  6. 6 Compactifications of a discrete space
    1. 6.1 The one-point compactification of a discrete space
    2. 6.2 The Stone-$\breve{C}$ech compactification of a discrete space

##  1\. Introduction

The work by Marshall Stone [8] associated to every Boolean algebra $B$ a
totally disconnected compact Hausdorff space $X_{B}$ called the Stone’s space
from which $B$ can be restored. Conversely, every Stone’s space $X$ can be
reconstructed from an associated Boolean algebra $B_{X}$. These two facts are
the core of the Stone’s Representation Theorem for Boolean algebras, and the
Stone’s Representation Theorem for Stone’s spaces, respectively. The main aim
of this self-contained exposition is to provide detailed proofs of the Stone’s
Representation Theorems using modern tools. The exposition can be outlined in
the following points:
$\bullet$ The family of all clopen subsets of a topological space $X$ forms a
Boolean algebra $Clop(X)$ 2.13. If $B$ is a Boolean algebra and $X_{B}$ is its
assigned Stone’s space, then $B$ is the clopen Boolean algebra of $X_{B}$ 4.1.
Furthermore, a homomorphism $\varphi:B_{1}\to B_{2}$ between Boolean algebras
induces a continuous map $\widetilde{\varphi}:X_{B_{2}}\to X_{B_{1}}$ between
their associated Stone’s spaces; if $\varphi:B_{1}\to B_{2}$ is an
isomorphism, then $\widetilde{\varphi}:X_{B_{2}}\to X_{B_{1}}$ is a
homeomorphism 3.29. The Stone’s space $X_{B}$ associated to a given Boolean
algebra can be obtained in three equivalent ways:
($1$) As the space of all ultrafilters of $B$ and denoted by $\mathcal{U}(B)$.
($2$) As the space of all homomorphisms from $B$ to $\mathbb{Z}_{2}$ and
denoted by $Hom(B,\mathbb{Z}_{2})$.
($3$) As the spectrum of its induced Boolean ring $R_{B}$ and denoted by
$Spec(R_{B})$.
The Stone’s spaces $\mathcal{U}(B)$, $Hom(B,\mathbb{Z}_{2})$ and $Spec(R_{B})$
are homeomorphic 3.32, 5.11.
$\bullet$ A continuous map $f:X\to Y$ between topological spaces induces a
homomorphism $\widetilde{f}:Clop(Y)\to Clop(X)$ between their clopen Boolean
algebras. If $f:X\to Y$ is a homeomorphism, $\widetilde{f}:Clop(Y)\to Clop(X)$
is an isomorphism 2.13. Moreover, if $X$ is a Stone’s space, then it is the
Stone’s space associated to the Boolean algebra $Clop(X)$ 4.1.
In a categorical language, the above can be re-stated as that there is a
duality between the category of Boolean algebras and the category of Stone’s
spaces.
$\bullet$ When $X$ is a discrete space; a topological space $Y$ is a
compactification of $X$ if and only if it is homeomorphic to a Stone’s space
$\mathcal{U}(\mathcal{B})$ for some a sub-Boolean algebra $\mathcal{B}$ of
$\mathcal{P}(X)$ that separates points of $X$ 6.16; when
$\mathcal{B}=\mathcal{P}(X)$, $\mathcal{U}(\mathcal{B})$ is the
Stone-$\breve{C}$ech compactification of $X$. In particular,
$Hom(B,\mathbb{Z}_{2})$ and $Spec(R_{B})$, where $B=\mathcal{P}(X)$, are
distinct approaches to construct the Stone-$\breve{C}$ech compactification of
$X$ 6.12.

##  2\. Boolean algebras and their homomorphisms

######  Definition 2.1.

A Boolean algebra $(B,+,\cdot,\neg)$ is a set $B$ equipped with two binary
operations $+$, $\cdot$ and a unary operation $\neg$ called the complement,
where the following are satisfied:
$(1)$ $+$ and $\cdot$ are commutative and associative, that is:
$\bullet$ $x+y=y+x$ and $x\cdot y=y\cdot x$, for all $x,y\in B$.
$\bullet$ $x+(y+z)=(x+y)+z$ and $x\cdot(y\cdot z)=(x\cdot y)\cdot z$, for all
$x,y,z\in B$.
$(2)$ $+$ is distributive over $\cdot$, and $\cdot$ is distributive over $+$,
that is:
$\bullet$ $x+(y\cdot z)=(x+y)\cdot(x+z)$, for all $x,y,z\in B$.
$\bullet$ $x\cdot(y+z)=(x\cdot y)+(x\cdot z)$, for all $x,y,z\in B$.
$(3)$ $+$ and $\cdot$ are connected by the absorption law, that is:
$\bullet$ $x+(x\cdot y)=x\cdot(x+y)=x$, for all $x,y\in B$.
$(4)$ There exist two elements $0,1\in B$ satisfying:
$\bullet$ $x+(\neg x)=1$, for all $x\in B$;
$\bullet$ $x\cdot(\neg x)=0$, for all $x\in B$.

######  Example 2.2.

Consider the power set $\mathcal{P}(X)$ of $X$; then
$(\mathcal{P}(X),+,\cdot,\neg)$ is a Boolean algebra, where $+:=\cup$,
$\cdot:=\cap$, $0:=\emptyset$, $1:=X$ and $\neg A:=X\setminus A$ for all
$A\in\mathcal{P}(X)$. The Boolean algebra $(\mathcal{P}(X),\cup,\cap,\neg)$ is
called the power set Boolean algebra.

######  Definition 2.3.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, a subset $A\subset B$ is called a
sub-Boolean algebra if $(A,+,\cdot,\neg)$ is itself a Boolean algebra.

######  Example 2.4.

Let $X$ be a topological space. The family of all clopen subsets
$Clop(X):=\\{U\subset X:U$ is clopen$\\}$ is a sub-Boolean algebra of
$\mathcal{P}(X)$. The Boolean algebra $(Clop(X),+,\cdot,\neg)$ is called the
clopen Boolean algebra associated with $X$.

######  Example 2.5.

Let $G$ be a group acts on $X$; a subset $A\subset X$ is called
$G$-commensurated or $G$-almost invariant if $g\cdot A\Delta A$ is a finite
subset of $X$ for all $g\in G$. The family of all $G$-commensurated subsets of
$X$ is denoted by $Comm_{G}(X)$. $Comm_{G}(X)$ is a sub-Boolean algebra of
$\mathcal{P}(X)$ [2].

######  Example 2.6.

Let $G$ be a group acts on $X$; define a relation $\sim$ on $Comm_{G}(X)$ by:
for all $A,B\in Comm_{G}(X)$, $A\sim B$ if and only if $A\Delta B$ is finite.
Then:
($a$) The relation $\sim$ is an equivalence relation on $Comm_{G}(X)$.
($b$) The family of all equivalence classes
$\faktor{Comm_{G}(X)}{\sim}:=\\{[A]:A\in Comm_{G}(X)\\}$ of $Comm_{G}(X)$ is a
Boolean algebra with operations:
$\bullet$ $[A]+[B]:=[A\cup B]$;
$\bullet$ $[A]\cdot[B]:=[A\cap B]$;
$\bullet$ $\neg[A]:=[X\setminus A]$.

The fact that the binary relations $+$ and $\cdot$ being connected by the
absorption law gives us the following properties.

######  Proposition 2.7.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, the following are satisfied:
$(a)$ $\cdot$ is idempotent, that is $x^{2}:=x\cdot x=x$, for all $x\in B$.
$(b)$ $+$ is idempotent, that is $2x:=x+x=x$ for all $x\in B$.
$(c)$ $1\cdot x=x$, for all $x\in B$.
$(d)$ $1+x=1$, for all $x\in B$.
$(e)$ $0+x=x$, for all $x\in B$.
$(f)$ $0\cdot x=0$, for all $x\in B$.

###### Proof.

$(a)$ Let $y\in B$, then $x=x+(x\cdot y)$ and therefore
$x^{2}=x\cdot(x+(x\cdot y))=x$.
$(b)$ By part $a)$ we have $2x=x^{2}+x^{2}=x\cdot(x+x)=x$.
$(c)$ $x\cdot 1=x\cdot(x+(\neg x))=x$.
$(d)$ $1+x=(x+(-x))+x=2x+(\neg x)=x+(\neg x)=1$.
$(e)$ $0+x=x+0=x+(x\cdot(\neg x))=x$.
$(f)$ $0\cdot x=x\cdot 0=x\cdot(x\cdot(\neg x))=x^{2}\cdot(\neg x)=x\cdot(\neg
x)=0$. ∎

######  Proposition 2.8.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra and $x,y\in B$. Then:
$(a)$ If $x\cdot y=0$ and $x+y=1$, then $y=\neg x$.
$(b)$ $\neg(\neg x)=x$.
$(c)$ $\neg 1=0$ and $\neg 0=1$.

###### Proof.

Notice that part $(b)$ and $(c)$ follow from $(a)$. Let $x,y\in B$, such that
$x\cdot y=0$ and $x+y=1$. On one hand, one has $\neg x=\neg x\cdot 1=\neg
x\cdot(x+y)=(\neg x)\cdot x+(\neg x)\cdot y=0+(\neg x)\cdot y=\neg x\cdot y$.
On the other hand, $\neg x=\neg x+0=(\neg x\cdot y)+(x\cdot y)=(\neg x+x)\cdot
y=1\cdot y=y$. ∎

######  Proposition 2.9.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and define a relation $\leqslant$
on $B$ by: $x\leqslant y$ if and only if $x+y=y$ for all $x,y\in B$. Then the
relation $\leqslant$ is a partial order relation on $B$.

###### Proof.

$\bullet$ Reflexivity: for all $x\in B$, $x+x=x^{2}+x^{2}=x\cdot(x+x)=x$ and
hence $x\leqslant x$.
$\bullet$ Antisymmetry: for all $x,y\in B$, if $x\leqslant y$ and $y\leqslant
x$ then $x+y=y$ and $x+y=x$; thus, $x=y$.
$\bullet$ Transitivity: for all $x,y,z\in B$, if $x\leqslant y$ and
$y\leqslant z$, then $x+y=y$ and $y+z=z$. Thus, $x+z=x+(y+z)=(x+y)+z=y+z=z$
which implies that $x\leqslant z$. ∎

######  Proposition 2.10.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra and $x,y,x^{\prime},y^{\prime}\in
B$. Then:
$(a)$ $0\leqslant x\leqslant 1$.
$(b)$ $x\cdot y\leqslant x\leqslant x+y$ and $x\cdot y\leqslant y\leqslant
x+y$.
$(c)$ If $x\leqslant y$ and $x^{\prime}\leqslant y^{\prime}$, then $x\cdot
x^{\prime}\leqslant y\cdot y^{\prime}$.
$(d)$ If $x\leqslant y$ and $x^{\prime}\leqslant y^{\prime}$, then
$x+x^{\prime}\leqslant y+y^{\prime}$.
$(e)$ $\sup\\{x,y\\}=x+y$.
$(f)$ $\inf\\{x,y\\}=x\cdot y$.

###### Proof.

$(a)$ Let $x\in B$, since $x+0=x$ and $x+1=1$, so $0\leqslant x\leqslant 1$.
$(b)$ Let $x,y\in B$, since $x+(x\cdot y)=x$ and $x+(x+y)=2x+y=x+y$, hence
$x\cdot y\leqslant x\leqslant x+y$. Similarly, $x\cdot y\leqslant y\leqslant
x+y$.
$(c)$ Let $x,y,x^{\prime},y^{\prime}\in B$ and assume that $x\leqslant y$ and
$x^{\prime}\leqslant y^{\prime}$. Now, $y\cdot y^{\prime}\leqslant x\cdot
x^{\prime}+y\cdot y^{\prime}\leqslant x\cdot x^{\prime}+x\cdot
y^{\prime}+y\cdot x^{\prime}+y\cdot
y^{\prime}=x\cdot(x^{\prime}+y^{\prime})+y\cdot(x^{\prime}+y^{\prime})=(x+y)\cdot(x^{\prime}+y^{\prime})=y\cdot
y^{\prime}$ and hence $x\cdot x^{\prime}+y\cdot y^{\prime}=y\cdot y^{\prime}$
as desired.
($d$) Let $x,y,x^{\prime},y^{\prime}\in B$ and assume that $x\leqslant y$ and
$x^{\prime}\leqslant y^{\prime}$, then
$(x+x^{\prime})+(y+y^{\prime})=(x+y)+(x^{\prime}+y^{\prime})=y+y^{\prime}$.
($e$) Let $x,y,z\in B$ such that $x\leqslant z$ and $y\leqslant z$; then by
part $(b)$, $x+y\leqslant 2z=z$.
($f$) Let $x,y,z\in B$ such that $z\leqslant x$ and $z\leqslant y$; then by
part $(c)$, $z=z^{2}\leqslant x\cdot y$.
∎

2.10 tells us that every Boolean algebra is a distributive lattice.

######  Definition 2.11.

Let $(B,+,\cdot,\neg)$ and
$(B^{\prime},+^{\prime},\cdot^{\prime},\neg^{\prime})$ be Boolean algebras. A
map
$\varphi:B\to B^{\prime}$ is called a homomorphism if for all $x,y\in B$:
$\bullet$ $\varphi(x+y)=\varphi(x)+^{\prime}\varphi(y)$,
$\bullet$ $\varphi(x\cdot y)=\varphi(x)\cdot^{\prime}\varphi(y)$,
$\bullet$ $\varphi(0_{B})=0_{B^{\prime}}$ and $\varphi(1_{B})=1_{B^{\prime}}$.

Notice that for $x\in B$, since
$0_{B^{\prime}}=\varphi(0_{B})=\varphi(x\cdot(\neg
x))=\varphi(x)\cdot^{\prime}\varphi(\neg x)$ and
$1_{B^{\prime}}=\varphi(1_{B})=\varphi(x+(\neg
x))=\varphi(x)+^{\prime}\varphi(\neg x)$, hence by 2.8, $\varphi(\neg
x)=\neg^{\prime}\varphi(x)$.

######  Example 2.12.

The field $\mathbb{Z}_{2}$ is a Boolean algebra where the complement $\neg$ of
its elements is given by: $\neg 0:=1$ and $\neg 1:=0$. Moreover,
$\mathbb{Z}_{2}$ is unique up to isomorphism; that is, any Boolean algebra
with cardinality $2$ is isomorphic to $\mathbb{Z}_{2}$. $\mathbb{Z}_{2}$ is
called the two-element Boolean algebra.

Now we associate a Boolean algebra to any given topological space with the
property that any continuous map between two topological spaces induces a
homomorphism between their associated Boolean algebras; and if the two
topological spaces are homeomorphic, their associated Boolean algebras are
isomorphic.

######  Proposition 2.13.

Let $X$ be a topological space. Then:
($a$) If $f:X\to Y$ is a continuous map between topological spaces, it induces
a homomorphism $\widetilde{f}:Clop(Y)\to Clop(X)$.
($b$) If $f:X\to Y$ is a homeomorphism between topological spaces, it induces
an isomorphism $\widetilde{f}:Clop(Y)\to Clop(X)$.

###### Proof.

($a$) Let $f:X\to Y$ be a continuous map between topological spaces, and
define $\widetilde{f}:Clop(Y)\to Clop(X)$ by $\widetilde{f}(U)=f^{-1}(U)$ for
all $U\in Clop(Y)$. Clearly, the map $\widetilde{f}:Clop(Y)\to Clop(X)$ is a
homomorphism.
($b$) Let $g:Y\to X$ be a continuous map such that $g\circ f=id_{X}$ and
$f\circ g=id_{Y}$. By part ($b$), $\widetilde{g}:Clop(X)\to Clop(Y)$ is a
homomorphism. Moreover, $\widetilde{f}\circ\widetilde{g}=id_{Clop(X)}$ and
$\widetilde{g}\circ\widetilde{f}=id_{Clop(Y)}$. ∎

The clopen Boolean algebra associated with a topological space $X$ is the two-
elements Boolean algebra if and only if $X$ is connected. The more components
$X$ has, the more interesting the Boolean algebra $Clop(X)$ is. Hausdorff
compact spaces that are totally disconnected are of a particular interests as
every Boolean algebra is the clopen Boolean algebra associated with a such
topological space.

##  3\. Stone’s space of a Boolean algebra

In this section, out of a given Boolean algebra $B$, we construct a compact
Hausdorff totally disconnected topological space called the Stone’s space
associated with the Boolean algebra $B$ such that every homomorphism between
two Boolean algebras induces a continuous map between their associated Stone’s
spaces. One way to construct a such space is by taking the set of all
ultrafilters of $B$ as the underlying set of the Stone’s space associated with
a Boolean algebra $B$. Another way to construct the Stone’s space associated
with $B$ is by taking the family of all homomorphisms from $B$ to
$\mathbb{Z}_{2}$ as the underlying set of the Stone’s space associated with a
Boolean algebra $B$. In section 5, it will be shown that the Stone’s space
associated with $B$ can also be obtained as the spectrum of the Boolean ring
induced from $B$. We first give a characterization of a totally disconnected
space $X$, when $X$ is locally compact.

###  3.1. Characterization of locally compact totally disconnected spaces

######  Definition 3.1.

Let $X$ a Hausdorff topological space:
$\bullet$ $X$ is called totally disconnected if the connected components of
$X$ are exactly the singletons. Obviously, every discrete topological space is
totally disconnected; however, the converse does not hold. $\mathbb{Q}$ as a
subspace of $\mathbb{R}$ serves as an example of a totally disconnected space
that is not discrete.
$\bullet$ $X$ is called a Stone’s space if it is compact and totally
disconnected.
$\bullet$ $X$ is called zero-dimensional if $X$ admits a basis consists of
clopen subsets.

######  Definition 3.2.

Let $X$ be a non-empty set and $\mathcal{A}\subset\mathcal{P}(X)$.
$\mathcal{A}$ is said to satisfy the finite intersection property if the
intersection of any finitely many elements of $\mathcal{A}$ is non-empty. It
is well-known that a topological space is compact if and only if
$\bigcap\limits_{A\in\mathcal{A}}A\neq\emptyset$ for every family of closed
subsets $\mathcal{A}$ that satisfies the finite intersection property.

######  Proposition 3.3.

Let $X$ be a topological space. If $X$ is zero-dimensional then it is totally
disconnected.

###### Proof.

Let $A\subset X$ and $x,y\in A$ be distinct elements. Since $X$ is a $T_{1}$
space, we can find a clopen subset $U$ containing $x$ but not $y$. Let
$V:=X\setminus U$, then $U$ and $V$ are disjoint open subsets containing $x$
and $y$, respectively, and $X=U\cup V$. ∎

When $X$ is locally compact, the concept of being totally disconnected
coincide with the concept of being zero-dimensional.

######  Lemma 3.4.

Let $X$ be a Hausdorff compact space. For $x\in X$, let $C_{x}$ be the
component of $X$ containing $x$, and $\mathcal{A}_{x}$ be the family of all
clopen subsets of $X$ containing $x$. Then,
$C_{x}=\bigcap\limits_{A\in\mathcal{A}_{x}}A$.

###### Proof.

Notice that every component of $X$ that intersects a clopen subset of $X$ is
contained in that clopen subset, so
$C_{x}\subset\bigcap\limits_{A\in\mathcal{A}_{x}}A$. To prove that
$\bigcap\limits_{A\in\mathcal{A}_{x}}A\subset C_{x}$, it suffices to show that
$\bigcap\limits_{A\in\mathcal{A}_{x}}A$ is connected. Put
$O_{x}:=\bigcap\limits_{A\in\mathcal{A}_{x}}A$ and assume that $U_{1}$ and
$U_{2}$ are disjoint closed subsets of $O_{x}$ such that $O_{x}=U_{1}\cup
U_{2}$. We aim to show that either $U_{1}=\emptyset$ or $U_{2}=\emptyset$.
Since $X$ is normal, there exist disjoint open subsets $V_{1}$ and $V_{2}$ of
$X$ such that $U_{1}\subset V_{1}$ and $U_{2}\subset V_{2}$. Put $V:=V_{1}\cup
V_{2}$; we claim that there exist finitely many elements
$A_{1},...,A_{n}\in\mathcal{A}_{x}$ such that
$\bigcap\limits_{i=1}^{n}A_{i}\subset V$. Seeking contradiction, assume that
the intersection of any finitely many elements of $\mathcal{A}_{x}$ non-
trivially intersects the closed subset $X\setminus V$. The family
$\mathcal{A}_{x}\cup\\{X\setminus V\\}$ consists of closed subsets of the
compact space $X$ and it satisfies the Finite Intersection Property, and hence
$O_{x}\cap(X\setminus V)\neq\emptyset$ which contradicts the fact that
$O_{x}\subset V$. Let, $A_{1},...,A_{n}\in\mathcal{A}_{x}$ such that
$\bigcap\limits_{i=1}^{n}A_{i}\subset V=V_{1}\cup V_{2}$. Without loss of
generality, assume that $x\in V_{1}$, and notice that
$A:=V_{1}\cap\bigcap\limits_{i=1}^{n}A_{i}$ is an open subset of $X$.
Furthermore, as $A^{c}=(\bigcap\limits_{i=1}^{n}A_{i})^{c}\cup(V_{1}^{c}\cap
V_{2})=(\bigcap\limits_{i=1}^{n}A_{i})^{c}\cup V_{2}$ is open,
$A=V_{1}\cap\bigcap\limits_{i=1}^{n}A_{i}$ is a clopen subset of $X$
containing $x$, and hence $A\in\mathcal{A}_{x}$. This shows that $O_{x}\subset
A\subset V_{1}$ which implies that $V_{2}=\emptyset$ and, a fortiori,
$U_{2}=\emptyset$. Thus, $O_{x}$ is connected as desired. ∎

######  Proposition 3.5.

Let $X$ be a compact totally disconnected space. Then $X$ is zero-dimensional.

###### Proof.

Let $C_{x}$ and $\mathcal{A}_{x}$ be as in 3.4. Notice that
$C_{x}=\bigcap\limits_{A\in\mathcal{A}_{x}}A=\\{x\\}$. Let $U$ be an open
neighborhood of $x\in X$. We can find finitely many elements
$A_{1},...,A_{n}\in\mathcal{A}_{x}$ such that
$\bigcap\limits_{i=1}^{n}A_{i}\subset U$. Since
$\bigcap\limits_{i=1}^{n}A_{i}$ is a clopen subset containing $x$, $X$ admits
a basis consists of clopen subsets. ∎

######  Theorem 3.6.

Let $X$ be a Hausdorff locally compact space. $X$ is totally disconnected if
and only if $X$ is zero-dimensional.

###### Proof.

Assume that $X$ is totally disconnected. For $x\in X$, let $U$ be an open
neighborhood of $x\in X$. Since $X$ is a Hausdorff locally compact space, we
can find a relatively compact open neighborhood $V$ of $x$ such that $V\subset
cl(V)\subset U$. Now, $cl(V)$ itself is compact totally disconnected as a
subspace of $X$, and by 3.5, $cl(V)$ is zero-dimensional. Let $A$ be a clopen
subset of $cl(V)$ containing $x$, then as $A\subset V$, $A$ is clopen subset
of $X$ containing $x$ and contained in $U$. Hence, $X$ is zero-dimensional.
The other direction follows from 3.3. ∎

######  Corollary 3.7.

Let $X$ be a Hausdorff compact space. $X$ is a Stone’s space if and only if
$X$ is zero-dimensional.

###### Proof.

Apply 3.6. ∎

###  3.2. Stone’s space of a Boolean algebra as the family of its ultrafilters

Now we construct the Stone’s space that is associated with a Boolean algebra
using its ultrafilters as the building blocks.

######  Definition 3.8.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F$ be a nonempty subset of
$B$; $F$ is called a filter of $B$ if:
$\bullet$ For all $x,y\in F$, $x\cdot y\in F$.
$\bullet$ For all $x\in F$ and $y\in B$, if $x\leqslant y$, then $y\in F$.
$\bullet$ $F$ is called a proper filter if it is a filter contained properly
in $B$. Clearly, a filter $F$ is proper if and only if $0\notin F$.

######  Definition 3.9.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $A\subset B$, $A$ is said to
satisfy the finite product property if for any finitely many elements
$x_{1},...,x_{n}\in A$, then $x_{1}\cdot...\cdot x_{n}\neq 0$.

In the Boolean algebra $(\mathcal{P}(X),\cup,\cap,\neg)$, a subset of
$\mathcal{P}(X)$ satisfies the finite product property if and only if it
satisfies the finite intersection property. Furthermore, $F\in\mathcal{P}(X)$
is a filter of the Boolean algebra $\mathcal{P}(X)$ if and only if it is a
filter in the usual sense.

It can be evidently seen that every subset of a proper filter must satisfy the
finite product property. Moreover, every subset of of a Boolean algebra that
satisfies the finite product property is contained in a proper filter.

######  Proposition 3.10.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra and $A$ be a nonempty subset of
$B$ that satisfies the finite product property. Then, $A$ is contained in a
proper filter $F_{A}$. Furthermore, if $F$ is a filter containing $A$, then
$F_{A}\subset F$.

###### Proof.

Set $F_{A}:=\\{x\in B:x_{1}\cdot...\cdot x_{n}\leqslant x$ for some
$n\in\mathbb{N}$ and $x_{1},...,x_{n}\in A\\}$. Then:
$\bullet$ $F_{A}\neq\emptyset$ as $1\in F_{A}$.
$\bullet$ If $x,y\in F_{A}$, then there exist $n,m\in\mathbb{N}$ and
$x_{1},...,x_{n},y_{1},...,y_{n}\in A$ such that $x_{1}\cdot...\cdot
x_{n}\leqslant x$ and $y_{1}\cdot...\cdot y_{m}\leqslant y$. By 2.10,
$x_{1}\cdot...\cdot x_{n}\cdot y_{1}\cdot...\cdot y_{m}\leqslant x\cdot y$ and
hence $x\cdot y\in F_{A}$.
$\bullet$ By the definition of $F_{A}$, if $x\in F_{A}$ and $y\in B$ such that
$x\leqslant y$, then $y\in F_{A}$.
Finally, it is clear that if $F_{A}$ is contained in every filter containing
$A$. ∎

######  Definition 3.11.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra and $A$ be a nonempty subset of
$B$ that satisfies the finite product property. The proper filter $F_{A}$ is
called the filter generated by $A$. When $A=\\{x\\}$ is a singleton, $F_{x}$
is called the principal filter generated by $\\{x\\}$.

######  Definition 3.12.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F$ be a proper filter of
$B$;
$\bullet$ $F$ is called a prime filter if for all $x,y\in B$ with $x+y\in F$,
then either $x\in F$ or $y\in F$.
$\bullet$ $F$ is called a maximal filter if it is maximal with respect to the
inclusion.
$\bullet$ $F$ is called an ultrafilter if for all $x\in B$, either $x\in F$ or
$\neg x\in F$ and not both.
$\bullet$ The family of all ultrafilters of $B$ is denoted by
$\mathcal{U}(B)$.

######  Lemma 3.13.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F^{\prime}$ be a proper
filter of $B$. For every $x\in B\setminus F$, there exists a maximal filter
$M$ of $B$ containing $F^{\prime}$ such that $x\notin M$.

###### Proof.

The family $\mathcal{F}:=\\{F\subset B:$ F is a proper filter containing
$F^{\prime}$ and $x\notin F$ $\\}$ is partially ordered by the inclusion. Let
$\mathcal{C}\subset\mathcal{F}$ be a chain; then
$E=\bigcup\limits_{F\in\mathcal{C}}F$ is an upper bound of $\mathcal{C}$ in
$\mathcal{F}$. By Zorn’s Lemma, $\mathcal{F}$ contains a maximal element $M$.
∎

######  Lemma 3.14.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F$ be a proper filter of
$B$. $F$ is an ultrafilter if and only if it is a prime filter.

###### Proof.

Assume that $F$ is an ultrafilter of $B$. Let $x,y\in B\setminus F$. Since $F$
is an ultrafilter, one has $\neg x,\neg y\in F$. Seeking contradiction, assume
that $x+y\in F$, then $(\neg x)\cdot(x+y)=(\neg x)\cdot y\in F$ and hence
$((\neg x)\cdot y)\cdot(\neg y)=0\in F$, a contradiction.

Conversely, If $F$ is prime and $x\in B\setminus F$; then as $x+(\neg x)=1\in
F$, $\neg x$ must be in $F$ and hence $F$ is an ultrafilter. ∎

######  Lemma 3.15.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F$ be a proper filter of
$B$; $F$ is an ultrafilter if and only if it is a maximal filter.

###### Proof.

Assume that $F$ is an ultrafilter of $B$. Seeking contradiction, assume that
$F$ is not a maximal filter. Let $M$ be a maximal filter containing $F$ and
let $x\in M\setminus F$. Since $F$ is an ultrafilter, $\neg x\in F\subset M$.
Thus, $x,\neg x\in M$ and in particular $x\cdot\neg x=0\in M$, a
contradiction.

Conversely, let $F$ be maximal and $x\in B\setminus F$; notice that if $a\cdot
x\neq 0$ for all $a\in F$, then $A:=F\cup\\{x\\}$ satisfies the finite product
property. By 3.10, $F$ is contained in a proper filter $F_{A}$. Since $F$ is
maximal, $F=F_{A}$ which contradicts that fact that $x\in B\setminus F$.
Therefore, there exists $a\in F$ such that $a\cdot x=0$. Now, $a=a\cdot
1=a\cdot(x+(\neg x))=a\cdot(\neg x)\in F$, and since $a\cdot(\neg
x)\leqslant\neg x$, $\neg x$ must be in $F$ and hence $F$ is an ultrafilter. ∎

######  Corollary 3.16.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F$ be an ultrafilter of $B$;
if $x\in B$ such that $x\cdot a\neq 0$ for all $a\in F$, then $x\in F$.

###### Proof.

Notice that $A:=F\cup\\{x\\}$ satisfies the finite product property. Hence
$F_{A}$ is a proper filer containing $F$. By the maximality of $F$, one has
$F=F_{A}$ as wanted. ∎

######  Example 3.17.

Consider the power set Boolean algebra $(\mathcal{P}(X),\cup,\cap,\neg)$ and
$x\in X$; the principal filter $F_{x}$ generated by $x$ is an ultrafilter.
Moreover, as ultrafilters are maximal, if $F$ is an ultrafilter containing
$\\{x\\}$, then $F_{x}=F$.

######  Lemma 3.18.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F$ be a proper filter of
$B$; $F$ is an ultrafilter if and only if
$F=\varphi_{{\textstyle\mathstrut}F}^{-1}\\{1\\}$ for some homomorphism
$\varphi_{{\textstyle\mathstrut}F}:B\to\mathbb{Z}_{2}$.

###### Proof.

Assume that $F$ is an ultrafilter, and define
$\varphi_{{\textstyle\mathstrut}F}:B\to\mathbb{Z}_{2}$ by:

$\varphi_{{\textstyle\mathstrut}F}(x)=\begin{cases}0\quad&\text{if}\,x\notin
F\\\ 1\quad&\text{if}\,x\in F\\\ \end{cases}$

We claim that $\varphi_{{\textstyle\mathstrut}F}:B\to\mathbb{Z}_{2}$ is a
homomorphism.
$\bullet$ Clearly, $\varphi_{{\textstyle\mathstrut}F}(0_{B})=0$ and
$\varphi_{{\textstyle\mathstrut}F}(1_{B})=1$ as $1_{B}\in F$ and $0_{B}\notin
F$.
$\bullet$ Let $x,y\in B$; on one hand, if $x+y\in F$ then either $x\in F$ or
$y\in F$. Without loss of generality assume that $x\in F$, then
$\varphi_{{\textstyle\mathstrut}F}(x)=1$ and
$\varphi_{{\textstyle\mathstrut}F}(x)+\varphi_{{\textstyle\mathstrut}F}(y)=1+\varphi_{{\textstyle\mathstrut}F}(y)=1$.
Hence,
$1=\varphi_{{\textstyle\mathstrut}F}(x+y)=\varphi_{{\textstyle\mathstrut}F}(x)+\varphi_{{\textstyle\mathstrut}F}(y)$.
On the other hand, if $x+y\notin F$ then $x\notin F$ and $y\notin F$. Hence,
$0=\varphi_{{\textstyle\mathstrut}F}(x+y)=\varphi_{{\textstyle\mathstrut}F}(x)+\varphi_{{\textstyle\mathstrut}F}(y)$.
$\bullet$ Let $x,y\in B$; on one hand, if $x\cdot y\in F$, then $x\in F$ and
$y\in F$. Hence, $1=\varphi_{{\textstyle\mathstrut}F}(x\cdot
y)=\varphi_{{\textstyle\mathstrut}F}(x)\cdot\varphi_{{\textstyle\mathstrut}F}(y)$.
On the other hand, if $x\cdot y\notin F$, then either $x\notin F$ or $y\notin
F$. Without loss of generality assume that $x\notin F$, then
$0=\varphi_{{\textstyle\mathstrut}F}(x\cdot
y)=0\cdot\varphi_{{\textstyle\mathstrut}F}(y)=\varphi_{{\textstyle\mathstrut}F}(x)\cdot\varphi_{{\textstyle\mathstrut}F}(y)$.
Therefore, $\varphi_{{\textstyle\mathstrut}F}:B\to\mathbb{Z}_{2}$ is a
homomorphism and $F=\varphi_{{\textstyle\mathstrut}F}^{-1}\\{1\\}$.

Conversely, let $\varphi_{{\textstyle\mathstrut}F}:B\to\mathbb{Z}_{2}$ is a
homomorphism, and let $F:=\varphi_{{\textstyle\mathstrut}F}^{-1}\\{1\\}$.
Clearly, $F$ is a proper filter. If $x\in B\setminus F$, then
$\varphi_{{\textstyle\mathstrut}F}(\neg
x)=\neg\varphi_{{\textstyle\mathstrut}F}(x)=\neg 0=1$ and hence $\neg x\in F$.
∎

######  Definition 3.19.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra; and $I\subset B$:
$\bullet$ $I$ is called an ideal if for all $a,b\in I$ and $x\in R$, then
$x\cdot a,a+b\in I$.
$\bullet$ A proper ideal $I$ is called a prime ideal if for all $x,y\in B$
with $x\cdot y\in I$, then either $x\in I$ or $x\in I$.
$\bullet$ A proper ideal $I$ is called a maximal ideal if it is maximal with
respect the inclusion.

######  Lemma 3.20.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $I$ be a proper ideal of $B$;
$I$ is a prime ideal if and only if
$I=\varphi_{{\textstyle\mathstrut}I}^{-1}\\{0\\}$ for some homomorphism
$\varphi_{{\textstyle\mathstrut}I}:B\to\mathbb{Z}_{2}$.

###### Proof.

Assume that $I$ is a prime ideal, and define
$\varphi_{{\textstyle\mathstrut}I}:B\to\mathbb{Z}_{2}$ by:

$\varphi_{{\textstyle\mathstrut}I}(x)=\begin{cases}0\quad&\text{if}\,x\in I\\\
1\quad&\text{if}\,x\notin I\\\ \end{cases}$

We claim that $\varphi_{{\textstyle\mathstrut}I}:B\to\mathbb{Z}_{2}$ is a
homomorphism.
$\bullet$ Clearly, $\varphi_{{\textstyle\mathstrut}I}(0_{B})=0$ and
$\varphi_{{\textstyle\mathstrut}I}(1_{B})=1$ as $1_{B}\notin I$ and $0_{B}\in
I$.
$\bullet$ Let $x,y\in B$; notice that $x+y\in I$ if and only if $x,y\in I$.
Indeed, if $x\notin I$ and $x+y\in I$, then by the absorption law one has
$x\cdot(x+y)=x+x\cdot y=x\in I$, a contradiction. Now, if $x+y\in I$ then
$x,y\in I$ and hence
$0=\varphi_{{\textstyle\mathstrut}I}(x+y)=\varphi_{{\textstyle\mathstrut}I}(x)+\varphi_{{\textstyle\mathstrut}I}(y)$.
If $x+y\notin I$ then $x,y\notin I$ and hence
$1=\varphi_{{\textstyle\mathstrut}I}(x+y)=\varphi_{{\textstyle\mathstrut}I}(x)+\varphi_{{\textstyle\mathstrut}I}(y)$
as $1+1=1$.
$\bullet$ Let $x,y\in B$; if either $x\in I$ or $y\in I$, then $x\cdot y\in I$
and hence $0=\varphi_{{\textstyle\mathstrut}I}(x\cdot
y)=\varphi_{{\textstyle\mathstrut}I}(x)\cdot\varphi_{{\textstyle\mathstrut}I}(y)$.
If $x,y\notin I$, then as $I$ is prime $x\cdot y\notin I$ and hence
$1=\varphi_{{\textstyle\mathstrut}I}(x\cdot
y)=\varphi_{{\textstyle\mathstrut}I}(x)\cdot\varphi_{{\textstyle\mathstrut}I}(y)$.

Conversely, let $\varphi_{{\textstyle\mathstrut}I}:B\to\mathbb{Z}_{2}$ is a
homomorphism such that $I:=\varphi_{{\textstyle\mathstrut}I}^{-1}\\{0\\}$.
Clearly, $I$ is a proper ideal. Seeking contradiction, assume that $x,y\in
B\setminus I$ such that $x\cdot y\in I$, then
$0=\varphi_{{\textstyle\mathstrut}I}(x\cdot
y)=\varphi_{{\textstyle\mathstrut}I}(x)\cdot\varphi_{{\textstyle\mathstrut}I}(y)=1$,
a contradiction. ∎

######  Corollary 3.21.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F\subset B$. $F$ is an
ultrafilter of $B$ if and only if $B\setminus F$ is a prime ideal.

###### Proof.

Apply 3.18 and 3.20. ∎

From 3.14, 3.15, 3.18 and 3.20, we have:

######  Corollary 3.22.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra, and $F\subset B$ be a proper
filter. The following are equivalent:
($a$) $F$ is an ultrafilter.
($b$) $F$ is a prime filter.
($c$) $F$ is a maximal filter.
($d$) $B\setminus F$ is a prime ideal.

######  Notation 3.23.

Let $B$ be a Boolean algebra and $x\in B$, we put:

$U_{x}=\\{F\in\mathcal{U}(B):x\in F\\}.$

######  Lemma 3.24.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra. Then:
($a$) For all $x,y\in B$, $U_{x}\cap U_{y}=U_{x\cdot y}$.
($b$) If $x_{1},...,x_{n}\in B$, then
$\bigcap\limits_{i=1}^{n}U_{x_{i}}=\emptyset$ if and only if
$x_{1}\cdot...\cdot x_{n}=0$.
($c$) For all $x,y\in B$, $U_{x}\cup U_{y}=U_{x+y}$.
($d$) For all $x\in B$, $\mathcal{U}(B)\setminus U_{x}=U_{\neg x}$.
($e$) The family $\\{U_{x}:x\in B\\}$ is a basis for a topology
$\mathcal{T}_{\mathcal{U}(B)}$ on $\mathcal{U}(B)$ in which $U_{x}$ is a
clopen subset for all $x\in B$.

###### Proof.

($a$) Let $x,y\in B$, then $F\in U_{x}\cap U_{y}$ if and only if $x\cdot y\in
F$.
($b$) By 3.10, 3.13 and 3.15, $U_{x}=\emptyset$ if and only if $x=0$. Applying
part ($a$), $\bigcap\limits_{i=1}^{n}U_{x_{i}}=\emptyset$ if and only if
$x_{1}\cdot...\cdot x_{n}=0$.
($c$) Let $x,y\in B$. If $F\in U_{x}\cup U_{y}$, then either $x\in F$ or $y\in
F$; in either case, $x+y$ must be in $F$. Conversely, if $F\in U_{x+y}$, then
$x+y\in F$ and hence either $x\in F$ or $y\in F$ as $F$ is prime.
($d$) Let $x\in B$, as $F$ is an ultrafilter then $x\notin F$ if and only if
$\neg x\in F$.
($e$) The family $\\{U_{x}:x\in B\\}$ being basis follows from part ($a$) and
the fact that $U_{1}=\mathcal{U}(B)$. Part ($c$) shows that $U_{x}$ is a
clopen subset for all $x\in B$. ∎

######  Theorem 3.25.

Let $B$ be a Boolean algebra, the topological space
$(\mathcal{U}(B),\mathcal{T}_{\mathcal{U}(B)})$ is a Stone’s space.

###### Proof.

$\bullet$ $\mathcal{U}(B)$ is Hausdorff. Indeed, let
$F_{1},F_{2}\in\mathcal{U}(B)$ be distinct ultrafilters and let $x\in
F_{1}\setminus F_{2}$, then $\neg x\in F_{2}$ and hence $F_{1}$ and $F_{2}$
are contained in the disjoint open sets $U_{x}$ and $U_{\neg x}$,
respectively.
$\bullet$ $\mathcal{U}(B)$ is compact. To this end, it suffices to show that
for arbitrary collection $\mathcal{C}$ of closed sets of $\mathcal{U}(B)$ that
satisfies the finite intersection property, then
$\bigcap\limits_{C\in\mathcal{C}}C\neq\emptyset$. Notice that for every
$C\in\mathcal{C}$, there exists $A_{C}\subset B$ such that
$C=\bigcap\limits_{x\in A_{C}}U_{x}$. Put
$A:=\bigcup\limits_{C\in\mathcal{C}}A_{C}$, and notice that as $\mathcal{C}$
satisfies the finite intersection property, the family $\\{U_{x}:x\in A\\}$
satisfies the finite intersection property and hence, by 3.24 part ($b$), $A$
satisfies the finite product property. Let $F_{A}$ be an ultrafilter of $B$
containing $A$. Therefore, $F_{A}\in\bigcap\limits_{x\in
A}U_{x}=\bigcap\limits_{C\in\mathcal{C}}C$.
$\bullet$ Since $(\mathcal{U}(B),\mathcal{T}_{\mathcal{U}(B)})$ is locally
compact and it admits a basis consists of clopen subsets, by 3.6 it is totally
disconnected. ∎

It is worth mentioning that if $G$ is a group acts on $X$; the Stone’s space
of the Boolean algebra $\faktor{Comm_{G}(X)}{\sim}$, given in 2.6, is called
the space of ends of $X$ [2].

######  Notation 3.26.

For $A\subset X$, the family of all principal filters $F_{a}$ such that $a\in
A$ is denoted by $Pr(A)$; that is
$Pr(A):=\\{F_{a}\in\mathcal{U}(\mathcal{P}(X)):a\in A\\}$.

######  Proposition 3.27.

Let $F\in\mathcal{U}(\mathcal{P}(X))$ and $A\in\mathcal{P}(X)$. $A\in F$ if
and only $F\in cl(Pr(A))$ where the closure is taking in the topological space
$\mathcal{U}(\mathcal{P}(X))$.

###### Proof.

Assume that $A\in F$, and let $U_{B}$ be a neighborhood of $F$. Since $A,B\in
F$, $A\cap B\neq\emptyset$. Pick $a\in A\cap B$, then $F_{a}\in U_{B}\cap
Pr(A)$.

Conversely, assume that $F\in cl(Pr(A))$. By 3.16, to show that $A\in F$, it
suffices to show that for every $B\in F$, $A\cap B\neq\emptyset$. Indeed, if
$B\in F$, then $F\in U_{B}$ and hence $U_{B}\cap Pr(A)\neq\emptyset$.
Therefore, there exists $a\in A$ such that $F_{a}\in U_{B}$, and hence $a\in
A\cap B$ as desired. ∎

######  Corollary 3.28.

Let $B$ be a Boolean algebra and $C$ be a clopen subset of $\mathcal{U}(B)$.
There exists $x\in B$ such that $C=U_{x}$, where
$U_{x}=\\{F\in\mathcal{U}(B):x\in F\\}$.

###### Proof.

On one hand, $C$ is an open subset of $\mathcal{U}(B)$, so $C$ there exists a
subset $A\subset B$ such that $C=\bigcup\limits_{x\in A}U_{x}$. On the other
hand, $C$ is a closed subset of the Hausdorff compact space $\mathcal{U}(B)$
and hence it is compact. In particular, there exist $x_{1},...,x_{n}\in A$
such that $C=\bigcup\limits_{i=1}^{n}U_{x_{i}}$. Put $x=x_{1}+...+x_{n}$, then
$C=U_{x}$ as desired. ∎

######  Theorem 3.29.

Every homomorphism $\varphi:B_{1}\to B_{2}$ between Boolean algebras induces a
continuous map $\widetilde{\varphi}:\mathcal{U}(B_{2})\to\mathcal{U}(B_{1})$
between their associated Stone’s spaces. If, moreover, $\varphi:B_{1}\to
B_{2}$ is an isomorphism, then its induced map
$\widetilde{\varphi}:\mathcal{U}(B_{2})\to\mathcal{U}(B_{1})$ is a
homeomorphism.

###### Proof.

First notice that if $F$ is an ultrafilter of $B_{2}$, then $\varphi^{-1}(F)$
is an ultrafilter of $B_{1}$. Consider the map
$\widetilde{\varphi}:\mathcal{U}(B_{2})\to\mathcal{U}(B_{1})$ that sends
$F\in\mathcal{U}(B_{2})$ to $\varphi^{-1}(F)$. Now, if $U_{x}$ is a clopen
subset of $\mathcal{U}(B_{1})$, then
$\widetilde{\varphi}^{-1}(U_{x})=U_{\varphi(x)}$ as
$F\in\widetilde{\varphi}^{-1}(U_{x})$ if and only if
$\widetilde{\varphi}(F)\in(U_{x})$ if and only if $\varphi^{-1}(F)\in(U_{x})$
if and only if $\varphi(x)\in F$, an hence
$\widetilde{\varphi}:\mathcal{U}(B_{2})\to\mathcal{U}(B_{1})$ is continuous.
It is straightforward to see that
$\widetilde{\varphi}:\mathcal{U}(B_{2})\to\mathcal{U}(B_{2})$ is a
homeomorphism when $\varphi:B_{1}\to B_{2}$ is an isomorphism. ∎

###  3.3. Stone’s space of a Boolean algebra $B$ as the family of all
homomorphisms from $B$ to $\mathbb{Z}_{2}$

The aim of this section is to equip $Hom(B,\mathbb{Z}_{2})$ with a topology
such that $Hom(B,\mathbb{Z}_{2})$ is a Stone’s space homeomorphic to
$(\mathcal{U}(B),\mathcal{T}_{\mathcal{U}(B)})$. That goal is achieved by
embedding $Hom(B,\mathbb{Z}_{2})$ in a Stone’s space where the image of
$Hom(B,\mathbb{Z}_{2})$ under that embedding is a closed subset.

######  Lemma 3.30.

Let $X$ be a non-empty set.
($a$) The family $2^{X}:=\\{f:X\to\mathbb{Z}_{2}:$ $f$ is a function $\\}$
admits a topology $\mathcal{T}_{2^{X}}$ such that
$(2^{X},\mathcal{T}_{2^{X}})$ is homeomorphic to $\prod\limits_{x\in
X}\mathbb{Z}_{2}$ equipped with the product topology considering
$\mathbb{Z}_{2}$ as a discrete space.
($b$) The subsets $0_{a}:=\\{f\in 2^{X}:f(a)=0\\}$ and $1_{a}:=\\{f\in
2^{X}:f(a)=1\\}$ are clopen subsets of $(2^{X},\mathcal{T}_{2^{X}})$ for all
$a\in X$ and that the family $\mathcal{S}:=\\{0_{a},1_{a}:a\in X\\}$ is a
subbasis for the topology $\mathcal{T}_{2^{X}}$ on $2^{X}$.
($c$) $2^{X}$ is a Stone’s space.

###### Proof.

($a$) Define the map $\varphi:2^{X}\to\mathbb{Z}_{2}$ that sends each $f\in
2^{X}$ to $(f(x))_{x\in X}\in\prod\limits_{x\in X}\mathbb{Z}_{2}$. Since
$\varphi:2^{X}\to\mathbb{Z}_{2}$ is a bijection, the family
$\mathcal{T}_{2^{X}}:=\\{\varphi^{-1}(U):U\subset\prod\limits_{x\in
X}\mathbb{Z}_{2}$ is open$\\}$ is a topology on $2^{X}$ such that
$(2^{X},\mathcal{T}_{2^{X}})$ is homeomorphic to $\prod\limits_{x\in
X}\mathbb{Z}_{2}$.
($b$) Notice that for $a\in X$, a subset of $\prod\limits_{x\in
X}\mathbb{Z}_{2}$ that has the form $\prod\limits_{x\in X}U_{x}$, where
$U_{x}=\mathbb{Z}_{2}$ for all $x\neq a$ and $U_{x}=\\{0\\}$ or
$U_{x}=\\{1\\}$, if $x=a$, is a subbasis element and the family of all such
subsets consists a subbasis for $\prod\limits_{x\in X}\mathbb{Z}_{2}$. If
$U_{a}=\\{0\\}$, then $\varphi^{-1}(\prod\limits_{x\in X}U_{x})=0_{a}$; and if
$U_{a}=\\{1\\}$, then $\varphi^{-1}(\prod\limits_{x\in X}U_{x})=1_{a}$. Hence,
the family $\mathcal{S}:=\\{0_{a},1_{a}:a\in X\\}$ is a subbasis. Moreover, as
$2^{X}=0_{a}\cup 1_{a}$ and $0_{a}\cap 1_{a}=\emptyset$ for all $a\in X$, the
subsets $0_{a}:=\\{f\in 2^{X}:f(a)=0\\}$ and $1_{a}:=\\{f\in 2^{X}:f(a)=1\\}$
are clopen.
($c$) $2^{X}$ is Hausdorff as $\mathbb{Z}_{2}$ is Hausdorff. By Tychonoff’s
Theorem, $2^{X}$ is compact. From part($b$), $(2^{X},\mathcal{T}_{2^{X}})$ has
a basis consists of clopen subsets. By 3.6, $2^{X}$ is a Stone’s space. ∎

######  Theorem 3.31.

Let $Hom(B,\mathbb{Z}_{2}):=\\{f:B\to\mathbb{Z}_{2}:$ $f$ is a
homomorphism$\\}$ where $(B,+_{{\textstyle\mathstrut}B},\cdot,\neg)$ is a
Boolean algebra. Then $Hom(B,\mathbb{Z}_{2})$ is a closed subset of the
Stone’s space $2^{B}$. In particular, $Hom(B,\mathbb{Z}_{2})$ is a Stone’s
space.

###### Proof.

Let $a\in B$, and consider the projection map $\pi_{a}:\prod\limits_{x\in
X}\mathbb{Z}_{2}\to\mathbb{Z}_{2}$ and the map
$\varphi:2^{X}\to\mathbb{Z}_{2}$ defined in the proof of 3.30. Since
$\varphi\circ\pi_{a}:2^{B}\to\mathbb{Z}_{2}$ is continuous, then
$(\varphi\circ\pi_{a})^{-1}\\{0\\}=\\{f\in 2^{B}:f(a)=0\\}$ and
$(\varphi\circ\pi_{a})^{-1}\\{1\\}=\\{f\in 2^{B}:f(a)=1\\}$ are clopen for all
$a\in B$. For all $a,b\in B$, put, $A_{0}:=\\{f\in 2^{B}:f(0_{B})=0\\}$,
$A_{1}:=\\{f\in 2^{B}:f(1_{B})=1\\}$, $A_{a+b}:=\\{f\in
2^{B}:f(a+_{{\textstyle\mathstrut}B}b)=f(a)+f(b)\\}$, and $A_{a\cdot
b}:=\\{f\in 2^{B}:f(a\cdot_{{\textstyle\mathstrut}B}b)=f(a)\cdot f(b)\\}$.
Notice that $A_{0}$ and $A_{1}$ are closed subsets by 3.30($b$). Moreover:
$\bullet$ $A_{a+b}$ is the union of the closed sets:

$\\{f\in 2^{B}:f(a+_{{\textstyle\mathstrut}B}b)=0\\}\cap\\{f\in
2^{B}:f(a)=0\\}\cap\\{f\in 2^{B}:f(b)=0\\},$ $\\{f\in
2^{B}:f(a+_{{\textstyle\mathstrut}B}b)=1\\}\cap\\{f\in
2^{B}:f(a)=1\\}\cap\\{f\in 2^{B}:f(b)=0\\},$ $\\{f\in
2^{B}:f(a+_{{\textstyle\mathstrut}B}b)=1\\}\cap\\{f\in
2^{B}:f(a)=0\\}\cap\\{f\in 2^{B}:f(b)=1\\},$ $\\{f\in
2^{B}:f(a+_{{\textstyle\mathstrut}B}b)=1\\}\cap\\{f\in
2^{B}:f(a)=1\\}\cap\\{f\in 2^{B}:f(b)=1\\}.$

$\bullet$ $A_{a\cdot b}$ is the union of the closed sets:

$\\{f\in 2^{B}:f(a\cdot_{{\textstyle\mathstrut}B}b)=0\\}\cap\\{f\in
2^{B}:f(a)=0\\}\cap\\{f\in 2^{B}:f(b)=0\\},$ $\\{f\in
2^{B}:f(a\cdot_{{\textstyle\mathstrut}B}b)=0\\}\cap\\{f\in
2^{B}:f(a)=1\\}\cap\\{f\in 2^{B}:f(b)=0\\},$ $\\{f\in
2^{B}:f(a\cdot_{{\textstyle\mathstrut}B}b)=0\\}\cap\\{f\in
2^{B}:f(a)=0\\}\cap\\{f\in 2^{B}:f(b)=1\\},$ $\\{f\in
2^{B}:f(a\cdot_{{\textstyle\mathstrut}B}b)=1\\}\cap\\{f\in
2^{B}:f(a)=1\\}\cap\\{f\in 2^{B}:f(b)=1\\}.$

Therefore, $Hom(B,\mathbb{Z}_{2})=A_{0}\cap A_{1}\cap(\bigcap\limits_{a,b\in
B}A_{a+b})\cap(\bigcap\limits_{a,b\in B}A_{a\cdot b})$ is a closed subset of
the Stone’s space $2^{B}$ and hence it is a Stone’s space. ∎

######  Theorem 3.32.

Let $(B,+_{{\textstyle\mathstrut}B},\cdot,\neg)$ be Boolean algebra. The
Stone’s space $Hom(B,\mathbb{Z}_{2})$ is homeomorphic to $\mathcal{U}(B)$.

###### Proof.

By 3.18, every $f\in Hom(B,\mathbb{Z}_{2})$ determines an ultrafilter
$F_{f}=f^{-1}(\\{1\\})$, and for every ultrafilter $F\in\mathcal{U}(B)$ there
is a homomorphism $f_{F}\in Hom(B,\mathbb{Z}_{2})$ such that
$f^{-1}_{F}(\\{1\\})=F$. Define
$\varphi:Hom(B,\mathbb{Z}_{2})\to\mathcal{U}(B)$ by $\varphi(f):=F_{f}$ for
all $f\in Hom(B,\mathbb{Z}_{2})$. Given a basis element
$U_{x}=\\{F\in\mathcal{U}(B):x\in F\\}$, where $x\in B$, then:

$\begin{split}\varphi^{-1}(U_{x})&=\\{f\in Hom(B,\mathbb{Z}_{2}):F_{f}\in
U_{x}\\}\\\ &=\\{f\in Hom(B,\mathbb{Z}_{2}):x\in F_{f}=f^{-1}(\\{1\\})\\}\\\
&=\\{f\in Hom(B,\mathbb{Z}_{2}):f(x)=1\\}\end{split}$

which is a clopen subset of $Hom(B,\mathbb{Z}_{2})$, because $\\{f\in
Hom(B,\mathbb{Z}_{2}):f(x)=1\\}\\\ =Hom(B,\mathbb{Z}_{2})\cap\\{f\in
2^{B}:f(x)=1\\}$ and $\\{f\in 2^{B}:f(x)=1\\}$ is a clopen subset of $2^{B}$
as seen in 3.31. Therefore, $\varphi:Hom(B,\mathbb{Z}_{2})\to\mathcal{U}(B)$
is a bijective continuous map between compact Hausdorff spaces and hence it is
a homeomorphism as desired. ∎

##  4\. The Stone’s representation theorems

Every Boolean algebra (up to isomorphism) is the clopen Boolean algebra
associated with some Stone’s space; and conversely, every Stone’s space (up to
homeomorphism) is the space of all ultrafilters of some Boolean algebra. These
tow facts are the core of the Stone’s representation theorems.

######  Theorem 4.1.

(Stone’s representation theorem for Boolean algebras) Up to isomorphism, every
Boolean algebra is the clopen Boolean algebra for some Stone’s space. More
specifically, if $B$ is a Boolean algebra, then it is isomorphic to the
Boolean algebra $Clop(\mathcal{U}(B))$.

###### Proof.

Let $(B,+,\cdot,\neg)$ be a Boolean algebra; consider the map $\varphi:B\to
Clop(\mathcal{U}(B))$ that sends each $x\in B$ to $U_{x}\in
Clop(\mathcal{U}(B))$. It is straightforward to see that $\varphi:B\to
Clop(\mathcal{U}(B))$ is a homomorphism. It is, moreover, surjective by 3.28.
Now, let $x,y\in B$ be distinct elements, then either $x\nleqslant y$ or
$y\nleqslant x$. Without loss of generality, assume that $x\nleqslant y$, then
$y\notin F_{x}$, where $F_{x}$ is the principal filter generated by $x$. Let
$M_{x}$ be a maximal filter containing $F_{x}$ such that $y\notin M_{x}$.
Notice that $M_{x}\in U_{x}$ and $M_{x}\notin U_{y}$; hence $U_{x}\neq U_{y}$.
Consequently, $\varphi:B\to Clop(\mathcal{U}(B))$ is an isomorphism as
desired. ∎

######  Lemma 4.2.

Let $X$ be a Stone’s space. Then:
($a$) For every $x\in X$, the family $F^{x}:=\\{U\in Clop(X):x\in U\\}$ is an
ultrafilter of the clopen Boolean algebra $Clop(X)$.
($b$) If $F$ is an ultrafilter of $Clop(X)$, then $\bigcap\limits_{U\in
F}U=\\{x_{F}\\}$ for some $x_{F}\in X$.

###### Proof.

($a$) Since $X$ is totally disconnected, $Clop(X)$ is a basis for $X$. Since
$X$ is Hausdorff $F^{x}$ is a proper subset of $Clop(X)$. Evidently, $F^{x}$
is a filter; and if $V\in Clop(X)\setminus F^{x}$, then $U:=X\setminus V$ is a
clopen subset containing $x$ and hence it is contained in $F^{x}$. Thus,
$F^{x}$ is an ultrafilter.
($b$) Let $F$ be an ultrafilter of $Clop(X)$. Since $F$ is a family of closed
sets of the compact space $X$ and that it satisfies the finite intersection
property; hence $\bigcap\limits_{U\in F}U$ is non-empty. Seeking
contradiction, let $x,y\in\bigcap\limits_{U\in F}U$ be distinct elements. Let
$U_{x}$ be a clopen set containing $x$ such that $y\notin U_{x}$. Notice that
$U_{y}:=X\setminus U_{x}$ is a clopen subset containing $y$. Since $F$ is an
ultrafilter, either $U_{x}\in F$ or $U_{y}\in F$. Without loss of generality,
assume that $U_{x}\in F$, then as $y\notin U_{x}$,
$y\notin\bigcap\limits_{U\in F}U$, a contradiction. ∎

######  Theorem 4.3.

(Stone’s representation theorem for Stone’s spaces) Up to homeomorphism, every
Stone’s space is the ultrafilter space of some Boolean algebra. More
specifically, if $X$ is a Stone’s space then it is homeomorphic to
$\mathcal{U}(Clop(X))$.

###### Proof.

Let $X$ be a Stone’s space. Exploiting 4.2($a$), we define the map
$\psi:X\to\mathcal{U}(Clop(X))$ that sends each $x\in X$ to $F^{x}$. To show
that $\psi:X\to\mathcal{U}(Clop(X))$ is homeomorphism, it suffices to show
that it is continuous bijective map as $X$ and $\mathcal{U}(Clop(X))$ are
compact Hausdorff spaces.
$\bullet$ $\psi$ is injective. Indeed, let $x,y\in X$ be distinct elements.
Since $X$ is Hausdorff and $Clop(X)$ is a basis for $X$, there exists
$U_{x}\in Clop(X)$ containing $x$ such that $y\notin U_{x}$. Hence, $U_{x}\in
F^{x}\setminus F^{y}$.
$\bullet$ $\psi$ is surjective. Indeed, let $F$ be an ultrafilter of
$Clop(X)$, then $\bigcap\limits_{U\in F}U=\\{x_{F}\\}$, by 4.2($b$). Now, by
4.2($a$), $F^{x_{F}}$ is an ultrafilter and it is obvious that $F\subset
F^{x_{F}}$. Since ultrafilters are maximal, $F=F^{x_{F}}$.
$\bullet$ $\psi$ is continuous. To this end, notice that a basis element of
$\mathcal{U}(Clop(X))$ is of the form $U_{C}=\\{F\in\mathcal{U}(Clop(X)):C\in
F\\}$ for some a clopen subset $C$ of $X$. Now, $\psi^{-1}(U_{C})=\\{x\in
X:F^{x}\in U_{C}\\}=\\{x\in X:C\in F^{x}\\}=C$ and thus $\psi$ is continuous.
∎

######  Corollary 4.4.

($a$) Let $B_{1}$ and $B_{1}$ be Boolean algebras; then $B_{1}$ and $B_{1}$
are isomorphic if and only if their Stone’s spaces $\mathcal{U}(B_{1})$ and
$\mathcal{U}(B_{2})$ are homeomorphic.
($b$) Let $X$ and $Y$ be Stone’s spaces; then $X$ and $Y$ are homeomorphic if
and only if $Clop(X)$ and $Clop(Y)$ are isomorphic.

###### Proof.

($a$) Assume that the Stone’s spaces $\mathcal{U}(B_{1})$ and
$\mathcal{U}(B_{2})$ are homeomorphic. By 2.13, $Clop(\mathcal{U}(B_{1}))$ and
$Clop(\mathcal{U}(B_{2}))$ are isomorphic. By 4.1, $B_{1}$ and $B_{1}$ are
isomorphic. The other direction follows from 3.29.
($b$) If $X$ and $Y$ are homeomorphic, then by 2.13, $Clop(X)$ and $Clop(Y)$
are isomorphic. Conversely, assume that $Clop(X)$ and $Clop(Y)$ are
isomorphic, then by 3.29, $\mathcal{U}(Clop(X))$ and $\mathcal{U}(Clop(Y))$
are homeomorphic. Therefore, by 4.3, $X$ and $Y$ are homeomorphic. ∎

##  5\. Stone’s space of a Boolean algebra as the spectrum of its induced
Boolean ring

Alternatively the Stone’s space
$(\mathcal{U}(B),\mathcal{T}_{\mathcal{U}(B)})$ can be obtained as the
spectrum of some Boolean ring induced from the Boolean algebra $B$.

######  Definition 5.1.

Let $R$ be a commutative ring with identity $1$; and $I\subset R$:
$\bullet$ $I$ is called an ideal if it is closed under addition and $R\cdot
I=I$; that is, for all $a,b\in I$ and $r\in R$, then $r\cdot a+b\in I$.
$\bullet$ A proper ideal $I$ is called a prime ideal if for all $x,y\in R$
with $x\cdot y\in I$, then either $x\in I$ or $x\in I$.
$\bullet$ A proper ideal $I$ is called a maximal ideal if it is maximal with
respect to the inclusion.
$\bullet$ The family of all prime ideals of $R$ is called the spectrum of $R$
and is denoted by $Spec(R)$.

######  Lemma 5.2.

Let $R$ be a commutative ring with identity $1$. Then:
($a$) Let $A\subset R$, the collection $I_{A}:=\\{r_{1}\cdot
a_{1}+...+r_{n}\cdot a_{n}:n\in\mathbb{N},a_{i}\in A,r_{i}\in R\\}$ of all
finite linear combinations of elements of $A$ is an ideal called the ideal
generated by $A$.
($b$) Every proper ideal is contained in a maximal ideal.
($c$) Every maximal ideal is prime.

###### Proof.

($a$) Straightforward.
($b$) Let $I$ be a proper ideal of $R$. Apply the Zorn’s Lemma to the family
$\\{J\subset R:J$ is a proper ideal containing $I\\}$.
($c$) Let $M$ be a maximal ideal and $x,y\in R$ such that $x\cdot y\in M$.
Assume that $x\notin M$, then $M$ is contained properly in the ideal $I_{A}$
generated by $A:=M\cup\\{x\\}$. By the maximality of $M$, $I_{A}=R$. Since
$1\in I_{A}$, $1=a+r\cdot x$ for some $a\in M$. Now, $y=y\cdot
1=y\cdot(a+r\cdot x)=y\cdot a+y\cdot r\cdot x$. Since, $a,x\in M$, any linear
combination of $a,x$ is in $M$ and hence $y\in M$. ∎

Now we equip the spectrum of a ring with a topology called the Zariski
topology that makes it a compact space. If, moreover, the ring is Boolean,
then it spectrum space is a Stone’s space.

######  Lemma 5.3.

Let $R$ be a commutative ring with identity $1$. For $x\in R$, set

$V_{x}=\\{P\in Spec(R):x\notin P\\}.$

($a$) For all $x,y\in R$, $V_{x\cdot y}=V_{x}\cap V_{y}$.
($b$) The family $\\{V_{x}:x\in R\\}$ is a basis for a topology
$\mathcal{T}_{Spec(R)}$ on $Spec(R)$ called the Zariski topology.

###### Proof.

($a$) Follows from the fact that for any prime ideal $P$ and $x,y\in R$, then
$x\cdot y\notin P$ if and only if $x,y\notin P$.
($b$) Notice that an ideal $I$ is proper if and only if $1\notin I$ and hence
$V_{1}=Spec(R)$. Now, by part ($a$), $\\{V_{x}:x\in R\\}$ is a basis. ∎

######  Theorem 5.4.

Let $R$ be a commutative ring with identity $1$. The topological space
$(Spec(R),\mathcal{T}_{Spec(R)})$ is compact.

###### Proof.

Let $\\{O_{\alpha}:\alpha\in J\\}$ be an open cover of $Spec(R)$. Since
$\\{V_{x}:x\in R\\}$ is a basis for $Spec(R)$, there exists $A\subset R$ such
that $\bigcup\limits_{\alpha\in J}O_{\alpha}=\bigcup\limits_{x\in A}V_{x}$.
Let $I_{A}$ be the ideal generated by $A$. We claim that $I_{A}=R$. Seeking
contradiction, assume that $I_{A}$ is a proper ideal. By 5.2, we can choose a
maximal ideal $M$ containing $I_{A}$. Since $M$ is prime, $M\in
Spec(R)=\bigcup\limits_{x\in A}V_{x}$, so $M\in V_{x}$ for some $x\in A$ which
contradicts the fact that $A\subset I_{A}\subset M$. Now, since $1\in
R=I_{A}$, $1=r_{1}\cdot a_{1}+...+r_{n}\cdot a_{n}$ for some $n\in\mathbb{N}$,
$a_{i}\in A$ and $r_{i}\in R$. We claim that
$Spec(R)=\bigcup\limits_{i=1}^{n}V_{a_{i}}$. Indeed, for $P\in Spec(R)$, since
$1=r_{1}\cdot a_{1}+...+r_{n}\cdot a_{n}\notin P$, there exists $1\leq i\leq
n$ such that $r_{i}\cdot a_{i}\notin P$ and therefore $P\in V_{a_{i}}$. ∎

######  Definition 5.5.

Let $R$ be a ring with identity $1$ and $x\in R$:
$\bullet$ $x$ is called an idempotent element if $x^{2}=x$.
$\bullet$ $R$ is called a Boolean ring if every element of $R$ is an
idempotent element.

One can see that an element $x$ of a ring $R$ with identity $1$ is idempotent
if and only if $x\cdot(1-x)=0$. In particular, for any prime ideal $P$ and
idempotent element $x$, then as $x\cdot(1-x)=0\in P$, either $x\in P$ or
$1-x\in P$, and not both (as if $x,1-x\in P$ then $1=x+1-x\in P$). Moreover, a
Boolean ring is automatically a commutative with characteristic 2, that is
$2x=x+x=0$ and hence $x=-x$ for all $x\in R$. This follows from the
observation:
$\bullet$ $2x=4x-2x=4x^{2}-2x=(2x)^{2}-2x=0$.
$\bullet$ $x\cdot y=x\cdot y+2(x+y)=x\cdot y+x+y+(x+y)^{2}=x\cdot
y+x+y+x^{2}+y^{2}+x\cdot y+y\cdot x\\\ =2(x\cdot y+x+y)+y\cdot x=y\cdot x$.

######  Lemma 5.6.

Let $R$ be a commutative ring with identity $1$, and $x\in R$ be idempotent
element; then $V_{x}$ is a clopen subset of $(Spec(R),\mathcal{T}_{Spec(R)})$.

###### Proof.

Notice $Spec(R)\setminus V_{x}=V_{1-x}$. ∎

######  Theorem 5.7.

Let $R$ be a Boolean ring; then $(Spec(R),\mathcal{T}_{Spec(R)})$ is a Stone’s
space.

###### Proof.

By 5.4, $(Spec(R),\mathcal{T}_{Spec(R)})$ is compact, and by 5.6,
$(Spec(R),\mathcal{T}_{Spec(R)})$ admits a basis consists of clopen subsets.
Hence, it suffices to show that $(Spec(R),\mathcal{T}_{Spec(R)})$ is Hausdorff
to conclude that $(Spec(R),\mathcal{T}_{Spec(R)})$ is a Stone’s space. Let
$P_{1},P_{2}\in Spec(R)$ be distinct prime ideals and let $x\in P_{1}\setminus
P_{2}$. Notice that $V_{1-x}$ and $V_{x}$ are open disjoint neighborhoods of
$P_{1}$ and $P_{2}$, respectively. ∎

The relation between Boolean algebras and Boolean rings is demonstrated in
following proposition.

######  Proposition 5.8.

($a$) Every Boolean algebra induces a Boolean ring.
($b$) Every Boolean ring induces a Boolean algebra.

###### Proof.

($a$) Let $(B,+_{B},\cdot,\neg)$ be Boolean algebra; the $3$-tuple
$(R_{B},+_{R_{B}},\cdot)$ is a Boolean ring, where $R_{B}:=B$, and for all
$x,y\in R_{B}$, $x+_{R_{B}}y:=x\cdot(\neg y)+_{B}y\cdot(\neg x)$.
$(R_{B},+_{R_{B}},\cdot)$ is called the Boolean ring induced from the Boolean
algebra $(B,+_{B},\cdot,\neg)$.
($b$) Let $(R,+_{R},\cdot)$ be a Boolean ring; the $4$-tuple
$(B_{R},+_{B_{R}},\cdot,\neg)$ is a Boolean algebra, where $B_{R}:=R$, and for
all $x,y\in B_{R}$, $x+_{B_{R}}y:=x+_{R}y+_{R}x\cdot y$ and $\neg x:=1+_{R}x$.
$(B_{R},+_{B_{R}},\cdot,\neg)$ is called the Boolean algebra induced from the
Boolean ring $(R,+_{R},\cdot)$. ∎

Ideals (prime ideals) of a Boolean algebra are ideals (prime ideals) of its
induced Boolean ring.

######  Lemma 5.9.

Let $(B,+_{B},\cdot,\neg)$ be Boolean algebra and $(R_{B},+_{R_{B}},\cdot)$ be
induced Boolean ring:
($a$) If $I$ is an ideal of $B$, then $I$ is an ideal of $R_{B}$.
($a$) If $I$ is a prime ideal of $B$, then $I$ is a prime ideal of $R_{B}$.

###### Proof.

Assume that $I$ is an ideal (prime) of $B$. Since $(B,+_{B},\cdot,\neg)$ and
$(R_{B},+_{R_{B}},\cdot)$ are equipped with the same multiplication operation,
it suffices to show that if $x,y\in I$, then $x+_{R}y\in I$. Notice that
$x\cdot(\neg y),y\cdot(\neg x)\in I$ and hence $x\cdot(\neg y)+_{B}y\cdot(\neg
x)=x+_{R_{B}}y\in I$. ∎

######  Lemma 5.10.

Let $R$ be a Boolean ring;
($a$) For all $x,y\in R$, $V_{x+y+x\cdot y}=V_{x}\cup V_{y}$.
($b$) If $V$ is a clopen subset of $Spec(R)$, then there exists $x\in R$ such
that $V=V_{x}$.

###### Proof.

($a$) Let $P\in V_{x+y+x\cdot y}$; if $x,y\in P$, then $x+y+x\cdot y\in P$, a
contradiction. Without loss of generality, we may assume that $x\notin P$ and
hence $P\in V_{x}$ which implies that $V_{x+y+x\cdot y}\subset V_{x}\cup
V_{y}$. Conversely, assume that $P\in V_{x}\cup V_{y}$. Without loss of
generality, we may assume that $P\in V_{x}$ and hence $x\notin P$. Since
$2x=0$, $1-x=1+x\in P$ and hence $(1+x)\cdot(1+y)=1+(x+y+x\cdot y)\in P$.
Since $(x+y+x\cdot y)$ is idempotent, $(x+y+x\cdot y)\notin P$, and therefore
$P\in V_{x+y+x\cdot y}$, as wanted.
($b$) Assume that $V$ is a clopen subset of $Spec(R)$. On one hand, since $V$
is open there exists a subset $A\subset R$ such that $V:=\bigcup\limits_{a\in
A}V_{a}$. On the other hand, since $V$ is a closed subset of a compact space,
it is compact and hence there exist $a_{1},...,a_{n}\in A$ such that
$V=\bigcup\limits_{i=1}^{n}V_{a_{i}}$. By part ($a$), there exists $x\in R$
such that $V=\bigcup\limits_{i=1}^{n}V_{a_{i}}=V_{x}$. ∎

######  Theorem 5.11.

Let $(B,+_{{\textstyle\mathstrut}B},\cdot,\neg)$ be Boolean algebra and
$(R_{B},+_{R_{B}},\cdot)$ be the Boolean ring induced from
$(B,+_{B},\cdot,\neg)$. Then, $Spec(R_{B})$ is homeomorphic to
$\mathcal{U}(B)$.

###### Proof.

By 4.1 and 4.3, it suffices to show that the Boolean algebras
$Clop(Spec(R_{B}))$ and $Clop(\mathcal{U}(B))$ are isomorphic. To that end,
define $\varphi:Clop(\mathcal{U}(B))\to Clop(Spec(R_{B}))$ that maps $U_{x}$
to $V_{x}$ for all $x\in B$.
$\bullet$ $\varphi$ is homomorphism. Indeed, let $x,y\in B$, then:
\- $\varphi(U_{x}\cup
U_{y})=\varphi(U_{x+_{{\textstyle\mathstrut}B}y})=V_{x+_{{\textstyle\mathstrut}B}y}=V_{x+_{R_{B}}y+_{R_{B}}x\cdot
y}=V_{x}\cup V_{y}$, by 5.10 part ($a$).
\- $\varphi(U_{x}\cap U_{y})=\varphi(U_{x\cdot y})=V_{x\cdot y}=V_{x}\cap
V_{y}$, by 3.24 and 5.3.
\- Since $U_{0}=V_{0}=\emptyset$ and $U_{1}=V_{1}=B$, $\varphi$ is
homomorphism.
$\bullet$ $\varphi$ is injective. Indeed, let $U_{x},U_{y}\in
Clop(\mathcal{U}(B))$ be distinct and let $F\in U_{x}\setminus U_{y}$. Let
$P:=B\setminus F$, then by 5.9, $P\in Spec(R_{B})$. Moreover, since $x\in F$
and $y\notin F$, $x\notin P$ and $y\in P$ and hence $P\in V_{x}\setminus
V_{y}$.
$\bullet$ The surjectivity of $\varphi$ follows from 5.10 part ($b$).
Therefore, the homomorphism $\varphi:Clop(\mathcal{U}(B))\to
Clop(Spec(R_{B}))$ is an isomorphism as desired. ∎

##  6\. Compactifications of a discrete space

Let $X$ and $Y$ be topological spaces. $Y$ is called a compactification of $X$
if $Y$ is a compact Hausdorff space that contains a dense subset homeomorphic
to $X$. In the case of discrete spaces, every compactification of a discrete
space $X$ is determined by a sub-Boolean algebra
$\mathcal{B}\subset\mathcal{P}(X)$ that separates points of $X$; and
conversely, every sub-Boolean algebra $\mathcal{B}\subset\mathcal{P}(X)$ that
separates points of $X$ determines a compactification of $X$. Every
compactification of $X$ is totally disconnected. The domination relation
between any two compactifications of $X$ is determined by the relation between
their corresponding sub-Boolean algebras.

######  Lemma 6.1.

Let $X$ be a non-empty set and $\mathcal{B}\subset\mathcal{P}(X)$ be a sub-
Boolean algebra. The family $\\{F_{x}:x\in X\\}$, where
$F_{x}:=\\{A\in\mathcal{B}:x\in A\\}$, is a dense discrete subspace of
$\mathcal{U}(\mathcal{B})$.

###### Proof.

Notice that a basis element of $\mathcal{U}(\mathcal{B})$ is of the form
$U_{A}=\\{F\in\mathcal{U}(\mathcal{B}):A\in F\\}$, and that for $x\in X$,
$F_{x}:=\\{A\in\mathcal{B}:x\in A\\}$ is an ultrafilter of $\mathcal{B}$. Now,
for any $a\in A$, $F_{a}\in U_{A}\cap\\{F_{x}:x\in X\\}$ and hence
$\\{F_{x}:x\in X\\}$ is a dense subset of $\mathcal{U}(\mathcal{B})$.
Moreover, for any $x\in X$, the singleton $\\{F_{x}\\}$ is an open set as
$\\{F_{x}\\}=U_{\\{x\\}}$ is a singleton. Consequently, $\\{F_{x}:x\in X\\}$
is a discrete subspace of $\mathcal{U}(\mathcal{B})$. ∎

######  Definition 6.2.

Let $X$ be a non-empty set and $\mathcal{B}\subset\mathcal{P}(X)$ be a subset.
$\mathcal{B}$ is said to separate points of $X$ if for all $x,y\in X$ such
that $x\neq y$, there exists $A\in\mathcal{B}$ such that $x\in A$ and $y\in
X\setminus A$.

######  Theorem 6.3.

Let $X$ be a non-empty set and $\mathcal{B}\subset\mathcal{P}(X)$ be a sub-
Boolean algebra. If $\mathcal{B}$ separates points of $X$, then
$\mathcal{U}(\mathcal{B})$ is a compactification of $X$.

###### Proof.

Define $\alpha:X\to\mathcal{U}(\mathcal{B})$ that sends each $x\in X$ to
$F_{x}\in\mathcal{U}(\mathcal{B})$.
$\bullet$ $\alpha:X\to\mathcal{U}(\mathcal{B})$ is continuous as $X$ is
discrete.
$\bullet$ It is injective since $\mathcal{B}$ separates points of $X$. Indeed,
if $x,y\in X$ are distinct, then there exists $A\in\mathcal{B}$ such that
$x\in A$ and $y\in X\setminus A$. Because $A\in F_{x}$ and $X\setminus A\in
F_{y}$, $\alpha(x)\neq\alpha(y)$.
$\bullet$ $\alpha:X\to\mathcal{U}(\mathcal{B})$ is open as for all $x\in X$,
$\alpha(\\{x\\})=\\{F_{x}\\}=U_{\\{x\\}}$ by 6.1.
$\bullet$ $\alpha(X)=\\{F_{x}:x\in X\\}$ which is a dense subset of
$\mathcal{U}(\mathcal{B})$. Therefore, $\mathcal{U}(\mathcal{B})$ is a
compactification of $X$. ∎

######  Example 6.4.

Let $G$ be a group acts on $X$; then $Comm_{G}(X)$ is a sub-Boolean algebra of
$\mathcal{P}(X)$ that separates points of $X$. The compactification
$\mathcal{U}(Comm_{G}(X))$ of $X$ is called the end compactification.

######  Example 6.5.

Let $X$ be a non-empty set, the family
$\mathcal{B}_{0}:=\\{A\in\mathcal{P}(X):$ $A$ is finite or $X\setminus A$ is
finite$\\}$ is a sub-Boolean algebra of $\mathcal{P}(X)$ that separates points
of $X$. We will prove that the compactification $\mathcal{U}(\mathcal{B}_{0})$
is the one-point compactification of $X$ [2].

###  6.1. The one-point compactification of a discrete space

Let $Y:=X\cup\\{\infty\\}$ be the one-point compactification of the discrete
space $X$. It can be easily seen that a subset $A\subset Y$ is clopen if and
only if either $A$ is a finite subset of $X$ or $Y\setminus A$ is finite. We
aim to show that the one-point compactification of $X$ is the Stone’s space of
the Boolean algebra $\mathcal{B}_{0}:=\\{A\in\mathcal{P}(X):$ $A$ is finite or
$X\setminus A$ is finite$\\}$.

######  Theorem 6.6.

Given a non-empty set $X$, the family
$\mathcal{B}_{0}:=\\{A\in\mathcal{P}(X):$ $A$ is finite or $X\setminus A$ is
finite$\\}$ is a sub-Boolean algebra of $\mathcal{P}(X)$ that separates points
of $X$. Moreover, as a discrete space $\mathcal{U}(\mathcal{B}_{0})$ is the
one-point compactification of $X$.

###### Proof.

The first part is straightforward. Let $Y:=X\cup\\{\infty\\}$ be the one-point
compactification of $X$. As noticed a subset $A\subset Y$ is clopen if and
only if either $A$ is a finite subset of $X$ or $Y\setminus A$ is finite. In
particular, if $A\in Clop(Y)$, then
$A\setminus\\{\infty\\}\in\mathcal{B}_{0}$. Define the map
$\varphi:Clop(Y)\to\mathcal{B}_{0}$ that sends $A\in Clop(Y)$ to
$A\setminus\\{\infty\\}$. It is straightforward to see that
$\varphi:Clop(Y)\to\mathcal{B}_{0}$ is an isomorphism and $Y$ is a Stone’s
space. Hence, by 4.4, $Y=\mathcal{U}(Clop(Y))$ is homeomorphic to the Stone’s
space $\mathcal{U}(\mathcal{B}_{0})$. ∎

###  6.2. The Stone-$\breve{C}$ech compactification of a discrete space

Given a compactification $Y$ of $X$ with an embedding $\alpha:X\to Y$ such
that $\alpha(X)$ is dense in $Y$. The compactification $Y$ is called the
Stone-$\breve{C}$ech compactification if for any compact Hausdorff space $K$,
every continuous map $f:\alpha(X)\to K$ admits a continuous extension
$\widetilde{f}:Y\to K$. For the discrete space $X$, the topological space
$\mathcal{U}(\mathcal{P}(X))$ is a compactification of $X$; we aim to show
that $\mathcal{U}(\mathcal{P}(X))$ is the Stone-$\breve{C}$ech
compactification of $X$. We first give a characterization for the continuity
of a map between topological spaces in terms of convergence of filters.

######  Definition 6.7.

Let $X$ be a topological space, and $F$ be a filter of $\mathcal{P}(X)$. We
say that $F$ converges to $x\in X$ if for every neighborhood $U$ of $X$, there
exists $A\in F$ such that $A\subset U$.

######  Example 6.8.

Let $X$ be a topological space, and $x\in X$. The family $\mathcal{N}(x)$ of
all neighborhoods of $x$ is a filter called the neighborhood filter of $x$.
Clearly, $\mathcal{N}(x)$ converges to $x$. Moreover, a filter $F$ converges
to $x$ if and only if $\mathcal{N}(x)\subset F$.

A function $f:X\to Y$ between sets maps filters (ultrafilters) of
$\mathcal{P}(X)$ to filters (ultrafilters) of $\mathcal{P}(Y)$.

######  Lemma 6.9.

Let $f:X\to Y$ be a function. Then:
($a$) If $F$ is a filter of $\mathcal{P}(X)$, then $F_{f}:=\\{f(A):A\in F\\}$
is a filter of $\mathcal{P}(Y)$.
($b$) If $F$ is an ultrafilter of $\mathcal{P}(X)$, then $F_{f}:=\\{f(A):A\in
F\\}$ is an ultrafilter of $\mathcal{P}(Y)$.

###### Proof.

($a$) Let $F$ be a filter of $\mathcal{P}(X)$; observe that for any
$B\in\mathcal{P}(Y)$, $f(f^{-1}(B))=B$. Now:
$\bullet$ Let $B_{1},B_{2}\in F_{f}$, then there exist $A_{1},A_{2}\in F$ such
that $f(A_{1})=B_{1}$ and $f(A_{2})=B_{2}$. Since $A_{1}\cap A_{2}\in F$ and
$A_{1}\cap A_{2}\subset f^{-1}(B_{1}\cap B_{2})$ and hence $f^{-1}(B_{1}\cap
B_{2})\in F$. Therefore, $f(f^{-1}(B_{1}\cap B_{2}))=B_{1}\cap B_{2}\in
F_{f}$.
$\bullet$ $B\in F_{f}$ and $C\subset Y$ such that $B\subset C$. Let $A\in F$
such that $f(A)=B$; since $A\subset f^{-1}(C)$, then $f^{-1}(C)\in F$ and
hence $f(f^{-1}(C))=C\in F_{f}$.
($b$) Let $F$ be an ultrafilter of $\mathcal{P}(X)$, then $F_{f}$ is a filter
of $\mathcal{P}(Y)$. Moreover, if $B\subset Y$, then either $f^{-1}(B)\in F$
or $X\setminus f^{-1}(B)=f^{-1}(Y\setminus B)\in F$ and therefore either
$B=f(f^{-1}(B))\in F_{f}$ or $Y\setminus B=f(f^{-1}(Y\setminus B))\in F_{f}$.
∎

######  Lemma 6.10.

A map $f:X\to Y$ between topological spaces is continuous if and only if for
every $x\in X$ and every filter $F$ of $\mathcal{P}(X)$ that converges to $x$,
the filter $F_{f}$ converges to $f(x)$.

###### Proof.

Assume that $f:X\to Y$ is continuous, and $F$ is a filter of $\mathcal{P}(X)$
that converges to $x$. Let $V$ be a neighborhood of $f(x)$, then $f^{-1}(V)$
is a neighborhood of $x$ and consequently $f^{-1}(V)\in F$. Hence,
$V=f(f^{-1}(V))\in F_{f}$ which implies that $F_{f}$ converges to $f(x)$.

Conversely, assume that for every $x\in X$ and every filter $F$ of
$\mathcal{P}(X)$ that converges to $x$, the filter $F_{f}$ converges to
$f(x)$. Since the neighborhood filter $F=\mathcal{N}(x)$ of $x$ converges to
$x$, the filter $F_{f}$ converges to $f(x)$. Let $V\subset Y$ be a
neighborhood of $f(x)$, then $V\in F_{f}$ and therefore there exists $U\in F$
such that $f(U)=V$ which proves that $f:X\to Y$ is continuous. ∎

######  Theorem 6.11.

Let $X$ be a non-empty set equipped with the discrete topology. The
compactification $\mathcal{U}(\mathcal{P}(X))$ of $X$ is the
Stone-$\breve{C}$ech compactification.

###### Proof.

By 6.3, $\mathcal{U}(\mathcal{P}(X))$ is a compactification of $X$. Let $K$ be
a compact Hausdorff space, and $f:\alpha(X)\to K$ be a continuous map, where
$\alpha:X\to\mathcal{U}(\mathcal{P}(X))$ is the embedding defined in the proof
of 6.3.
Claim 1: For every $F\in\mathcal{U}(\mathcal{P}(X))$, $\bigcap\limits_{A\in
F}cl(f(\alpha(A)))$ is a singleton.
Proof of claim 1: Put $g:=f\circ\alpha$. Notice that the family $F$ satisfies
the finite intersection property and hence the family $\\{g(A):A\in F\\}$
satisfies the finite intersection property. Therefore, the family of the
closed sets $\\{cl(g(A)):A\in F\\}$ satisfies the finite intersection
property. Consequently, $\bigcap\limits_{A\in F}cl(g(A))\neq\emptyset$, as $K$
is compact. Since $F$ is an ultrafilter, for every $O\subset X$, either $O\in
F$ or $X\setminus O\in F$; in particular, if $O_{1},O_{2}\subset X$ are
disjoint subsets of $X$, then there exists $A\in F$ such that either $A\cap
O_{1}=\emptyset$ and $A\cap O_{2}\neq\emptyset$ or $A\cap O_{1}\neq\emptyset$
and $A\cap O_{2}=\emptyset$. Seeking contradiction assume that
$a,b\in\bigcap\limits_{A\in F}cl(g(A))$ are distinct, and choose two disjoint
neighborhoods $U_{a}$ and $U_{b}$ of $a$ and $b$, respectively. Since
$g^{-1}(U_{a})$ and $g^{-1}(U_{b})$ are disjoint, we may choose, without loss
of generality, $A\in F$ such that $A\cap g^{-1}(U_{a})\neq\emptyset$ and
$A\cap g^{-1}(U_{b})=\emptyset$. Hence, $g(A)\cap U_{a}\neq\emptyset$ and
$g(A)\cap U_{b}=\emptyset$, and thus $b\notin cl(g(A))$, a contradiction.
Therefore, there exists $a_{F}\in K$ such that $\bigcap\limits_{A\in
F}cl(g(A))=\\{a_{{\textstyle\mathstrut}F}\\}$.
Claim 2: Define $\widetilde{f}:\mathcal{U}(\mathcal{P}(X))\to K$ by
$\widetilde{f}(F)=a_{{\textstyle\mathstrut}F}$. Then $\widetilde{f}$ is an
extension of $f$.
Proof of claim 2: For $x\in X$, then
$\widetilde{f}(F_{x})=a_{{\textstyle\mathstrut}F_{x}}$. On the other hand,
$F_{x}\in\alpha(A)$ for all $A\in F_{x}$ and hence
$f(F_{x})\in\bigcap\limits_{A\in F}f(\alpha(A)))\subset\bigcap\limits_{A\in
F}cl(f(\alpha(A)))=\\{a_{{\textstyle\mathstrut}F_{x}}\\}$.
Claim 3: $cl(\widetilde{f}(U_{A}))=cl(g(A))$ for all $A\in\mathcal{P}(X)$;
where:

$U_{A}=\\{F\in\mathcal{U}(\mathcal{P}(X)):A\in F\\}.$

Proof of claim 3: Let $a\in\widetilde{f}(U_{A})$, then there exists $F\in
U_{A}$ such that $a=a_{{\textstyle\mathstrut}F}$ where
$\\{a_{{\textstyle\mathstrut}F}\\}=\bigcap\limits_{A\in F}cl(g(A))\subset
cl(g(A))$, and hence $cl(\widetilde{f}(U_{A}))\subset cl(g(A))$. On the other
hand, since $F_{a}\in U_{A}$ for all $a\in A$, thus
$g(a)=f(F_{a})=\widetilde{f}(F_{a})\in\widetilde{f}(U_{A})$. Therefore,
$cl(g(A))\subset cl(\widetilde{f}(U_{A}))$.
Claim 4: $\widetilde{f}:\mathcal{U}(\mathcal{P}(X))\to K$ is continuous.
Proof of claim 4: By 6.10, it suffices to show that if $\mathcal{F}$ is a
filter of $\mathcal{P}(\mathcal{U}(\mathcal{P}(X)))$ that converges to
$F\in\mathcal{U}(\mathcal{P}(X))$, then $\mathcal{F}_{\widetilde{f}}$
converges to $\widetilde{f}(F)$. Notice that for all $A\in F$, $F\in U_{A}$,
and since $\mathcal{F}$ converges to $F$ and $U_{A}$ is a neighborhood of $F$,
$U_{A}$ must be in $\mathcal{F}$. Consequently, $\\{\widetilde{f}(U_{A}):A\in
F\\}\subset\mathcal{F}_{\widetilde{f}}$. Since $\widetilde{f}(U_{A})\subset
cl(\widetilde{f}(U_{A}))=cl(g(A))$ and $\mathcal{F}_{\widetilde{f}}$ is a
filter, so $\\{cl(g(A)):A\in F\\}\subset\mathcal{F}_{\widetilde{f}}$. Now, let
$O$ be a neighborhood of $\widetilde{f}(F)=a_{{\textstyle\mathstrut}F}$.
Choose a relatively compact neighborhood $V$ of
$a_{{\textstyle\mathstrut}F}\in\bigcap\limits_{A\in F}cl(g(A))$ such that
$V\subset cl(V)\subset O$. As $V\cap g(A)\neq\emptyset$ for all $A\in F$ and
that $F_{g}$ is an ultrafilter, hence, by 3.16, $V\in F_{g}$ and therefore
there exists $B\in F$ such that $V=g(B)$. Thus,
$cl(V)\in\mathcal{F}_{\widetilde{f}}$ and $cl(V)\subset O$ which implies that
$\mathcal{F}_{\widetilde{f}}$ converges to $\widetilde{f}(F)$. As a result,
$\mathcal{U}(\mathcal{P}(X))$ is the Stone-$\breve{C}$ech compactification of
$X$. ∎

######  Corollary 6.12.

Let $X$ be a non-empty set equipped with the discrete topology. The
Stone-$\breve{C}$ech compactification $\beta X$ of $X$ is totally
disconnected. Moreover, $\beta X$ is homeomorphic to each of the following
Stone’s spaces:
($1$)
$(\mathcal{U}(\mathcal{P}(X)),\mathcal{T}_{\mathcal{U}(\mathcal{P}(X))})$.
($2$) $Hom(\mathcal{P}(X),\mathbb{Z}_{2})$.
($3$) $Spec(R_{\mathcal{P}(X)})$, such that $(R_{\mathcal{P}(X)},+,\cdot)$ is
the ring where $R_{\mathcal{P}(X)}=\mathcal{P}(X)$ and for all
$A,B\in\mathcal{P}(X)$:
$\bullet$ $A+B:=A\Delta B=(A\setminus B)\cup(B\setminus A)$;
$\bullet$ $A\cdot B:=A\cap B$.

###### Proof.

Apply 6.11, 3.32 and 5.11. ∎

######  Corollary 6.13.

Let $X$ be a non-empty set equipped with the discrete topology, and $Y$ be a
compactification of $X$, then $Y$ is totally disconnected.

###### Proof.

Let $\beta X$ be the Stone-$\breve{C}$ech compactification of $X$, then $Y$ is
a quotient space of the totally disconnected space $\beta X$, and therefore
$Y$ is totally disconnected. ∎

When $X$ is infinite, its Stone-$\breve{C}$ech compactification is non-
metrizable.

######  Theorem 6.14.

Let $X$ be an infinite discrete space. The Stone-$\breve{C}$ech
compactification $\beta X$ is non-metrizable.

###### Proof.

Seeking contradiction, assume that $\beta X$ is metrizable. Choose
$\omega\in\beta X\setminus X$ and a sequence $(x_{n})_{n\in\mathbb{N}}$ of
distinct elements of $X$ that converges to $\omega$. Let
$A:=\\{x_{2n}:n\in\mathbb{N}\\}$, and $\chi_{A}:X\to\\{0,1\\}$ be the
characteristic function of $A$. Since $\chi_{A}$ is continuous, it admits a
continuous extension $\widetilde{\chi_{A}}:\beta X\to\\{0,1\\}$. On one hand,
we have $\widetilde{\chi_{A}}(cl_{\beta X}A)=\\{1\\}$, and hence
$\widetilde{\chi_{A}}(\omega)=1$. On the other hand, we have
$\widetilde{\chi_{A}}(x_{2n-1})=0$ for all $n\in\mathbb{N}$ and
$(x_{2n-1})_{n\in\mathbb{N}}$ converges to $\omega$, a contradiction. ∎

Now we prove the converse of 6.3.

######  Theorem 6.15.

Let $Y$ be a compactification of the discrete space $X$; there exists a sub-
Boolean algebra $\mathcal{B}\subset\mathcal{P}(X)$ that separates points of
$X$ such that $Y$ is homeomorphic to $\mathcal{U}(\mathcal{B})$.

###### Proof.

Without loss of generality we may assume that $X\subset Y$. Put
$\mathcal{B}:=\\{A\cap X:A\in Clop(Y)\\}$. It is straightforward to see that
$\mathcal{B}\subset\mathcal{P}(X)$ is a sub-Boolean algebra that separates
points of $X$. Therefore, $\mathcal{U}(\mathcal{B})$ is a compactification of
$X$. Consider the a map $\varphi:Clop(Y)\to\mathcal{B}$ that sends every $A\in
Clop(Y)$ to $A\cap X$. It is evident to see that
$\varphi:Clop(Y)\to\mathcal{B}$ is a surjective homomorphism. Furthermore, let
$A,B\in Clop(Y)$ be distinct. Without loss of generality assume that
$A\setminus B$ is non-empty. Since $A\setminus B$ is an open subset of $Y$ and
hence $(A\setminus B)\cap X\neq\emptyset$. Thus, $A\cap X\neq B\cap X$, which
shows that $\varphi:Clop(Y)\to\mathcal{B}$ is injective. By 4.4, $Y$ is
homeomorphic to $\mathcal{U}(\mathcal{B})$. ∎

######  Corollary 6.16.

Let $X$ be a non-empty set equipped with the discrete topology. A topological
space $Y$ is a compactification of $X$ if and only if there exists a sub-
Boolean algebra $\mathcal{B}\subset\mathcal{P}(X)$ that separates points of
$X$ such that $Y$ is homeomorphic to $\mathcal{U}(\mathcal{B})$.

###### Proof.

Apply 6.3 and 6.15. ∎

Recall if $Y$ and $Z$ are compactifications of $X$, we say that $Y$ dominates
$Z$, or $Z$ is dominated by $Y$, if there exists a continuous function $f:Y\to
Z$ such that $f|_{X}=id_{X}$. The continuous function $f:Y\to Z$ is called a
domination.

######  Theorem 6.17.

Let $Y:=\mathcal{U}(\mathcal{B}_{Y})$ and $Z:=\mathcal{U}(\mathcal{B}_{Z})$ be
compactifications of the discrete space $X$, where $\mathcal{B}_{Y}$ and
$\mathcal{B}_{Z}$ sub-Boolean algebras of $\mathcal{P}(X)$ that separate
points of $X$. $Y$ dominates $Z$ if and only if $\mathcal{B}_{Z}$ is
isomorphic to a sub-Boolean algebra of $\mathcal{B}_{Y}$.

###### Proof.

By 2.13, a continuous function
$f:\mathcal{U}(\mathcal{B}_{Y})\to\mathcal{U}(\mathcal{B}_{Z})$ induces a
homomorphism $\widetilde{f}:Clop(\mathcal{U}(\mathcal{B}_{Z}))\to
Clop(\mathcal{U}(\mathcal{B}_{Y}))$ given by $\widetilde{f}(A)=f^{-1}(A)$ for
all $A\in Clop(\mathcal{U}(\mathcal{B}_{Z}))$. It can be easily seen that if
$f:\mathcal{U}(\mathcal{B}_{Y})\to\mathcal{U}(\mathcal{B}_{Z})$ is surjective,
then $\widetilde{f}:Clop(\mathcal{U}(\mathcal{B}_{Z}))\to
Clop(\mathcal{U}(\mathcal{B}_{Y}))$ is injective. On the other hand, we may
assume that $\mathcal{B}_{Z}\subset\mathcal{B}_{Y}$. Now the inclusion
$\varphi:\mathcal{B}_{Z}\to\mathcal{B}_{Y}$ is a homomorphism. By 3.29, it
induces a continuous map
$\widetilde{\varphi}:\mathcal{U}(\mathcal{B}_{Y})\to\mathcal{U}(\mathcal{B}_{Z})$
that sends $F\in\mathcal{U}(\mathcal{B}_{Y})$ to
$\varphi^{-1}(F)=F\cap\mathcal{B}_{Z}$. Notice that every ultrafilter $F_{Z}$
of $\mathcal{B}_{Z}$ is contained in an ultrafilter $F_{Y}$ of
$\mathcal{B}_{Y}$ with $F_{Z}=F_{Y}\cap\mathcal{B}_{Z}$ which shows that
$\widetilde{\varphi}:\mathcal{U}(\mathcal{B}_{Y})\to\mathcal{U}(\mathcal{B}_{Z})$
is surjective. Moreover, since $X$ is homeomorphic to $\\{F^{Y}_{x}:x\in X\\}$
and homeomorphic to $\\{F^{Z}_{x}:x\in X\\}$ where
$F^{Y}_{x}=\\{A\in\mathcal{B}_{Y}:x\in A\\}$ and
$F^{Z}_{x}=\\{A\in\mathcal{B}_{Z}:x\in A\\}$, so
$\widetilde{\varphi}:\mathcal{U}(\mathcal{B}_{Y})\to\mathcal{U}(\mathcal{B}_{Z})$
is a domination. ∎

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