# Topological Representation of Double Boolean Algebras111Declarations of
interest: none

Prosenjit Howlader<EMAIL_ADDRESS>Mohua Banerjee
<EMAIL_ADDRESS>Department of Mathematics and Statistics, Indian Institute of
Technology Kanpur, Kanpur 208016, Uttar Pradesh, India

###### Abstract

In formal concept analysis, the collection of protoconcepts of any context
forms a double Boolean algebra (dBa) which is fully contextual. Semiconcepts
of a context form a pure dBa. The present article is a study on topological
representation results for dBas, and in particular, those for fully contextual
and pure dBas. The representation is in terms of object oriented protoconcepts
and semiconcepts of a context, notions that were defined by the authors
earlier for introducing negation in the study of object oriented concepts. In
this work, a context on topological spaces (CTS) is considered, and the focus
is on a special kind of CTS in which the relation defining the context as well
as the converse of the relation are continuous with respect to the topologies.
Such CTS are denoted as “CTSCR”. It is observed that clopen object oriented
protoconcepts of a CTSCR form a fully contextual dBa, while clopen object
oriented semiconcepts form a pure dBa. Every dBa is shown to be quasi-
embeddable into the dBa of clopen object oriented protoconcepts of a
particular CTSCR. The quasi-embedding turns into an embedding in case of a
contextual dBa, and into an isomorphism, when the dBa is fully contextual. For
pure dBas, one obtains an isomorphism with the algebra of clopen object
oriented semiconcepts of the CTSCR. Representation of finite dBas and Boolean
algebras are also addressed in the process. Abstraction of properties of this
CTSCR leads to the definition of “Stone contexts”. Stone contexts and CTSCR-
homeomorphisms are seen to form a category, denoted Scxt. Furthermore,
correspondences are observed between dBa isomorphisms and CTSCR-
homeomorphisms. This motivates a study of categorical duality of dBas,
constituting the second part of the article. Pure dBas and fully contextual
dBas along with dBa isomorphisms form categories, denoted PDBA and FCDBA
respectively. It is established that PDBA is equivalent to FCDBA, while FCDBA
and PDBA are dually equivalent to Scxt.

###### keywords:

Formal concept analysis, protoconcept, semiconcept, object oriented concept,
double Boolean algebra, continuous relation, categorical duality.

###### MSC:

[2010] 06B15 , 06D50 , 06E15, 06E75

##  1 Introduction

In the field of lattice theory, topological representation and categorical
duality have been a subject of study for many years. Well-known instances of
topological representation and duality results are those for Boolean algebras
by Stone [31, 32], and for distributive lattices by Stone [33] and Priestley
[27]. Duality for general lattices was investigated by Hartonas and Dunn [13],
while that for bounded lattices was studied by Urquhart [34] and Hartung [14].
[14] is of particular relevance to the present work, as formal concept
analysis (FCA) [10] was made use of to obtain the results. This article
addresses topological representation results for a class of algebraic
structures that arise in FCA, namely the double Boolean algebras. Furthermore,
categorical duality is investigated for the classes of fully contextual and
pure double Boolean algebras.

FCA was introduced by Wille [38] and has given rise to a rich body of work,
both in theory and applications. In particular, a lot of theoretical
development has been made in the direction of algebraic and category-theoretic
studies related to FCA (see e.g. [40, 16, 15, 46, 22, 12, 19]). The central
objects in FCA are contexts and concepts. A context is a triple $(G,M,R)$,
where $G,M$ are sets of objects and properties respectively, and $R$ is a
relation between them with $gRm$ signifying that object $g$ has the property
$m$. A concept of a context is a pair $(A,B),A\subseteq G,B\subseteq M$, such
that $B$ contains exactly the properties that all objects of $A$ have, while
$A$ contains exactly the objects having all the properties in $B$. The set of
all concepts of a context forms a complete lattice, called the concept lattice
of the context. Moreover, any complete lattice is isomorphic to a concept
lattice of some context. Wille initiated the study of the negation of a
concept [39, 40], to make formal concept analysis more useful for
representation, processing, and acquisition of conceptual knowledge. If set-
complement is used to define the negation of a concept, one encounters the
problem of closure. So the notion of a concept was generalized to that of a
semiconcept and a protoconcept [40]. The set of all protoconcepts fails to
form a lattice. However, it leads to the algebraic structure of a double
Boolean algebra (dBa) [40]. In particular, protoconcepts form fully contextual
dBas [37]. In comparison, the set of all semiconcepts forms a subalgebra of
the algebra of protoconcepts. Abstraction of properties of this subalgebra
yields a pure dBa [40]. In this work, we obtain topological representation
results for dBas in general, as well as for fully contextual and pure dBas.

A concept lattice is also dually isomorphic to the lattice formed by the set
of all object oriented concepts, the latter defined in the context of rough
set theory. Proposed by Pawlak [26], rough set theory is a well-established
mathematical tool to handle incompleteness in data. The theory hinges on the
notions of approximation space and lower and upper approximation operators
defined on the space. The Pawlakian definition of approximation space has been
generalized; a generalized approximation space [45] consists of a set $W$ and
a binary relation $R$ on $W$. On such a space, different definitions of lower
and upper approximation operators are found in literature. In the ones used
here, the lower approximation of any subset $B$ of $W$ collects every element
of $W$ such that all elements $R$-related to it lie within $B$, while the
upper approximation of $B$ contains all elements of $W$ that are $R$-related
to at least one element of $B$. Many comparative studies between rough set
theory and FCA have been made, e.g. in [21, 20, 30, 8, 42, 43, 44, 24, 9, 17,
18]. The approximation operators in a generalized approximation space have
been imported into FCA and named as necessity and possibility operators [8].
Using these operators, Düntsch and Gediga [8] defined property oriented
concepts, and Yao [43] defined object oriented concepts. Besides property
oriented and object oriented concepts, various other types of concepts and
related notions have been defined in rough set theory and studied from the
algebraic and categorical points of view (see e.g. [23, 41, 11, 17, 18]). The
present authors introduced negation in the study of object oriented concepts
of a context, in the lines of Wille’s study on negation of concepts. Notions
of object oriented semiconcepts and object oriented protoconcepts of a context
were defined in [17, 18]. It was shown that the algebra of protoconcepts is
isomorphic to that of object oriented protoconcepts [18], and the algebra of
semiconcepts is dually isomorphic to that of object oriented semiconcepts
[17]. The representation results proved in this work involve object oriented
protoconcepts and object oriented semiconcepts of certain special contexts.

Wille [40] constructed a standard context for each dBa D, denoted
$\mathbb{K}(\textbf{D})$. The context consists of the sets of all primary
filters and ideals of D, and a relation $\Delta$ such that $F\Delta I$ if and
only if $F\cap I\neq\emptyset$ for any primary filter $F$ and primary ideal
$I$. It has been proved that every dBa D is quasi-embedded into the algebra of
protoconcepts of $\mathbb{K}(\textbf{D})$. The quasi-embedding becomes an
embedding in case of pure dBas [3]. The special case of the representation
result for finite dBas was also given by Wille in [40]. In this work, we equip
the sets of all primary filters and ideals of D with certain topologies. The
resulting structure, denoted $\mathbb{K}_{pr}^{T}(\textbf{D})$, is an instance
of a context on topological spaces (CTS), that consists of a pair of
topological spaces and a relation between the domains. In fact,
$\mathbb{K}_{pr}^{T}(\textbf{D})$ is proved to be an instance of a special
kind of CTS, denoted as CTSCR, in which the relation and its converse are both
continuous.

Continuity (or hemicontinuity) of a relation was introduced by Berge [4]. We
work with a more general definition of continuity that is considered in [1]. A
relation $R$ is said to be continuous when the upper and lower inverses of any
open set under $R$ are both open; we observe that the upper and lower inverses
of a set are just the images of the set under the necessity and possibility
operators (respectively) in FCA. Our approach for proving the representation
results presented in this paper is based on the approach adopted by Hartung
[14] to deal with bounded lattices, and the above observation.

That $\mathbb{K}_{pr}^{T}(\textbf{D})$ is a CTSCR, is proved by using the
prime ideal theorem [22, 19] for dBas. We consider clopen object oriented
semiconcepts and protoconcepts of a CTS, namely where the component sets of
the object oriented semiconcepts and protoconcepts are both clopen (that is,
closed and open) in the respective topologies. It is established that the set
of clopen object oriented protoconcepts (semiconcepts) of a CTSCR forms a
fully contextual dBa (pure dBa). The following representation results are then
obtained. Any dBa D is quasi-embeddable into the algebra of clopen object
oriented protoconcepts of $\mathbb{K}_{pr}^{T}(\textbf{D})$. In case D is
contextual, the quasi-embedding is an embedding. If D is fully contextual, the
quasi-embedding turns into an isomorphism. On the other hand, it is shown that
the largest pure subalgebra $\textbf{D}_{p}$ of any dBa D is isomorphic to the
algebra of clopen object oriented semiconcepts of
$\mathbb{K}_{pr}^{T}(\textbf{D})$. This results in an isomorphism theorem for
pure dBas, since $\textbf{D}_{p}=\textbf{D}$ if D is pure. The representation
theorems for fully contextual and pure dBas yield an isomorphism theorem for
Boolean algebras as well. It is observed that a representation result for
finite dBas can be obtained in terms of object oriented protoconcepts and
semiconcepts; we show that it is also obtained as a special case from the
above-mentioned representation result for dBas.

The second part of the paper focusses on categorical duality results for dBas.
For a dBa D, the CTSCR $\mathbb{K}_{pr}^{T}(\textbf{D})$ is observed to have
some special properties, which, on abstraction, lead to the definition of a
Stone context. Stone contexts and CTSCR-homeomorphisms form a category,
denoted Scxt. Fully contextual and pure dBas along with dBa isomorphisms also
form categories, denoted FCDBA and PDBA respectively. It is shown that FCDBA
is equivalent to PDBA, whereas PDBA is dually equivalent to Scxt. As a
consequence, one obtains a dual equivalence between FCDBA and Scxt.

The paper has been arranged as follows. Section 2 gives the preliminaries that
are required in this work. Some algebraic properties of dBas have been
obtained in Section 3 – these are used later in Sections 5 and 6. In Section
4, CTS, clopen object oriented semiconcepts and protoconcepts, and CTSCR are
defined and studied in relation to dBas. The representation theorems for dBas
are proved in Section 5. Section 6 presents the study on categorical duality
of fully contextual and pure dBas. Section 7 concludes the work.

In our presentation, the symbols $\forall$, $\Rightarrow$, $\Leftrightarrow$,
and, or and $not$ will be used with the usual meanings in the metalanguage.

##  2 Preliminaries

In the following subsections, we present preliminaries related to dBas, object
oriented concepts, semiconcepts and protoconcepts. Our primary references are
[10, 40, 8, 43, 42, 17, 18, 19].

###  2.1 Concept, semiconcept and protoconcept of a context

Let us recall the definitions and some properties of contexts, concepts,
semiconcepts and protoconcepts. The following are taken from [10].

######  Definition 1.

A context is a triple $\mathbb{K}:=(G,M,R)$, where $G$ is a set of objects,
$M$ a set of properties, and $R\subseteq G\times M$.
The complement of a context $\mathbb{K}:=(G,M,R)$ is the context
$\mathbb{K}^{c}:=(G,M,-R)$, where $-R:=(G\times M)\setminus R$.
For any $A\subseteq G,B\subseteq M$, consider the sets
$A^{\prime}:=\\{m\in M:\forall g\in G(g\in A\implies gRm)\\}$ and
$B^{\prime}:=\\{g\in G:\forall m\in M(m\in B\implies gRm)\\}$.
$(A,B)$ is a concept of $\mathbb{K}$, when $A^{\prime}=B$ and $B^{\prime}=A$.
The set of all concepts is denoted by $\mathfrak{B}(\mathbb{K})$.
A partial order relation $\leq$ is given on $\mathfrak{B}(\mathbb{K})$ as
follows. For concepts $(A_{1},B_{1})$ and $(A_{2},B_{2})$,
$(A_{1},B_{1})\leq(A_{2},B_{2})$ if and only if $A_{1}\subseteq A_{2}$
(equivalently $B_{2}\subseteq B_{1})$.

######  Definition 2.

Let $\mathbb{K}:=(G,M,R)$ be a context and $H\subseteq G$, $N\subseteq M$.
$\mathbb{S}:=(H,N,R\cap(H\times N))$ is called a subcontext of $\mathbb{K}$.

######  Definition 3.

Let $\mathbb{K}_{1}:=(G_{1},M_{1},R_{1})$ and
$\mathbb{K}_{2}:=(G_{2},M_{2},R_{2})$ be two contexts. A context homomorphism
$f:\mathbb{K}_{1}\rightarrow\mathbb{K}_{2}$ is a pair of maps
$(\alpha,\beta)$, where $\alpha:G_{1}\rightarrow G_{2}$,
$\beta:M_{1}\rightarrow M_{2}$ are such that
$gR_{1}m\iff\alpha(g)R_{2}\beta(m),~{}\mbox{for all}~{}g\in
G_{1}~{}\mbox{and}~{}m\in M_{1}.$
If $\alpha$ and $\beta$ are injective then $f$ is called a context embedding.
$f$ is called a context isomorphism if $\alpha$, $\beta$ are bijective – in
that case one says that $\mathbb{K}_{1}$ is isomorphic to $\mathbb{K}_{2}$.

For the context $\mathbb{K}:=(G,M,R)$ and the identity maps $id_{G},id_{M}$ on
$G,M$ respectively, $id_{\mathbb{K}}:\mathbb{K}\rightarrow\mathbb{K}$ denotes
the context isomorphism $(id_{G},id_{M})$.
Composition of context homomorphisms is defined component wise:

${\mathbb{K}^{T}_{1}}$${\mathbb{K}^{T}_{2}}$${\mathbb{K}^{T}_{3}}$$\scriptstyle{f_{1}:=(\alpha\mathbin{,}\beta)}$$\scriptstyle{f_{2}\circ
f_{1}=(\delta\circ\alpha\mathbin{,}\gamma\circ\beta)}$$\scriptstyle{f_{2}:=(\delta\mathbin{,}\gamma)}$

######  Proposition 1.

For a context isomorphism $f:=(\alpha,\beta)$ from $\mathbb{K}_{1}$ to
$\mathbb{K}_{2}$, $g:=(\alpha^{-1},\beta^{-1})$ is a context isomorphism from
$\mathbb{K}_{2}$ to $\mathbb{K}_{1}$ such that $f\circ g=id_{\mathbb{K}_{2}}$
and $g\circ f=id_{\mathbb{K}_{1}}$.

$g$ is called the inverse of $f$.

######  Notation 1.

For a relation $R\subseteq G\times M$, $R^{-1}$ denotes the converse of $R$,
that is $R^{-1}\subseteq M\times G$ and $yR^{-1}x$ if and only if $xRy$.
$R(x):=\\{y\in M~{}:~{}xRy\\}$, and $R^{-1}(y):=\\{x\in G~{}:~{}xRy\\}$, for
any $x\in G$, $y\in M$.
$X^{c}$ denotes the complement of a subset $X$ of $G$ (or $M$) and
$\mathcal{P}(X)$, the power set of any set $X$.
$\mathcal{B}(\mathbb{K})$ denotes the set of all concepts of the context
$\mathbb{K}$. For a concept $(A,B)$, $A:=ext((A,B))$ is its extent, its
intent, $int((A,B))$, is $B$.

While attempting to define the negation of a formal concept, it was noticed
that the set-complement could not be used as it resulted in a problem of
closure. So generalizations of the notion of a concept, namely semiconcepts
and protoconcepts were introduced [40].

######  Definition 4.

For a context $\mathbb{K}:=(G,M,R)$ and $A\subseteq G,B\subseteq M$, the pair
$(A,B)$ is called a semiconcept of $\mathbb{K}$ if and only if $A^{\prime}=B$
or $B^{\prime}=A$. $(A,B)$ is a protoconcept of $\mathbb{K}$ if and only if
$A^{\prime\prime}=B^{\prime}~{}(\mbox{equivalently}~{}A^{\prime}=B^{\prime\prime})$.

######  Notation 2.

$\mathfrak{H}(\mathbb{K})$ denotes the set of all semiconcepts and the set of
all protoconcepts is denoted by $\mathfrak{P}(\mathbb{K})$.

######  Observation 1.

$\mathfrak{H}(\mathbb{K})\subseteq\mathfrak{P}(\mathbb{K})$.

On $\mathfrak{P}(\mathbb{K})$, operations $\sqcap,\sqcup,\neg,\lrcorner,\top$
and $\bot$ are defined as follows. For any protoconcepts $(A_{1},B_{1}),\\\
(A_{2},B_{2}),(A,B)$,
$(A_{1},B_{1})\sqcap(A_{2},B_{2}):=(A_{1}\cap A_{2},(A_{1}\cap
A_{2})^{\prime}),$ $(A_{1},B_{1})\sqcup(A_{2},B_{2}):=((B_{1}\cap
B_{2})^{\prime},B_{1}\cap B_{2}),$
$\neg(A,B):=(A^{c},A^{c\prime}),~{}\lrcorner(A,B):=(B^{c\prime},B^{c}),~{}\top:=(G,\emptyset)$
and $\bot:=(\emptyset,M)$.

######  Notation 3.

$\underline{\mathfrak{P}}(\mathbb{K})$ denotes the algebra of protoconcepts of
a context $\mathbb{K}$, which is the abstract algebra
$(\mathfrak{P}(\mathbb{K}),\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$ of type
$(2,2,1,1,0,0)$ formed by $\mathfrak{P}(\mathbb{K})$.
$\underline{\mathfrak{H}}(\mathbb{K})$ denotes the algebra of semiconcepts of
$\mathbb{K}$, which is the algebra
$(\mathfrak{H}(\mathbb{K}),\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$.
$\underline{\mathfrak{H}}(\mathbb{K})$ is a subalgebra of
$\underline{\mathfrak{P}}(\mathbb{K})$.

###  2.2 Double Boolean algebras

The structure of a dBa was proposed [40] as an abstraction of the algebra of
protoconcepts of a context. A pure dBa is the result of abstraction of the
algebra of semiconcepts.

######  Definition 5.

[40] An abstract algebra
$\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$ satisfying the
following properties is called a double Boolean algebra (dBa). For any
$x,y,z\in D$,

$(x\sqcap x)\sqcap y=x\sqcap y$

$x\sqcap y=y\sqcap x$

$x\sqcap(y\sqcap z)=(x\sqcap y)\sqcap z$

$\neg(x\sqcap x)=\neg x$

$x\sqcap(x\sqcup y)=x\sqcap x$

$x\sqcap(y\vee z)=(x\sqcap y)\vee(x\sqcap z)$

$x\sqcap(x\vee y)=x\sqcap x$

$\neg\neg(x\sqcap y)=x\sqcap y$

$x\sqcap\neg x=\bot$

$\neg\bot=\top\sqcap\top$

$\neg\top=\bot$

$(x\sqcup x)\sqcup y=x\sqcup y$

$x\sqcup y=y\sqcup x$

$x\sqcup(y\sqcup z)=(x\sqcup y)\sqcup z$

$\lrcorner(x\sqcup x)=\lrcorner x$

$x\sqcup(x\sqcap y)=x\sqcup x$

$x\sqcup(y\wedge z)=(x\sqcup y)\wedge(x\sqcup z)$

$x\sqcup(x\wedge y)=x\sqcup x$

$\lrcorner\lrcorner(x\sqcup y)=x\sqcup y$

$x\sqcup\lrcorner x=\top$

$\lrcorner\top=\bot\sqcup\bot$

$\lrcorner\bot=\top$

  * (12)

$(x\sqcap x)\sqcup(x\sqcap x)=(x\sqcup x)\sqcap(x\sqcup x),$

where $x\vee y:=\neg(\neg x\sqcap\neg y)$ and $x\wedge y:=\lrcorner(\lrcorner
x\sqcup\lrcorner y)$.
A quasi-order relation $\sqsubseteq$ on $D$ is given as follows. For any
$x,y\in D$,
$x\sqsubseteq y\iff x\sqcap y=x\sqcap x~{}\mbox{and}~{}x\sqcup y=y\sqcup y$.

######  Definition 6.

[40, 37] Let D be a dBa.

  1. 1.

D is contextual if and only if the quasi-order $\sqsubseteq$ on D is a partial
order.

  2. 2.

D is fully contextual if and only if it is contextual and, for each $y\in
D_{\sqcap}$ and $x\in D_{\sqcup}$ with $y_{\sqcup}=x_{\sqcap}$, there is a
unique $z\in D$ with $z_{\sqcap}=y$ and $z_{\sqcup}=x$.

  3. 3.

D is called pure if, for all $x\in D$, either $x\sqcap x=x$ or $x\sqcup x=x$.

Contextual dBas are also referred to as regular dBas in literature [5].

######  Theorem 1.

[40, 37] $\underline{\mathfrak{P}}(\mathbb{K})$ is a fully contextual dBa,
while $\underline{\mathfrak{H}}(\mathbb{K})$ is a pure dBa.

In the following, let $\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$
be a dBa.

######  Notation 4.

$D_{\sqcap}:=\\{x\in D~{}:~{}x\sqcap x=x\\}$, $D_{\sqcup}:=\\{x\in
D~{}:~{}x\sqcup x=x\\}$, $D_{p}:=D_{\sqcap}\cup D_{\sqcup}$.
For $x\in D$, $x_{\sqcap}:=x\sqcap x$ and $x_{\sqcup}:=x\sqcup x$.

######  Proposition 2.

[36]

  1. 1.

$\textbf{D}_{\sqcap}:=(D_{\sqcap},\sqcap,\vee,\neg,\bot,\neg\bot)$ is a
Boolean algebra. The partial order relation in $\textbf{D}_{\sqcap}$ is the
restriction of the quasi-order $\sqsubseteq$ on $D$ to $D_{\sqcap}$ and is
denoted by $\sqsubseteq_{\sqcap}$.

  2. 2.

$\textbf{D}_{\sqcup}:=(D_{\sqcup},\sqcup,\wedge,\lrcorner,\top,\lrcorner\top)$
is a Boolean algebra. For $\textbf{D}_{\sqcup}$, the partial order relation is
the restriction of $\sqsubseteq$ to $D_{\sqcup}$ and is denoted by
$\sqsubseteq_{\sqcup}$.

  3. 3.

For any $x,y\in D$, $x\sqsubseteq y$ if and only if $x\sqcap x\sqsubseteq
y\sqcap y$ and $x\sqcup x\sqsubseteq y\sqcup y$, that is,
$x_{\sqcap}\sqsubseteq_{\sqcap}y_{\sqcap}$ and
$x_{\sqcup}\sqsubseteq_{\sqcup}y_{\sqcup}$.

######  Proposition 3.

[3] If D is a pure dBa, the quasi-order $\sqsubseteq$ on $D$ becomes a partial
order, that is, every pure dBa is contextual.

We state below some results giving further properties of dBas. In [22], Kwuida
obtained the following.

######  Proposition 4.

[22] Let $x,y,a\in D$.

  1. 1.

$x\sqcap\bot=\bot$ and $x\sqcup\bot=x\sqcup x$, that is $\bot\sqsubseteq x$.

  2. 2.

$x\sqcup\top=\top$ and $x\sqcap\top=x\sqcap x$, that is $x\sqsubseteq\top$.

  3. 3.

$x=y$ implies that $x\sqsubseteq y$ and $y\sqsubseteq x$.

  4. 4.

$x\sqsubseteq y$ and $y\sqsubseteq x$ if and only if $x\sqcap x=y\sqcap y$ and
$x\sqcup x=y\sqcup y$.

  5. 5.

$x\sqcap y\sqsubseteq x$ and $x\sqcap y\sqsubseteq y.$ Dually $y\sqsubseteq
x\sqcup y$ and $x\sqsubseteq x\sqcup y$.

  6. 6.

$x\sqsubseteq y$ implies $x\sqcap a\sqsubseteq y\sqcap a$ and $x\sqcup
a\sqsubseteq y\sqcup a$.

In [19], one observed the following.

######  Proposition 5.

Let $x,y\in D$.

  1. 1.

$\neg x=(\neg x)_{\sqcap}\in D_{\sqcap}$ and $\lrcorner x=(\lrcorner
x)_{\sqcup}\in D_{\sqcup}$.

  2. 2.

$x\sqsubseteq y$ if and only if $\neg y\sqsubseteq\neg x$ and $\lrcorner
y\sqsubseteq\lrcorner x$.

  3. 3.

$\neg\neg x=x\sqcap x$ and $\lrcorner\lrcorner x=x\sqcup x$.

  4. 4.

$x\sqcap x$, $x\vee y\in D_{\sqcap}$ and $x\sqcup x$, $x\wedge y\in
D_{\sqcup}$.

  5. 5.

$\neg(x\vee y)=\neg x\sqcap\neg y$ and $\neg(x\sqcap y)=\neg x\vee\neg y$.

  6. 6.

$\lrcorner(x\wedge y)=\lrcorner x\sqcup\lrcorner y$ and $\lrcorner(x\sqcup
y)=\lrcorner x\wedge\lrcorner y$.

  7. 7.

$x\sqsubseteq\lrcorner y$ if and only if $y\sqsubseteq\lrcorner x$.

  8. 8.

$\neg x\sqsubseteq y$ if and only if $\neg y\sqsubseteq x$.

  9. 9.

$\neg\neg\neg x=\neg x$.

Using the properties of dBas given above, one can show

######  Proposition 6.

[40] $\textbf{D}_{p}:=(D_{p},\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$ is the
largest pure subalgebra of D. Moreover, if D is pure,
$\textbf{D}_{p}=\textbf{D}$.

For the set $\mathfrak{P}(\mathbb{K})$ of all protoconcepts, Kwuida [22]
showed that
$(x\sqcup y)\sqcap(x\sqcup\lrcorner y)\leq x\sqcup x$ and $x\sqcap
x\leq(x\sqcap y)\sqcup(x\sqcap\neg y)$, for any
$x,y\in\mathfrak{P}(\mathbb{K})$.
He then redefined Wille’s dBa by adding these two properties as axioms to
Definition 5. These new algebras were also termed as ‘dBas’ by Kwuida. In
[19], it has been established that Kwuida’s class of dBas is equivalent to
that defined by Wille. The result (Corollary 1 below) is obtained by proving a
number of intermediate properties of a dBa that are given in the following
theorem.

######  Theorem 2.

[19] For all $x,y\in D$,

  1.   2. 1.

$x\sqcap\neg(x\sqcup y)=\bot$.

  3. 2.

$\neg(x\sqcup y)=\neg(x\sqcup y)\sqcap\neg x$.

  4. 3.

$x\sqcap y=x\sqcap\neg(x\sqcap\neg y)$.

  5. 4.

$x\sqcup(y\sqcap\neg x)=x\sqcup(y\sqcap y)$.

  6. 5.

$(x\sqcap y)\sqcup(x\sqcap\neg y)=(x\sqcap x)\sqcup(x\sqcap x)$.

  7. 6.

$x\sqcup\lrcorner(x\sqcap y)=\top$.

  8. 7.

$\lrcorner(x\sqcap y)=\lrcorner(x\sqcap y)\sqcup\lrcorner x$.

  9. 8.

$x\sqcup y=x\sqcup\lrcorner(x\sqcup\lrcorner y)$.

  10. 9.

$x\sqcap(y\sqcup\lrcorner x)=x\sqcap(y\sqcup y)$.

  11. 10.

$(x\sqcup y)\sqcap(x\sqcup\lrcorner y)=(x\sqcup x)\sqcap(x\sqcup x)$.

######  Corollary 1.

[19] For all $x,y\in D$, the following hold.

  1. 1.

$(x\sqcup y)\sqcap(x\sqcup\lrcorner y)\sqsubseteq x\sqcup x.$

  2. 2.

$x\sqcap x\sqsubseteq(x\sqcap y)\sqcup(x\sqcap\neg y)$.

We shall also require the notion of a dBa homomorphism.

######  Definition 7.

Let
$\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top_{\textbf{D}},\bot_{\textbf{D}})$
and
$\textbf{M}:=(M,\sqcup,\sqcap,\neg,\lrcorner,\top_{\textbf{M}},\bot_{\textbf{M}})$
be two dBas. A map $h:D\rightarrow M$ is called a dBa homomorphism from D to
M, if the following hold for all $a,b\in D$: $h(a\sqcap b)=h(a)\sqcap h(b)$,
$h(a\sqcup b)=h(a)\sqcup h(b)$, $h(\neg a)=\neg h(a)$, $h(\lrcorner
a)=\lrcorner h(a)$, and
$h(\top_{\textbf{D}})=\top_{\textbf{M}}$,
$h(\bot_{\textbf{D}})=\bot_{\textbf{M}}$.
The dBa homomorphism $h$ is called quasi-injective, when $x\sqsubseteq y$ if
and only if $h(x)\sqsubseteq h(y)$, for all $x,y\in D$. D is then said to be
quasi-embedded into M.
If $h$ is an injective dBa homomorphism, D is said to be embedded into M.
A dBa homomorphism from D to M that is quasi-injective and surjective is
called a dBa quasi-isomorphism, and D is said to be quasi-isomorphic to M.
If there is a dBa homomorphism from D to M that is bijective then the map is
called a dBa isomorphism, and the dBas $\textbf{D},\textbf{M}$ are said to be
isomorphic to each other.

Note that composition of two dBa homomorphisms is also a dBa homomorphism.

###  2.3 Filters, ideals and the prime ideal theorem for dBas

Kwuida proved the prime ideal theorem for his class of dBas by generalizing
the notion of a prime filter (ideal) of Boolean algebras to define a primary
filter (ideal) for dBas [22]. Due to Corollary 1, these notions and the prime
ideal theorem hold for Wille’s dBas. Let us give the basic definitions and
results in this regard. Again, let
$\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$ be a dBa.

######  Definition 8.

A filter in D is a subset $F$ of $D$ such that $x\sqcap y\in F$ for all
$x,y\in F$, and for all $z\in D$ and $x\in F,x\sqsubseteq z$ implies that
$z\in F$. An ideal in a dBa is defined dually.
The filter (ideal) generated by a set $X$, denoted $F(X)(I(X))$, is the
smallest filter (ideal) in D containing $X$.
A base for the filter $F$ is a subset $F_{0}(\subseteq D)$ such that
$F=\\{y\in D:x\sqsubseteq y~{}\mbox{for some}~{}x\in F_{0}\\}$. A base for an
ideal is dually defined.
A filter $F$ (ideal $I$) is proper if and only if $F\neq D$ ($I\neq D$).
A primary filter $F$ (ideal $I$) is a proper filter (ideal) such that $x\in
F~{}\mbox{or}~{}\neg x\in F~{}(x\in I~{}\mbox{or}~{}\lrcorner x\in I)$, for
all $x\in D$.

It can be shown that $F(X)=\\{x\in D~{}:~{}\sqcap_{i=1}^{n}a_{i}\sqsubseteq
x~{}\mbox{for some}~{}a_{i}\in X~{}\mbox{and}~{}i=1,\ldots,n\\}$, and dually,
$I(X)=\\{x\in D~{}:~{}x\sqsubseteq\sqcup_{i=1}^{n}a_{i}~{}\mbox{for
some}~{}a_{i}\in X~{}\mbox{and}~{}i=1,\ldots,n\\}$.

######  Theorem 3 (Prime ideal theorem for dBas).

Let $F$ be a filter and $I$ an ideal in D such that $F\cap I=\emptyset$. Then
there exists a primary filter $G$ and a primary ideal $J$ of D satisfying
$F\subseteq G,~{}I\subseteq J$ and $G\cap J=\emptyset$.

######  Notation 5.

$\mathcal{F}_{pr}(\textbf{D})$ denotes the set of all primary filters and
$\mathcal{I}_{pr}(\textbf{D})$, the set of all primary ideals.
$\mathcal{F}_{p}(\textbf{D}):=\\{F\subseteq D:F~{}\mbox{is a filter
of}~{}\textbf{D}~{}\mbox{and}~{}F\cap D_{\sqcap}~{}\mbox{is a prime filter
in}~{}\textbf{D}_{\sqcap}\\}$, while
$\mathcal{I}_{p}(\textbf{D}):=\\{I\subseteq D:I~{}\mbox{is an ideal
of}~{}\textbf{D}~{}\mbox{and}~{}I\cap D_{\sqcup}~{}\mbox{ is a prime ideal
in}~{}\textbf{D}_{\sqcup}\\}$.
$F_{x}:=\\{F\in\mathcal{F}_{p}(\textbf{D})~{}:~{}x\in F\\}$ and
$I_{x}:=\\{I\in\mathcal{I}_{p}(\textbf{D})~{}:~{}x\in I\\}$, for any $x\in D$.
$\mathbb{K}(\textbf{D}):=(\mathcal{F}_{p}(\textbf{D}),\mathcal{I}_{p}(\textbf{D}),\Delta)$
is defined as the standard context, where $F\Delta I$ if and only if $F\cap
I\neq\emptyset$, for any
$F\in\mathcal{F}_{p}(\textbf{D}),~{}I\in\mathcal{I}_{p}(\textbf{D})$.

######  Lemma 1.

[40] Let $x\in\textbf{D}$.

  1. 1.

$F_{x}^{\prime}=I_{x_{\sqcap\sqcup}}$ and
$I_{x}^{\prime}=F_{x_{\sqcup\sqcap}}$.

  2. 2.

$(F_{x})^{c}=F_{\neg x}$ and $(I_{x})^{c}=I_{\lrcorner x}$.

  3. 3.

$I_{x}\cap I_{y}=I_{x\sqcup y}$ and $I_{x_{\sqcup}}=I_{x}$.

  4. 4.

$F_{x}\cap F_{y}=F_{x\sqcap y}$ and $F_{x_{\sqcap}}=F_{x}$.

######  Lemma 2.

Let $F$ be a filter and $I$ be an ideal of D.

  1. 1.

$F\cap D_{\sqcap}$ and $F\cap D_{\sqcup}$ are filters of the Boolean algebras
$\textbf{D}_{\sqcap}$, $\textbf{D}_{\sqcup}$ respectively.

  2. 2.

Each filter $F_{0}$ of the Boolean algebra $\textbf{D}_{\sqcap}$ is the base
of some filter $F$ of D such that $F_{0}=F\cap D_{\sqcap}$. Moreover if
$F_{0}$ is a prime filter of $\textbf{D}_{\sqcap}$,
$F\in\mathcal{F}_{p}(\textbf{D})$.

  3. 3.

$I\cap D_{\sqcap}$ and $I\cap D_{\sqcup}$ are ideals of the Boolean algebras
$\textbf{D}_{\sqcap}$, $\textbf{D}_{\sqcup}$ respectively.

  4. 4.

Each ideal $I_{0}$ of the Boolean algebra $\textbf{D}_{\sqcup}$ is the base of
some ideal $I$ of D such that $I_{0}=I\cap D_{\sqcup}$. Moreover if $I_{0}$ is
prime, $I\in\mathcal{I}_{p}(\textbf{D})$.

(i) and (ii) of Lemma 2 have been proved in [40]. (iii) and (iv) for ideals
can be proved dually.

######  Observation 2.

For each filter $F$ and ideal $I$ of D, $F\cap D_{\sqcap}$ is a base of $F$,
$I\cap D_{\sqcup}$ is a base of $I$.

In [19], one observed that

######  Proposition 7.

[19]

  1. 1.

$\mathcal{F}_{pr}(\textbf{D})$=$\mathcal{F}_{p}(\textbf{D}).$

  2. 2.

$\mathcal{I}_{pr}(\textbf{D})$=$\mathcal{I}_{p}(\textbf{D})$.

Proposition 7 implies that there is a one-one and onto correspondence between
the set of primary filters (ideals) of D and the set of prime filters (ideals)
of $\textbf{D}_{\sqcap}(\textbf{D}_{\sqcup}$).

###  2.4 Object oriented concept, semiconcept and protoconcept of a context

In the following, let $\mathbb{K}:=(G,M,R)$ be a context, and $A\subseteq G$,
$B\subseteq M$. Operators $\lozenge,\square,\blacklozenge,\blacksquare$ are
introduced on the power sets of $G,M$ as follows. $B_{R}^{\lozenge}:=\\{x\in
G:R(x)\cap
B\neq\emptyset\\},~{}~{}~{}~{}~{}~{}~{}~{}~{}B_{R}^{\square}:=\\{x\in
G:R(x)\subseteq B\\}$, $A_{R^{-1}}^{\blacklozenge}:=\\{y\in M:R^{-1}(y)\cap
A\neq\emptyset\\},~{}~{}~{}A_{R^{-1}}^{\blacksquare}:=\\{y\in
M:R^{-1}(y)\subseteq A\\}$.

If the relation involved is clear from the context, we shall omit the
subscripts and denote $B_{R}^{\lozenge}$ by $B^{\lozenge}$, $B_{R}^{\square}$
by $B^{\square}$, and similarly for the case of $A$. Let us recall the notions
of closure and interior operators.

######  Definition 9.

[7] An operator $C$ on the power set $\mathcal{P}(X)$ of a set $X$ is called a
closure operator on $X$, if for all $A,B\in\mathcal{P}(X)$,

  1. C1

$A\subseteq C(A)$,

  2. C2

$A\subseteq B$ implies $C(A)\subseteq C(B)$, and

  3. C3

$C(C(A))=C(A)$.

$A\in\mathcal{P}(X)$ is called closed if and only if $C(A)=A$.
An interior operator $I$ on the set $X$ is defined dually.
$A\in\mathcal{P}(X)$ is open if and only if $I(A)=A$.

We next list some properties of the operators
$\square,\lozenge,\blacklozenge,\blacksquare$.

######  Theorem 4.

[8, 42] Let $A,A_{1},A_{2}\subseteq G$ and $B,B_{1},B_{2}\subseteq M$.

  1. 1.

$A_{1}\subseteq A_{2}$ implies that $A_{1}^{\blacksquare}\subseteq
A_{2}^{\blacksquare}$ and $A_{1}^{\blacklozenge}\subseteq
A_{2}^{\blacklozenge}$.

  2. 2.

$B_{1}\subseteq B_{2}$ implies that $B_{1}^{\square}\subseteq B_{2}^{\square}$
and $B_{1}^{\lozenge}\subseteq B_{2}^{\lozenge}$.

  3. 3.

$(B_{1}\cap B_{2})^{\square}=B_{1}^{\square}\cap B_{2}^{\square}$ and
$(B_{1}\cup B_{2})^{\lozenge}=B_{1}^{\lozenge}\cup B_{2}^{\lozenge}$ .

  4. 4.

$(A_{1}\cap A_{2})^{\blacksquare}=A_{1}^{\blacksquare}\cap
A_{2}^{\blacksquare}$ and $(A_{1}\cup
A_{2})^{\blacklozenge}=A_{1}^{\blacklozenge}\cup A_{2}^{\blacklozenge}$ .

  5. 5.

$B^{\square}=B^{c\lozenge c}$ and $A^{\blacksquare}=A^{c\blacklozenge c}$ .

  6. 6.

$A^{\blacksquare}_{R}=A^{c\prime}_{-R};B^{\square}_{R}=B^{c\prime}_{-R}$ and
$A^{\blacklozenge}_{R}=A^{\prime c}_{-R};B^{\lozenge}_{R}=B^{\prime c}_{-R}$ .

  7. 7.

$A^{\blacksquare\lozenge\blacksquare}=A^{\blacksquare}$ and
$B^{\square\lozenge\square}=B^{\square}$.

  8. 8.

$A^{\blacklozenge\square\blacklozenge}=A^{\blacklozenge}$ and
$B^{\lozenge\square\lozenge}=B^{\lozenge}$.

  9. 9.

$\blacksquare\lozenge$ is interior operator on $G$ and $\lozenge\blacksquare$
is closure operator on $M$.

In 2004, Yao defined object oriented concepts [42].

######  Definition 10.

[42] $(A,B)$ is an object oriented concept of the context $\mathbb{K}$ if
$A^{\blacksquare}=B$ and $B^{\lozenge}=A$.
A partial order $\leq$ is given by the following relation defined on the set
of all object oriented concepts. For any object oriented concepts
$(A_{1},B_{1}),(A_{2},B_{2})$,
$(A_{1},B_{1})\leq(A_{2},B_{2})~{}\mbox{if and only if}~{}A_{1}\subseteq
A_{2}$ (equivalently, $B_{1}\subseteq B_{2}$).

In [17], object oriented semiconcepts were introduced in order to bring the
notion of negation into the study.

######  Definition 11.

[17] $(A,B)$ is an object oriented semiconcept of $\mathbb{K}$ if
$A^{\blacksquare}=B$ or $B^{\lozenge}=A$.

######  Notation 6.

$RO-L(\mathbb{K})$ denotes the set of all object oriented concepts of
$\mathbb{K}$, while the set of all object oriented semiconcepts is denoted by
$\mathfrak{S}(\mathbb{K})$.

In [17], one observes the following.

######  Proposition 8.

  1. 1.

$(A,B)\in\mathfrak{S}(\mathbb{K})$ if and only if either
$(A,B)=(A,A^{\blacksquare})$ or $(A,B)=(B^{\lozenge},B).$

  2. 2.

$RO-L(\mathbb{K})\subseteq\mathfrak{S}(\mathbb{K}).$

  3. 3.

$(A,B)$ is a semiconcept of $\mathbb{K}$ if and only if $(A^{c},B)$ is an
object oriented semiconcept of the context $\mathbb{K}^{c}$.

Operations $\sqcap,\sqcup,\lrcorner,\neg,\top,\bot$ are defined in
$\mathfrak{S}(\mathbb{K})$ as follows. Let $(A_{1},B_{1}),(A_{2},B_{2}),(A,B)$
be any object oriented semiconcepts in $\mathfrak{S}(\mathbb{K})$.
$(A,B)\sqcap(C,D):=(A\cup C,(A\cup C)^{\blacksquare})$,
$(A,B)\sqcup(C,D):=((B\cap D)^{\lozenge},B\cap D)$,
$\lrcorner(A,B):=(B^{c\lozenge},B^{c}),~{}\neg(A,B):=(A^{c},A^{c\blacksquare}),~{}\top:=(\emptyset,\emptyset),~{}\bot:=(G,M).$

######  Notation 7.

$\mathcal{S}(\mathbb{K})$ denotes the algebra formed by
$\mathfrak{S}(\mathbb{K})$ with respect to the above operations.

Recall the algebra of semiconcepts $\underline{\mathfrak{H}}(\mathbb{K})$ (cf.
Notation 3).

######  Theorem 5.

[18]

  1. 1.

$\mathcal{S}(\mathbb{K}):=(\mathfrak{S}(\mathbb{K}),\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$
is a pure dBa.

  2. 2.

$\underline{\mathfrak{H}}(\mathbb{K})$ is isomorphic to
$\mathcal{S}(\mathbb{K}^{c})$.

The dBa isomorphism in Theorem 5(ii) above is obtained using Proposition
8(iii). As mentioned in Section 1, it was shown in [17] that
$\underline{\mathfrak{H}}(\mathbb{K})$ is dually isomorphic to the algebra of
object oriented semiconcepts. This is because the operations on the set of
object oriented semiconcepts considered in [17] are dual to those defined
above. As we would like to prove representation results and such results are
usually obtained in terms of isomorphisms (and not dual isomorphisms), we
consider the operations in the form given above. With reference to
$\mathfrak{S}(\mathbb{K})$ and the quasi-order $\sqsubseteq$ given in
Definition 5, we get the following.

######  Proposition 9.

For any $(A,B),(C,D)\in\mathfrak{S}(\mathbb{K})$, $(A,B)\sqsubseteq(C,D)$ if
and only if $C\subseteq A$ and $D\subseteq B$.

We next define and give some properties of object oriented protoconcepts which
were introduced in [18].

######  Definition 12.

[18] $(A,B)$ is an object oriented protoconcept of $\mathbb{K}$ if
$A^{\blacksquare\lozenge}=B^{\lozenge}$.

######  Notation 8.

$\mathfrak{R}(\mathbb{K})$ denotes the set of all object oriented
protoconcepts.

Observe that an object oriented semiconcept is an object oriented
protoconcept, that is
$\mathfrak{S}(\mathbb{K})\subseteq\mathfrak{R}(\mathbb{K})$. Moreover, the
following yields an equivalent definition of object oriented protoconcepts.

######  Observation 3.

$A^{\blacksquare\lozenge}=B^{\lozenge}$ if and only if
$A^{\blacksquare}=B^{\lozenge\blacksquare}$.

A characterization of object oriented protoconcepts of $\mathbb{K}$ was
established in [18], using a notion of ‘approximation’ by object oriented
concepts.

Similar to Propositions 8(iii) and 9, one obtains

######  Proposition 10.

$(A,B)$ is a protoconcept of $\mathbb{K}$ if and only if $(A^{c},B)$ is an
object oriented protoconcept of $\mathbb{K}^{c}$.

$\mathfrak{R}(\mathbb{K})$ is closed with respect to the operations
$\sqcup,\sqcap,\neg,\lrcorner,\top,\bot$ defined on $\mathfrak{S}(\mathbb{K})$
and we have

######  Notation 9.

$\underline{\mathfrak{R}}(\mathbb{K})$ denotes the algebra formed by
$\mathfrak{R}(\mathbb{K})$ with respect to these operations.

######  Proposition 11.

  1. 1.

$\mathcal{S}(\mathbb{K})$ is a subalgebra of
$\underline{\mathfrak{R}}(\mathbb{K})$.

  2. 2.

$\mathcal{S}(\mathbb{K})=\underline{\mathfrak{R}}(\mathbb{K})_{p}$.

(ii) of the proposition can be easily proved using definitions of object
oriented semiconcepts and the set $\mathfrak{R}(\mathbb{K})_{p}$. Now recall
the algebra of protoconcepts $\underline{\mathfrak{P}}(\mathbb{K})$ (cf.
Notation 3). One obtains

######  Theorem 6.

  1. 1.

$\underline{\mathfrak{R}}(\mathbb{K}):=(\mathfrak{R}(\mathbb{K}),\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$
is a dBa.

  2. 2.

$\underline{\mathfrak{P}}(\mathbb{K})$ is isomorphic to
$\underline{\mathfrak{R}}(\mathbb{K}^{c})$.

Theorem 6(ii) is obtained by using Proposition 10. From Theorem 6(ii) we get

######  Corollary 2.

$\underline{\mathfrak{R}}(\mathbb{K})$ is a fully contextual dBa.

Again we have

######  Proposition 12.

For any $(A,B),(C,D)\in\mathfrak{R}(\mathbb{K})$, $(A,B)\sqsubseteq(C,D)$ if
and only if $C\subseteq A$ and $D\subseteq B$.

A detailed example was given in [18] to motivate and illustrate both the
notions of object oriented semiconcepts and object oriented protoconcepts. We
re-present the example here to demonstrate that a fully contextual dBa need
not be pure and vice versa.

######  Example 1.

Let $G:=\\{q_{1},q_{2},q_{3},q_{4},q_{5},q_{6}\\}$ be a set of objects and
consider a set of properties
$S:=\\{s_{1},s_{2},s_{3},s_{4},s_{5},s_{6},s_{7},s_{8},s_{9},s_{10},s_{11}\\}$.
Table 1 below represents the context $\mathbb{K}:=(G,S,\Gamma)$, where
$(i,j)$-th cell containing a cross indicates that the object $q_{i}$ is
related to the property $s_{j}$ by $\Gamma$.

Table 1: Context $\mathbb{K}$ | $s_{1}$ | $s_{2}$ | $s_{3}$ | $s_{4}$ | $s_{5}$ | $s_{6}$ | $s_{7}$ | $s_{8}$ | $s_{9}$ | $s_{10}$ | $s_{11}$
---|---|---|---|---|---|---|---|---|---|---|---
$q_{1}$ | $\times$ |  | $\times$ |  | $\times$ |  | $\times$ |  |  | $\times$ |
$q_{2}$ |  | $\times$ | $\times$ | $\times$ |  | $\times$ |  | $\times$ |  |  |
$q_{3}$ | $\times$ | $\times$ |  | $\times$ |  | $\times$ |  |  | $\times$ |  |
$q_{4}$ | $\times$ |  | $\times$ |  | $\times$ |  |  |  |  |  |
$q_{5}$ |  |  |  |  | $\times$ |  |  | $\times$ | $\times$ |  | $\times$
$q_{6}$ |  |  |  |  |  |  |  |  |  |  | $\times$

Consider $A_{1}:=\\{q_{1},q_{2},q_{4},q_{6}\\}$ and $B_{1}:=\\{s_{3}\\}$. Then
$A_{1}^{\blacksquare\lozenge}=B_{1}^{\lozenge}$, so that $(A_{1},B_{1})$ is an
object oriented protoconcept of $\mathbb{K}$. Observe that
$B_{1}^{\lozenge}=\\{q_{1},q_{2},q_{4}\\}$ and so $B_{1}^{\lozenge}\neq
A_{1}$; $A_{1}^{\blacksquare}=\\{s_{3},s_{7},s_{10}\\}$, so
$A_{1}^{\blacksquare}\neq B_{1}$. This means
$(A_{1},B_{1})\in\mathfrak{R}(\mathbb{K})$ but
$(A_{1},B_{1})\notin\mathfrak{S}(\mathbb{K})$. Now we consider the
semiconcepts $(A_{1},A_{1}^{\blacksquare})$ and $(B_{1}^{\lozenge},B_{1})$.
Since $(A_{1},B_{1})$ is an object oriented protoconcept, the following
equations hold:
$(A_{1},A_{1}^{\blacksquare})_{\sqcup}=(A_{1}^{\blacksquare\lozenge},A_{1}^{\blacksquare})=(B_{1}^{\lozenge},B_{1}^{\lozenge\blacksquare})=(B_{1}^{\lozenge},B_{1})_{\sqcap}$.
Let us assume that there exists a semiconcept
$(A,B)\in\mathfrak{S}(\mathbb{K})$ such that
$(A,B)_{\sqcap}=(A,A^{\blacksquare})=(A_{1},A_{1}^{\blacksquare})$ and
$(A,B)_{\sqcup}=(B^{\lozenge},B)=(B_{1}^{\lozenge},B_{1})$, which implies that
$A=A_{1}$ and $B=B_{1}$ – a contradiction. Therefore the subalgebra
$\mathcal{S}(\mathbb{K})$ of $\underline{\mathfrak{R}}(\mathbb{K})$ is a pure
dBa but not fully contextual. On the other hand, the fully contextual dBa
$\underline{\mathfrak{R}}(\mathbb{K})$ is not pure, as
$(A_{1},B_{1})_{\sqcap}\neq(A_{1},B_{1})$ and
$(A_{1},B_{1})_{\sqcup}\neq(A_{1},B_{1})$.

######  Observation 4.

The class K of fully contextual dBas does not form a variety [6]: Example 1
gives $\underline{\mathfrak{R}}(\mathbb{K})\in\textbf{K}$ such that the
subalgebra $\mathcal{S}(\mathbb{K})\notin\textbf{K}$.

###  2.5 Representation results

Let us recollect in this section, existing representation results related to
arbitrary dBas as well as contextual, fully contextual and pure dBas. In the
process, we also indicate the kind of representation results that are obtained
in this work. Recall Definition 7, the standard context
$\mathbb{K}(\textbf{D}):=(\mathcal{F}_{p}(\textbf{D}),\mathcal{I}_{p}(\textbf{D}),\Delta)$
for any dBa D, and the sets $F_{x},I_{x}$ defined in Notation 5. The following
theorem is proved by Wille in [40].

######  Theorem 7.

[40] The map $h:D\rightarrow\mathfrak{P}(\mathbb{K}(\textbf{D}))$ defined by
$h(x):=(F_{x},I_{x})$ for all $x\in D$, is a quasi-injective dBa homomorphism
from D to $\underline{\mathfrak{P}}(\mathbb{K}(\textbf{D}))$.

Using Theorems 7 and 6 and Lemma 1(ii), one gets a representation result for
dBas in terms of object oriented protoconcepts also.

######  Theorem 8.

[18] For a dBa D, the map
$h:D\rightarrow\mathfrak{R}(\mathbb{K}^{c}(\textbf{D}))$ defined by
$h(x):=(F_{\neg x},I_{x})$ for any $x\in D$, is a quasi-injective dBa
homomorphism from D to $\underline{\mathfrak{R}}(\mathbb{K}^{c}(\textbf{D}))$.

In this work we show that, in fact, the map $h$ of Theorem 8 is a quasi-
embedding into a subalgebra of
$\underline{\mathfrak{R}}(\mathbb{K}^{c}(\textbf{D}))$. Moreover, if D is a
contextual dBa, this quasi-embedding turns into an embedding.

Let D be a finite dBa. Wille [40] has a representation result for this special
case. It may be observed that by Lemma 2 and Proposition 7, the elements of
$\mathcal{F}_{pr}(\textbf{D})$ in this case are just the primary filters whose
bases are the principal filters of the Boolean algebra $\textbf{D}_{\sqcap}$
generated by its atoms, while the elements of $\mathcal{I}_{pr}(\textbf{D})$
are the primary ideals whose bases are the principal ideals of the Boolean
algebra $\textbf{D}_{\sqcup}$ generated by its coatoms. In other words,
$F\in\mathcal{F}_{pr}(\textbf{D})$ if and only if $F=\\{x\in
D~{}:~{}a\sqsubseteq x\\}$ for some atom
$a\in\mathcal{A}(\textbf{D}_{\sqcap})$, the set of all atoms of the Boolean
algebra $\textbf{D}_{\sqcap}$. $I\in\mathcal{I}_{pr}(\textbf{D})$ if and only
if $I=\\{x\in D~{}:~{}x\sqsubseteq b\\}$ for some
$b\in\mathcal{C}(\textbf{D}_{\sqcup})$, the set of all coatoms of the Boolean
algebra $\textbf{D}_{\sqcup}$. The result proved in [40] is as follows.

######  Theorem 9.

[40] Let D be a finite dBa and
$\mathbb{K}_{\mathcal{AC}}:=(\mathcal{A}(\textbf{D}_{\sqcap}),\mathcal{C}(\textbf{D}_{\sqcup}),\sqsubseteq)$.
Then the map $h:D\rightarrow\mathfrak{P}(\mathbb{K}_{\mathcal{AC}})$ defined
by $h(x):=(\\{a\in\mathcal{A}(\textbf{D}_{\sqcap})~{}:~{}a\sqsubseteq
x\\},\\{b\in\mathcal{C}(\textbf{D}_{\sqcup})~{}:~{}x\sqsubseteq b\\})$ for all
$x\in D$, is a quasi-injective dBa homomorphism from D to
$\underline{\mathfrak{P}}(\mathbb{K}_{\mathcal{AC}})$. Moreover,
$\textbf{D}_{p}$ is isomorphic to
$\underline{\mathfrak{H}}(\mathbb{K}_{\mathcal{AC}})$.

From Theorems 9, 6, and 5, we have a representation theorem for finite dBas in
terms of object oriented protoconcepts and object oriented semiconcepts.

######  Corollary 3.

If D is a finite dBa, the map
$h:D\rightarrow\mathfrak{R}(\mathbb{K}^{c}_{\mathcal{AC}})$ defined by
$h(x):=(\\{a\in\mathcal{A}(\textbf{D}_{\sqcap})~{}:~{}a\not{\sqsubseteq}x\\},\\{b\in\mathcal{C}(\textbf{D}_{\sqcup})~{}:~{}x\sqsubseteq
b\\})$ for all $x\in D$, is a quasi-injective dBa homomorphism from D to
$\underline{\mathfrak{R}}(\mathbb{K}^{c}_{\mathcal{AC}})$. Moreover,
$\textbf{D}_{p}$ is isomorphic to
$\underline{\mathcal{S}}(\mathbb{K}^{c}_{\mathcal{AC}})$.

We shall show that the representation result for dBas obtained in this work
yields the above as a special case when D is finite. For a pure dBa, the
following is established in [3].

######  Theorem 10.

[3] If $\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$ is a pure dBa,
the map $h:D\rightarrow\mathfrak{H}(\mathbb{K}(\textbf{D}))$ defined by
$h(x):=(F_{x},I_{x})$ for all $x\in D$, is an injective dBa homomorphism from
D to $\underline{\mathfrak{H}}(\mathbb{K}(\textbf{D}))$.

From Lemma 1(ii), Theorems 5 and 10, we then have a representation result in
terms of object oriented semiconcepts.

######  Theorem 11.

[18] For a pure dBa D, the map
$h:D\rightarrow\mathfrak{S}(\mathbb{K}^{c}(\textbf{D}))$ defined by
$h(x):=(F_{\neg x},I_{x})$ for all $x\in D$ is an injective dBa homomorphism
from D to $\mathcal{S}(\mathbb{K}^{c}(\textbf{D}))$.

A dBa D is complete [37] if and only if the Boolean algebras
$\textbf{D}_{\sqcup}$ and $\textbf{D}_{\sqcap}$ are complete. Vormbrock [37]
proved that any complete pure dBa D for which $\textbf{D}_{\sqcap}$ and
$\textbf{D}_{\sqcup}$ are atomic Boolean algebras, is isomorphic to the
algebra of semiconcepts of some context. Furthermore, any complete fully
contextual dBa for which $\textbf{D}_{\sqcap}$ and $\textbf{D}_{\sqcup}$ are
atomic Boolean algebras is isomorphic to the algebra of protoconcepts of some
context. However, all dBas are clearly not complete – one may simply consider
Boolean algebras that are not complete. What about an isomorphism theorem of
the above kind for (fully contextual/pure) dBas in general? An attempt in this
regard was made in [5], for contextual dBas. However, a counterexample to the
proof of the representation result of [5] was given in [19]. In this paper, we
establish isomorphism theorems for fully contextual and pure dBas in terms of
algebras of certain protoconcepts and semiconcepts (respectively) of contexts
equipped with topologies.

##  3 Some further results on dBas

In this section, we derive some further results on dBas that will be used in
Sections 5 and 6. In particular, we give a relationship between Boolean
algebras and dBas in Theorem 12 below. To prove this theorem, we require the
following.

######  Proposition 13.

Let $\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$ be a dBa. For any
$x,y\in D$, the following hold.

  1. 1.

$x\sqcap y\sqsubseteq x\vee y\sqsubseteq x\sqcup y$.

  2. 2.

$x\sqcap y\sqsubseteq x\wedge y\sqsubseteq x\sqcup y$.

###### Proof.

The proof of (ii) is dual to the proof of (i), and we only prove (i). For any
$x,y\in D$, $\neg x\sqcap\neg y\sqsubseteq\neg x$ and $\neg x\sqcap\neg
y\sqsubseteq\neg y.$ So by Proposition 5(ii), $\neg\neg x\sqsubseteq\neg(\neg
x\sqcap\neg y)$ and $\neg\neg y\sqsubseteq\neg(\neg x\sqcap\neg y).$ Then
Proposition 4(vi) gives $\neg\neg x\sqcap\neg\neg y\sqsubseteq\neg(\neg
x\sqcap\neg y)\sqcap\neg\neg y$ and $\neg\neg y\sqcap\neg(\neg x\sqcap\neg
y)\sqsubseteq\neg(\neg x\sqcap\neg y)\sqcap\neg(\neg x\sqcap\neg y).$
Therefore $\neg\neg x\sqcap\neg\neg y\sqsubseteq\neg(\neg x\sqcap\neg
y)\sqcap\neg(\neg x\sqcap\neg y).$ By Proposition 5(i), $\neg\neg
x\sqcap\neg\neg y\sqsubseteq\neg(\neg x\sqcap\neg y)$, that is $(x\sqcap
x)\sqcap(y\sqcap y)\sqsubseteq x\vee y.$ By axiom $(1a)$ and $(3a)$, $x\sqcap
y\sqsubseteq x\vee y$.

We know that $x,y\sqsubseteq x\sqcup y.$ Proposition 5(ii) gives $\neg(x\sqcup
y)\sqsubseteq\neg x,\neg y.$ Therefore by Proposition 4(vi), $\neg(x\sqcup
y)\sqcap\neg y\sqsubseteq\neg x\sqcap\neg y$ and $\neg(x\sqcup
y)\sqcap\neg(x\sqcup y)\sqsubseteq\neg(x\sqcup y)\sqcap\neg y.$ So
$\neg(x\sqcup y)\sqcap\neg(x\sqcup y)\sqsubseteq\neg x\sqcap\neg y.$ By
Proposition 5(i), $\neg(x\sqcup y)\sqsubseteq\neg x\sqcap\neg y,$ and by
Proposition 5(ii), $\neg(\neg x\sqcap\neg y)\sqsubseteq\neg\neg(x\sqcup
y)=(x\sqcup y)\sqcap(x\sqcup y)\sqsubseteq x\sqcup y.$ Hence $x\vee
y\sqsubseteq x\sqcup y.$ ∎

######  Theorem 12.

Any Boolean algebra $(D,\sqcap,\sqcup,\neg,\top,\bot)$ forms a fully
contextual as well as pure dBa
$\textbf{D}:=(D,\sqcap,\sqcup,\neg,\lrcorner,\top,\bot),$ where for all $a\in
D$, $\lrcorner a:=\neg a$. On the other hand, a dBa
$\textbf{D}:=(D,\sqcap,\sqcup,\neg,\lrcorner,\top,\bot)$ forms a Boolean
algebra $(D,\sqcap,\sqcup,\neg,\top,\bot),$ if for all $a\in D$, $\neg
a=\lrcorner a$ and $\neg\neg a=a$.

###### Proof.

It is easy to see that in a Boolean algebra
$(D,\sqcap,\sqcup,\neg,\top,\bot)$, if we set $\lrcorner a:=\neg a$ for all
$a\in D$, $(D,\sqcap,\sqcup,\neg,\lrcorner,\top,\bot)$ forms a dBa. The dBa is
also pure, due to the idempotence of the operators $\sqcap$ and $\sqcup$ in a
Boolean algebra. Further, note that in this case $D_{\sqcap}=D_{\sqcup}=D$,
and the dBa is fully contextual as well. We also have $\neg\neg a=a$ for all
$a\in D$.
Now let D be a dBa such that for all $a\in D$, $\neg a=\lrcorner a$ and
$\neg\neg a=a$. Let $x,y\in D$ such that $x\sqsubseteq y$ and $y\sqsubseteq
x$. By Proposition 4(iv), $x\sqcap x=y\sqcap y$ and $x\sqcup x=y\sqcup y$.
Using Proposition 5(iii), $\neg\neg x=\neg\neg y$ and so $x=y$. Therefore
$(D,\sqsubseteq)$ is a partially ordered set. From Definition 5(2a and 2b) it
follows that $\sqcap,\sqcup$ is commutative, while Definition 5(3a and 3b)
gives that $\sqcap,\sqcup$ is associative. Using Definition 5(5a) and
Proposition 5(iii), $x\sqcap(x\sqcup y)=x\sqcap x=\neg\neg x$. So
$x\sqcap(x\sqcup y)=x$. Again using Definition 5(5b) and Proposition 5(iii),
$x\sqcup(x\sqcap y)=x$. Therefore
$(D,\sqcap,\sqcup,\neg,\top,\bot,\sqsubseteq)$ is a bounded complemented
lattice. To show it is a distributive lattice, let $x,y,z\in D.$ Proposition
13 implies that $x\sqcap y\sqsubseteq x\wedge y$ and $x\vee y\sqsubseteq
x\sqcup y$. Using Proposition 4(v) and Proposition 5(ii), $\neg y\sqcap\neg
z\sqsubseteq\neg y,\neg z$. So $\neg\neg y\sqsubseteq\neg(\neg y\sqcap\neg z)$
and $\neg\neg z\sqsubseteq\neg(\neg y\sqcap\neg z)$. Therefore
$y\sqsubseteq\neg(\neg y\sqcap\neg z)=y\vee z$ and $z\sqsubseteq\neg(\neg
y\sqcap\neg z)=y\vee z$. Proposition 4(vi) gives $y\sqcup z\sqsubseteq y\vee
z$, as $(y\vee z)\sqcup(y\vee z)=\lrcorner\lrcorner(y\vee z)=\neg\neg(y\vee
z)=y\vee z$. So $y\sqcup z=y\vee z$. Dually we can show that $y\sqcap
z=y\wedge z$. From Definition 5(6a and 6b) it follows that
$(D,\sqcap,\sqcup,\neg,\top,\bot,\sqsubseteq)$ is a complemented distributive
lattice and hence a Boolean algebra. ∎

For each dBa D there is a pure subalgebra $\textbf{D}_{p}$, by Proposition 6.
In the case of fully contextual dBas, these subalgebras play a special role.

######  Theorem 13.

Let D and M be fully contextual dBas. Then D is isomorphic to M if and only if
$\textbf{D}_{p}$ is isomorphic to $\textbf{M}_{p}$. Moreover, every dBa
isomorphism from $\textbf{D}_{p}$ to $\textbf{M}_{p}$ can be uniquely extended
to a dBa isomorphism from D to M.

###### Proof.

Let $f$ be a dBa isomorphism from D to M. We show that $f|_{D_{p}}$, the
restriction of $f$ to $\textbf{D}_{p}$, is the required dBa isomorphism from
$\textbf{D}_{p}$ to $\textbf{M}_{p}$. For that, it is enough to prove that
$f|_{D_{p}}(D_{p})=M_{p}$. Let $x\in D_{p}$. Then either $x\sqcap x=x$ or
$x\sqcup x=x$, which implies that either $f|_{D_{p}}(x)\sqcap
f|_{D_{p}}(x)=f(x)\sqcap f(x)=f(x\sqcap x)=f(x)=f|_{D_{p}}(x)$ or
$f|_{D_{p}}(x)\sqcup f|_{D_{p}}(x)=f(x)\sqcup f(x)=f(x\sqcup
x)=f(x)=f|_{D_{p}}(x)$, as $f$ is a homomorphism. So
$f|_{D_{p}}(D_{p})\subseteq M_{p}$. Conversely, let $y\in M_{p}$. As $f$ is
surjective, there exists a $c\in D$ such that $f(c)=y$. Now $f(c\sqcap
c)=y\sqcap y$ and $f(c\sqcup c)=y\sqcup y$. Since $y\in M_{p}$, either
$y\sqcap y=y$ or $y\sqcup y=y$, which implies that either $f(c\sqcap
c)=y=f(c)$ or $f(c\sqcup c)=y=f(c)$. So either $c\sqcap c=c$ or $c\sqcup c=c$,
as $f$ is injective. So $c\in D_{p}$. Therefore $M_{p}=f|_{D_{p}}(D_{p})$.

Let $\textbf{D}_{p}$ be isomorphic to $\textbf{M}_{p}$ and $h$ be a dBa
isomorphism from $\textbf{D}_{p}$ to $\textbf{M}_{p}$. Let $x\in D$. Then
$x\sqcap x\in D_{\sqcap}\subseteq D_{p}$, $x\sqcup x\in D_{\sqcup}\subseteq
D_{p}$ and $(x\sqcap x)\sqcup(x\sqcap x)=(x\sqcup x)\sqcap(x\sqcup x)$ by
Definition 5(12). As $h$ is a dBa isomorphism, $h(x\sqcap x)\in
M_{\sqcap}\subseteq M_{p}$, $h(x\sqcup x)\in M_{\sqcup}\subseteq M_{p}$ and
$h(x\sqcap x)\sqcup h(x\sqcap x)=h(x\sqcup x)\sqcap h(x\sqcup x)$. As M is
fully contextual, there exists a unique $c^{x}\in M$ such that
$c^{x}_{\sqcap}=h(x\sqcap x)$ and $c^{x}_{\sqcup}=h(x\sqcup x)$. Let us define
a map $f:D\rightarrow M$ by $f(x):=c^{x}$, for all $x\in D$. From the
definition of $c^{x}$, it follows that $f$ is well-defined. We show that $f$
is the required dBa isomorphism.
(i) Let $x,y\in D$ and $f(x)=c^{x},f(y)=c^{y}$. We will show that $f(x\sqcap
y)=c^{x\sqcap y}=c^{x}\sqcap c^{y}$. Now $c^{x}_{\sqcap}=h(x\sqcap x)$,
$c^{x}_{\sqcup}=h(x\sqcup x)$ and $c^{y}_{\sqcap}=h(y\sqcap y)$,
$c^{y}_{\sqcup}=h(y\sqcup y)$. So $c^{x}_{\sqcap}\sqcap
c^{y}_{\sqcap}=h(x\sqcap x)\sqcap h(y\sqcap y)=h(x\sqcap y)$, using Definition
5(1a) and the fact that $h$ is a homomorphism. This gives $h(x\sqcap y)\sqcap
h(x\sqcap y)=(c^{x}_{\sqcap}\sqcap c^{y}_{\sqcap})_{\sqcap}$ and $h(x\sqcap
y)\sqcup h(x\sqcap y)=(c^{x}_{\sqcap}\sqcap c^{y}_{\sqcap})_{\sqcup}$. By the
uniqueness property of $c^{x\sqcap y}$, $c^{x\sqcap y}=c^{x}_{\sqcap}\sqcap
c^{y}_{\sqcap}=c^{x}\sqcap c^{y}$ (again, using Definition 5(1a)). Hence
$f(x\sqcap y)=f(x)\sqcap f(y)$.
Dually, one gets $f(x\sqcup y)=f(x)\sqcup f(y)$.
(ii) In order to establish that $f$ preserves the other dBa operations, we
first show that for all $x\in D_{p}$, $f(x)=h(x)$, that is, $f|_{D_{p}}=h$.
Let $x\in D_{p}$. Then either $x\sqcap x=x$ or $x\sqcup x=x$. Let us assume
that $x\sqcap x=x$. Then, using (i), $f(x)=f(x\sqcap x)=f(x)\sqcap
f(x)=c^{x}\sqcap c^{x}=c^{x}_{\sqcap}=h(x\sqcap x)=h(x)$ and if $x\sqcup x=x$,
$f(x)=f(x\sqcup x)=f(x)\sqcup f(x)=c^{x}\sqcup c^{x}=c^{x}_{\sqcup}=h(x\sqcup
x)=h(x)$. Therefore $f$ is an extension of $h$.
Now by Propositions 4 and 5, it follows that $\top,\bot,\neg x,\lrcorner x\in
D_{p}$ for all $x\in D$. So, using Definition 5(4a) and (ii), $f(\neg
x)=f(\neg(x\sqcap x))=h(\neg(x\sqcap x))=\neg h(x\sqcap x)=\neg f(x\sqcap
x)=\neg(f(x)\sqcap f(x))=\neg f(x)$, as $h$ is a dBa homomorphism and
$f|_{D_{p}}=h$. Similarly, one gets $f(\lrcorner x)=\lrcorner
f(x),f(\top)=\top$, and $f(\bot)=\bot$. So $f$ is a dBa homomorphism.
(iii) $f$ is injective: let $x,y\in D$, such that $c^{x}=c^{y}$. Then
$h(x\sqcap x)=h(y\sqcap y)$ and $h(x\sqcup x)=h(y\sqcup y)$. So $x\sqcap
x=y\sqcap y$ and $x\sqcup x=y\sqcup y$, as $h$ is injective. By Proposition
4(iv), $x\sqsubseteq y$ and $y\sqsubseteq x$. $\sqsubseteq$ is a partial
order, as D is a contextual dBa. So $x=y$.
(iv) To show $f$ is surjective, let $a\in M$. Since $h$ is a dBa isomorphism
from $\textbf{D}_{p}$ to $\textbf{M}_{p}$, as done above, one can find a
unique $x^{a}\in D$ such that $x^{a}_{\sqcap}=h^{-1}(a\sqcap a)$ and
$x^{a}_{\sqcup}=h^{-1}(a\sqcup a)$. So
$c^{x^{a}}_{\sqcap}=h(x^{a}_{\sqcap})=a\sqcap a$ and
$c^{x^{a}}_{\sqcup}=h(x^{a}_{\sqcup})=a\sqcup a$. By Proposition 4(iv),
$c^{x^{a}}\sqsubseteq a$ and $a\sqsubseteq c^{x^{a}}$. As D is a contextual
dBa, $\sqsubseteq$ is a partial order on $D$. So $f(x^{a})=c^{x^{a}}=a$. Hence
$f$ is surjective.
Therefore $f$ is a dBa isomorphism from D to M such that $h=f|_{D_{p}}$.

If possible, assume that there exists another dBa isomorphism $f_{1}$ from D
to M such that $f_{1}|_{D_{p}}=h$. Let $x\in D$. Then $f(x)\sqcap
f(x)=f(x\sqcap x)=h(x\sqcap x)=f_{1}(x\sqcap x)=f_{1}(x)\sqcap f_{1}(x)$.
Dually, one can show that $f(x)\sqcup f(x)=f_{1}(x)\sqcup f_{1}(x)$. So
$f(x)\sqsubseteq f_{1}(x)$ and $f_{1}(x)\sqsubseteq f(x)$, which implies that
$f(x)=f_{1}(x)$, as D is contextual. ∎

In the following, let $\textbf{D}:=(D,\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$
be a dBa.

######  Notation 10.

For a non-empty finite subset $B$ of $D$,
$\sqcap B:=\sqcap_{a\in B}a$, $\sqcup B:=\sqcup_{a\in B}a$, $\vee
B:=\vee_{a\in B}a$, $\wedge B:=\wedge_{a\in B}a$.

######  Note 1.

Using Proposition 5(iv) and induction on the cardinality of $B$, one can show
that for each non-empty finite subset $B$ of $D$, $\vee B\in D_{\sqcap}$ and
$\wedge B\in D_{\sqcup}$.

######  Proposition 14.

For each non-empty finite subset $B$ of $D,$ the following hold.

  1. 1.

$\sqcap B\sqsubseteq\vee B\sqsubseteq\sqcup B$.

  2. 2.

$\sqcap B\sqsubseteq\wedge B\sqsubseteq\sqcup B$.

###### Proof.

The proof of (ii) is similar to (i). (i) can be easily proved using induction
on the cardinality of $B$ and Proposition 13. ∎

######  Theorem 14.

The following hold for D.

  1. 1.

If $I$ is a proper ideal in D then there exists a primary ideal $I^{1}$ in D
such that $I\subseteq I^{1}$.

  2. 2.

If $F$ is a proper filter in D then there exists a primary filter $F^{1}$ in D
such that $F\subseteq F^{1}$.

###### Proof.

(i) Let $I$ be a proper ideal in $\textbf{D}.$ Then by Lemma 2(iii), $I\cap
D_{\sqcup}$ is an ideal in the Boolean algebra $\textbf{D}_{\sqcup}$. Since
$I$ is proper in D, $I\cap D_{\sqcup}$ must also be proper in
$\textbf{D}_{\sqcup}$. (For, if not, $\top\in I\cap D_{\sqcup}$ and so
$\top\in I$, which is not possible in $\textbf{D}_{\sqcup}$.) Then there
exists a prime ideal $I_{0}$ in $\textbf{D}_{\sqcup}$ such that $I\cap
D_{\sqcup}\subseteq I_{0}$. We define $I^{1}:=\\{x\in D~{}:~{}x\sqsubseteq
y~{}\mbox{for some}~{}y\in I_{0}\\},$ and let $x_{1},x_{2}\in I^{1}.$ So there
are $y_{1},y_{2}\in I_{0}$ such that $x_{1}\sqsubseteq y_{1},x_{2}\sqsubseteq
y_{2}$. Using Proposition 4(vi), we get $x_{1}\sqcup x_{2}\sqsubseteq
y_{1}\sqcup y_{2}$. Since $y_{1},y_{2}\in I_{0}$ and $I_{0}$ is an ideal in
$\textbf{D}_{\sqcup}$, $y_{1}\sqcup y_{2}\in I_{0}$. Thus $x_{1}\sqcup
x_{2}\in I^{1}$. Let $x\in D$ such that $x\sqsubseteq x_{0}$ for some
$x_{0}\in I^{1}$. Then there exists $y\in I_{0}$ such that $x\sqsubseteq
x_{0}\sqsubseteq y$ and so $x\in I^{1}$. Therefore $I_{0}$ is an ideal in D.

Now we will show that $I^{1}\cap D_{\sqcup}=I_{0}$. Let $x\in I^{1}\cap
D_{\sqcup}.$ Then $x\in I^{1}$ implies that there exists $y\in I_{0}\subseteq
D_{\sqcup}$ such that $x\sqsubseteq y$. Since $x,y\in D_{\sqcup}$ and $I_{0}$
is an ideal of $\textbf{D}_{\sqcup}$, by Proposition 2 we get $x\in I_{0}.$
Therefore $I^{1}\cap D_{\sqcup}\subseteq I_{0}.$ By definition of $I^{1}$ it
is clear that $I_{0}\subseteq I^{1}.$ Therefore $I^{1}\cap D_{\sqcup}=I_{0}.$
Thus by Proposition 7 it follows that $I^{1}$ is a primary ideal. Now we claim
that $I\subseteq I^{1}$. Let $x\in I.$ Then $x\sqcup x\in I$, as $I$ is an
ideal. Therefore $x\sqcup x\in I\cap D_{\sqcup}\subseteq I_{0}$. This implies
that $x\sqcup x\in I_{0}$. So $x\sqcup x\in I^{1}$. Since $x\sqsubseteq
x\sqcup x$ and $I^{1}$ is an ideal, $x\in I^{1}.$ Hence $I\subseteq I^{1}$.
Dually one can prove (ii). ∎

######  Proposition 15.

Let M and D be two dBas, and let $h:\textbf{M}\rightarrow\textbf{D}$ be a dBa
homomorphism. The following hold.

  1. 1.

If $a\sqsubseteq b$ then $h(a)\sqsubseteq h(b)$, for all $a,b\in M.$

  2. 2.

If $I$ is a primary ideal in D then $h^{-1}(I)$ is primary ideal in
$\textbf{M}.$

  3. 3.

If $F$ is a primary filter in D then $h^{-1}(F)$ is a primary filter in
$\textbf{M}.$

Moreover, if $h$ is a dBa quasi-isomorphism, the following hold.

  1. (iv)

If $I$ is a primary ideal in M then $h(I)$ is a primary ideal in D.

  2. (v)

If $F$ is a primary filter in M then $h(F)$ is a primary filter in D.

###### Proof.

(i) Let $a,b\in M$ such that $a\sqsubseteq b.$ Then $a\sqcap b=a\sqcap a$ and
$a\sqcup b=b\sqcup b.$ Therefore $h(a)\sqcap h(b)=h(a\sqcap b)=h(a\sqcap
a)=h(a)\sqcap h(a)$ and $h(a\sqcup b)=h(b\sqcup b)=h(b)\sqcup h(b).$ Hence
$h(a)\sqsubseteq h(b).$
The proof of (iii) is dual to the proof of (ii) and the proof of (v) is dual
to that of (iv). We prove (ii) and (iv).
(ii) Let $I$ be a primary ideal in $\textbf{D},$ and let $a,b\in h^{-1}(I).$
Then $h(a\sqcup b)=h(a)\sqcup h(b)\in I$, as $h(a),h(b)\in I$ and $I$ is an
ideal. Therefore $a\sqcup b\in h^{-1}(I)$. Now let $a\in h^{-1}(I)$ and
$x\sqsubseteq a$ for some $x\in M.$ Then by (1), $h(x)\sqsubseteq h(a)$.
Therefore $h(x)\in I,$ which implies $x\in h^{-1}(I)$. If possible, suppose
$h^{-1}(I)=M.$ Then $\top_{M}\in h^{-1}(I)$ implies that
$\top_{D}=h(\top_{M})\in I$, which is not possible. Hence $h^{-1}(I)$ is a
proper ideal in M. Now let $x\in M.$ Then either $h(x)\in I$ or $\lrcorner
h(x)\in I$, as $I$ is primary. That is, either $h(x)\in I$ or $h(\lrcorner
x)\in I.$ So either $x\in h^{-1}(I)$ or $\lrcorner x\in h^{-1}(I)$. Hence
$h^{-1}(I)$ is a primary ideal.
(iv) Let $I$ be a primary ideal in M and $h(a),h(b)\in h(I)$ for some $a,b\in
I$. Since $I$ is an ideal and $a,b\in I$, $h(a)\sqcup h(b)=h(a\sqcup b)\in
h(I)$. Now let $x\sqsubseteq h(a)$ for some $x\in D$. As $h$ is surjective,
there exists a $d\in M$ such that $h(d)=x$. So $h(d)\sqsubseteq h(a)$.
Therefore $d\sqsubseteq a$, as $h$ is quasi-injective. So $d\in I$. Therefore
$x=h(d)\in h(I)$. If possible, suppose $h(I)=D$. Then $\top\in h(I)$, and
there exists $y\in I$ such that $h(y)=\top=h(\top),$ which implies that
$h(\top)\sqsubseteq h(y)$. So $\top\sqsubseteq y$, as $h$ is quasi-injective.
Then $\top\in I$, which is a contradiction. Hence $h(I)$ is a proper ideal in
D. Now let $z\in D$. As $h$ is surjective, there exists $e\in M$ such that
$h(e)=z$. As $I$ is a primary ideal, either $e\in I$ or $\lrcorner e\in I$.
Therefore $z=h(e)\in h(I)$ or $\lrcorner z=\lrcorner h(e)=h(\lrcorner e)\in
h(I)$. Hence $h(I)$ is a primary ideal. ∎

##  4 Contexts on topological spaces

As mentioned in Section 1, for proving the topological representation theorem
for (fully contextual/pure) dBas, we enhance the standard context defined by
Wille by adding topologies on the sets $\mathcal{F}_{pr}(\textbf{D})$ of all
primary filters and $\mathcal{I}_{pr}(\textbf{D})$ of all primary ideals. The
resulting structure is an instance of a context on topological spaces (CTS),
that we define now. It will be shown in the sequel that the structure is, in
fact, an instance of a special kind of CTS, denoted as “CTSCR” (Definition 18
below).

######  Definition 13.

$\mathbb{K}^{T}:=((G,\rho),(M,\tau),R)$ is called a context on topological
spaces (CTS) if

  1. 1.

$(G,\rho)$ and $(M,\tau)$ are topological spaces, and

  2. 2.

$\mathbb{K}:=(G,M,R)$ is a context.

It may be noted that a CTS is a generalization of the topological context of
[14]. Utilizing the presence of the topologies in a CTS, special kinds of
object oriented protoconcepts and object oriented semiconcepts of a CTS shall
be considered now.

######  Definition 14.

A clopen object oriented protoconcept $(A,B)$ of a CTS $\mathbb{K}^{T}$ is an
object oriented protoconcept of $\mathbb{K}$ such that $A$ is clopen (closed
and open) in $(G,\rho)$ and $B$ is clopen in $(M,\tau)$. The set of all clopen
object oriented protoconcepts of $\mathbb{K}^{T}$ is denoted by
$\mathfrak{R}^{T}(\mathbb{K}^{T})$.

Similarly,

######  Definition 15.

A clopen object oriented semiconcept $(A,B)$ of $\mathbb{K}^{T}$ is an object
oriented semiconcept of $\mathbb{K}$ such that $A$ is clopen in $(G,\rho)$ and
$B$ is clopen in $(M,\tau)$. $\mathfrak{S}^{T}(\mathbb{K}^{T})$ denotes the
set of all clopen object oriented semiconcepts of $\mathbb{K}^{T}$.

Let us give an example.

######  Example 2.

Consider the CTS $\mathbb{K}^{T}:=((G,\tau),(M,\rho),R),$ where
$G:=\\{a,b,c,d,e\\}$, $M:=\\{1,2,3,4\\}$,
$\tau:=\\{\\{a,b,c\\},\\{d,e\\},G,\emptyset\\}$,
$\rho:=\\{\\{2\\},\\{1,3,4\\},M,\emptyset\\}$ and
$R$:=$\\{(a,2),(b,2),(c,4),(d,1),(d,4),(e,1),\\\ (e,3)\\}$.
$(\\{a,b,c\\},\\{2\\})$ is a clopen object oriented semiconcept (hence also an
object oriented protoconcept) of $\mathbb{K}^{T}$. Note that
$\\{a,b,c\\}^{\blacksquare}=\\{2\\}$.

######  Observation 5.

For the subset $A:=\\{a,b\\}$ of $G$ in Example 2, $(A,A^{\blacksquare})$ is
an object oriented semiconcept of $\mathbb{K}:=(G,M,R)$, but not a clopen
object oriented semiconcept of the CTS $\mathbb{K}^{T}$.

Different kinds of homomorphisms of CTS are defined as follows.

######  Definition 16.

Let $\mathbb{K}^{T}_{1}:=((X_{1},\tau_{1}),(Y_{1},\rho_{1}),R_{1})$ and
$\mathbb{K}^{T}_{2}:=((X_{2},\tau_{2}),(Y_{2},\rho_{2}),R_{2})$ be two CTS. A
CTS-homomorphism
$(\alpha,\beta):\mathbb{K}^{T}_{1}\rightarrow\mathbb{K}^{T}_{2}$ consists of a
pair of maps such that

  1. 1.

$(\alpha,\beta):\mathbb{K}_{1}\rightarrow\mathbb{K}_{2}$, is a context
homomorphism, and

  2. 2.

$\alpha:X_{1}\rightarrow X_{2}$ and $\beta:Y_{1}\rightarrow Y_{2}$ are
continuous functions.

Denote $f:=(\alpha,\beta)$. If $\alpha$ and $\beta$ are injective, the
homomorphism $f:\mathbb{K}^{T}_{1}\rightarrow\mathbb{K}^{T}_{2}$ is called a
CTS-embedding.
A CTS-embedding $f$ is called a CTS-isomorphism if $\alpha$ and $\beta$ are
surjective.
If $f:\mathbb{K}^{T}_{1}\rightarrow\mathbb{K}^{T}_{2}$ is a CTS-isomorphism
and $\alpha$, $\beta$ are homeomorphisms, $f$ is called a CTS-homeomorphism.
We say $\mathbb{K}^{T}_{1}$ is homeomorphic to $\mathbb{K}^{T}_{2}$.

It is easy to see the following.

######  Proposition 16.

If $(\alpha,\beta):\mathbb{K}^{T}_{1}\rightarrow\mathbb{K}^{T}_{2}$ is a CTS-
homeomorphism then
$(\alpha^{-1},\beta^{-1}):\mathbb{K}^{T}_{2}\rightarrow\mathbb{K}^{T}_{1}$ is
also a CTS-homeomorphism.

$(\alpha^{-1},\beta^{-1})$ in Proposition 16 is called the inverse of
$(\alpha,\beta)$. Let $(X,\rho)$ and $(Y,\tau)$ be two topological spaces, and
$R$ be a binary relation between $X$ and $Y$. In [4, 1], a relation $R$ is
called a many-valued mapping  or correspondence from $X$ into $Y$, as $R$ maps
each $x\in X$ to a subset of $Y$. $X^{*}:=\\{x\in X:R(x)\neq\emptyset\\}$ is
called the domain of the mapping and $Y^{*}:=\cup_{x\in X}R(x)$, the range or
set of values of $R$. The lower inverse for $R$ of a subset $B$ of $Y$ is
defined by $R^{-}(B):=\\{x\in X:R(x)\cap B\neq\emptyset\\}$. On the other
hand, its upper inverse is defined by $R^{+}(B):=\\{x\in X:R(x)\subseteq
B\\}.$ Next we recall the definition of continuity of a many-valued mapping
(relation) [1]. In the following, let $(X,\rho)$ and $(Y,\tau)$ be two
topological spaces and $R$ be a binary relation between $X$ and $Y$.

######  Definition 17.

[1] For any $x_{0}\in X$, a relation $R$ is lower semi-continuous at $x_{0}$
if for each open set $O$ in $(Y,\tau)$ with $R(x_{0})\cap O\neq\emptyset,$
there exists a neighbourhood $U(x_{0})$ in $(X,\rho)$ such that

$x\in U(x_{0})~{}\mbox{implies that}~{}R(x)\cap O\neq\emptyset.$

$R$ is upper semi-continuous at $x_{0}\in X$, if for each open set $O$ in
$(Y,\tau)$ containing $R(x_{0})$ there is a neighbourhood $U(x_{0})$ in
$(X,\rho)$ such that

$x\in U(x_{0})~{}\mbox{implies that}~{}R(x)\subseteq O.$

$R$ is continuous at $x_{0}\in X$, if it is both lower and upper semi-
continuous at $x_{0}$. $R$ is lower semi-continuous in $X$ if $R$ is lower
semi-continuous at each point of $X$. $R$ is upper semi-continuous in $X$ if
$R$ is upper semi-continuous at each point of $X$, and $R$ is continuous in
$X$ if $R$ is both lower and upper semi-continuous in $X$.

######  Observation 6.

Let us recall the definitions of $\square,\lozenge,\blacksquare,\blacklozenge$
given in Section 2 and the definitions of upper and lower inverse of a
relation $R$ given above.

  1. 1.

It can be seen that $B^{\lozenge}=R^{-}(B),B^{\square}=R^{+}(B)$ and
$A^{\blacklozenge}=(R^{-1})^{-}(A),A^{\blacksquare}=(R^{-1})^{+}(A).$

  2. 2.

If $R$ is a function, $B^{\lozenge}=B^{\square}=R^{-1}(B)$. Moreover, if R is
a bijection, $A^{\blacklozenge}=A^{\blacksquare}=R(A)$.

We now define a CTSCR.

######  Definition 18.

A CTSCR is a CTS $\mathbb{K}^{T}:=((G,\rho),(M,\tau),R)$ where $R$ and
$R^{-1}$ are continuous in $G$ and $M$ respectively.

It will be shown in Theorem 17 below that the set of all clopen object
oriented protoconcepts of a CTSCR forms a fully contextual dBa, while the set
of all clopen object oriented semiconcepts forms a pure dBa.

We shall give an example of a CTSCR, but before that let us note the following
results which will be useful while demonstrating semi-continuity and
continuity of relations and their converses.

######  Theorem 15.

[1]

  * I.

The following are equivalent.

    1. (a)

$R$ is upper semi-continuous in $X$.

    2. (b)

For each open set $O$ in $(Y,\tau)$, $O^{\square}$ is open in $(X,\rho)$.

    3. (c)

For each closed set $A$ in $(Y,\tau)$, $A^{\lozenge}$ is closed in $(X,\rho)$.

  * II.

The following are equivalent.

    1. (a)

$R$ is lower semi-continuous in $X$.

    2. (b)

For each open set $O$ in $(Y,\tau)$, $O^{\lozenge}$ is open in $(X,\rho)$.

    3. (c)

For each closed set $B$ in $(Y,\tau)$, $B^{\square}$ is closed in $(X,\rho)$.

######  Corollary 4.

The following are equivalent.

  1. 1.

$R$ is continuous in $X$.

  2. 2.

if $B$ is open in $(Y,\tau)$ then both $B^{\lozenge}$ and $B^{\square}$ are
open in $(X,\rho)$.

  3. 3.

if $B$ is closed in $(Y,\tau)$ then both $B^{\lozenge}$ and $B^{\square}$ are
closed in $(X,\rho)$.

For the converse $R^{-1}$ of $R$, one can similarly derive

######  Theorem 16.

  * I.

The following are equivalent.

    1. (a)

$R^{-1}$ is lower semi-continuous in $Y$.

    2. (b)

For each open set $O$ in $(X,\rho)$, $O^{\blacklozenge}$ is open in
$(Y,\tau)$.

    3. (c)

For each closed set $A$ in $(X,\rho)$, $O^{\blacklozenge}$ is closed in
$(Y,\tau)$.

  * II.

The following are equivalent.

    1. (a)

$R^{-1}$ is upper semi-continuous in $Y$.

    2. (b)

For each open set $O$ in $(X,\rho)$, $O^{\blacksquare}$ is open in $(Y,\tau)$.

    3. (c)

For each closed set $A$ in $(X,\rho)$, $A^{\blacksquare}$ is closed in
$(Y,\tau)$.

######  Corollary 5.

The following are equivalent.

  1. 1.

$R^{-1}$ is continuous in $Y$.

  2. 2.

if $A$ is open in $(X,\rho)$ then both $A^{\blacklozenge}$ and
$A^{\blacksquare}$ are open in $(Y,\tau)$.

  3. 3.

if $A$ is closed in $(X,\rho)$ then both $A^{\blacklozenge}$ and
$A^{\blacksquare}$ are closed in $(Y,\tau)$.

Now let us give an example of a CTSCR.

######  Example 3.

Let $(X,\tau_{1})$ and $(Y,\tau_{2})$ be two non-empty totally disconnected
spaces, and $C$ be a fixed non-empty clopen subset of $(Y,\tau_{2}).$ Then
$\mathbb{K}_{+}^{T}:=((X,\tau_{1}),(Y,\tau_{2}),R),$ where $xRy$ if and only
if $y\in C$, is a CTSCR.

###### Proof.

One needs to show that $R$ and $R^{-1}$ are continuous. For establishing
continuity of $R$, utilizing Corollary 4, we just show that for any open set
$O$ in $(Y,\tau_{2})$, $O^{\square},O^{\lozenge}$ are open in $(X,\tau_{1})$.
Observe that for any $B\subseteq Y$,

$B^{\square}=\begin{cases}X&\mbox{if }C\subseteq B\\\ \emptyset&\mbox{if
}C\nsubseteq B\end{cases}~{}\mbox{and}~{}B^{\lozenge}=\begin{cases}X&\mbox{if
}B\nsubseteq C^{c}\\\ \emptyset&\mbox{if }B\subseteq C^{c}\end{cases}$

Therefore for any open set $O$ in $(Y,\tau_{2})$, $O^{\square}$ and
$O^{\lozenge}$ are always open in $(X,\tau_{1})$ and hence $R$ is continuous.
Now consider the case of $R^{-1}$. From the definition of $R$ it is clear that

$R^{-1}(y)=\begin{cases}\emptyset&\mbox{if}~{}y\notin C\\\ X&\mbox{if}~{}y\in
C\end{cases}$

So for any $A\subseteq X$,

$A^{\blacksquare}=\begin{cases}C^{c}&\mbox{if}~{}A\subset X\\\
Y&\mbox{if}~{}A=X\end{cases}~{}\mbox{and}~{}A^{\blacklozenge}=C$

Thus for any open set $O$ of $(X,\tau_{1})$,
$O^{\blacksquare},O^{\blacklozenge}$ are open in $(Y,\tau_{2})$, making
$R^{-1}$ continuous by Corollary 5. ∎

For CTSCRs, we rename the maps given in Definition 16.

######  Definition 19.

In case of CTSCRs, a CTS-homomorphism, CTS-embedding, CTS-isomorphism and CTS-
homeomorphism are respectively called CTSCR-homomorphism, CTSCR-embedding,
CTSCR-isomorphism, and CTSCR-homeomorphism.

######  Theorem 17.

Let $\mathbb{K}^{T}:=((X,\tau_{1}),(Y,\tau_{2}),R)$ be a CTSCR.

  1. 1.

$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}):=(\mathfrak{R}^{T}(\mathbb{K}^{T}),\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$
is a fully contextual dBa.

  2. 2.

$\mathcal{S}^{T}(\mathbb{K}^{T}):=(\mathfrak{S}^{T}(\mathbb{K}^{T}),\sqcup,\sqcap,\neg,\lrcorner,\top,\bot)$
is a pure dBa.

###### Proof.

(i) Let $(A,B)$ and $(C,D)$ belong to $\mathfrak{R}^{T}(\mathbb{K}^{T}).$ Then
$(A,B)\sqcap(C,D)=(A\cup C,(A\cup
C)^{\blacksquare}),\neg(A,B)=(A^{c},A^{c\blacksquare})$ and
$(A,B)\sqcup(C,D)=((B\cap D)^{\lozenge},B\cap
D),\lrcorner(A,B)=(B^{c\lozenge},B^{c})$. Since $A,C$ are clopen in
$(X,\tau_{1})$, $A\cup C$ and $A^{c}$ are also clopen in $(X,\tau_{1})$.
Similarly $B,D$ are clopen in $(Y,\tau_{2})$ implies that $B\cap D$ and
$B^{c}$ are clopen in $(Y,\tau_{2})$. Since $R^{-1}$ is continuous, by
Corollary 5, $(A\cup C)^{\blacksquare}$ and $A^{c\blacksquare}$ are closed and
open in $(Y,\tau_{2})$. Similarly, continuity of $R$ implies that $(B\cap
D)^{\lozenge}$ and $B^{c\lozenge}$ are clopen in $(X,\tau_{1})$, by Corollary
4. Therefore $\mathfrak{R}^{T}(\mathbb{K}^{T})$ is closed under
$\sqcup,\sqcap,\neg,\lrcorner$. It is clear that $\top=(\emptyset,\emptyset)$
and $\bot=(G,M)$ both belong to $\mathfrak{R}^{T}(\mathbb{K}^{T})$. It is
routine to verify that the set of all clopen object oriented protoconcepts
satisfies the axioms of dBa with respect to
$\sqcup,\sqcap,\neg,\lrcorner,\top,\bot$. Since
$\mathfrak{R}^{T}(\mathbb{K}^{T})\subseteq\mathfrak{R}(\mathbb{K})$, by
Proposition 12, $\sqsubseteq$ is a partial order on
$\mathfrak{R}^{T}(\mathbb{K}^{T}).$ Therefore
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T})$ is a contextual dBa. Next we
show that $\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T})$ is fully contextual.
Let $(A,B),(C,D)\in\mathfrak{R}^{T}(\mathbb{K}^{T})$ such that
$(A,B)=(A,B)_{\sqcap}=(A,A^{\blacksquare})$ and
$(C,D)=(C,D)_{\sqcup}=(D^{\lozenge},D)$. Let us also assume that
$(A,B)_{\sqcup}=(C,D)_{\sqcap}$. Then
$(A^{\blacksquare\lozenge},A^{\blacksquare})=(D^{\lozenge},D^{\lozenge\blacksquare})$,
which implies that $A^{\blacksquare\lozenge}=D^{\lozenge}$. So
$(A,D)\in\mathfrak{R}^{T}(\mathbb{K}^{T})$ such that $(A,D)_{\sqcap}=(A,B)$
and $(A,D)_{\sqcup}=(C,D)$ and it is clearly unique.
(ii) Similar to the proof in (i), it can be shown that
$\mathfrak{S}^{T}(\mathbb{K}^{T})$ is closed with respect to the operations
$\sqcup,\sqcap,\neg,\lrcorner,\top,\bot$ and forms a pure dBa. ∎

######  Corollary 6.

Let $\mathbb{K}^{T}:=((X,\tau_{1}),(Y,\tau_{2}),R)$ be a CTSCR such that $R$
is a homeomorphism from $(X,\tau_{1})$ to $(Y,\tau_{2})$. Then
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T})=\mathcal{S}^{T}(\mathbb{K}^{T})$
is a Boolean algebra.

###### Proof.

Since $R$ is a bijection, it follows from Observation 6(ii) that for all
$A\subseteq X$ and $B\subseteq Y$, $A^{\blacksquare}=R(A)$ and
$B^{\lozenge}=R^{-1}(B)$, respectively. Let
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T})$. Then
$A^{\blacksquare\lozenge}=B^{\lozenge}$, which is equivalent to
$R^{-1}(R(A))=R^{-1}(B)$. As $R$ is a bijection, $R(A)=B$, which implies that
$(A,R(A))=(A,A^{\blacksquare})=(A,B)\in\mathfrak{S}^{T}(\mathbb{K}^{T})$. So
$\mathfrak{S}^{T}(\mathbb{K}^{T})=\mathfrak{R}^{T}(\mathbb{K}^{T})$. From
Theorem 17 it follows that
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T})=\mathcal{S}^{T}(\mathbb{K}^{T})$
is a fully contextual pure dBa. Let
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T})$. As $R$ and $R^{-1}$ are
bijections, we get the following equalities.
$\neg(A,B)=(A^{c},R(A^{c}))=(A^{c},(R(A))^{c})=((R^{-1}(B))^{c},B^{c})=(R^{-1}(B^{c}),B^{c})=\lrcorner(A,B)$
and $\neg\neg(A,B)=(A,B)$. By Theorem 12,
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T})$ is a Boolean algebra. ∎

Theorem 18 below gives us a relation between the algebras
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{1}^{T}),\underline{\mathfrak{R}}^{T}(\mathbb{K}_{2}^{T})$
and $\mathcal{S}^{T}(\mathbb{K}_{1}^{T}),\mathcal{S}^{T}(\mathbb{K}_{2}^{T})$
when the CTSCRs $\mathbb{K}^{T}_{1},\mathbb{K}^{T}_{2}$ are homeomorphic. For
that, we use the following results about isomorphic contexts.

######  Proposition 17.

Let $\mathbb{K}_{1}:=(X_{1},Y_{1},R_{1}),\mathbb{K}_{2}:=(X_{2},Y_{2},R_{2})$
be two contexts and $(\alpha,\beta)$ a context isomorphism from
$\mathbb{K}_{1}$ to $\mathbb{K}_{2}$. Then the following hold.

  1. 1.

For $A\subseteq X_{1}$, $\alpha(A)^{\blacksquare}=\beta(A^{\blacksquare})$ and
$\beta(A^{\blacklozenge})=\alpha(A)^{\blacklozenge}.$

  2. 2.

For $B\subseteq Y_{1}$, $\alpha(B^{\square})=\beta(B)^{\square}$ and
$\beta(B)^{\lozenge}=\alpha(B^{\lozenge}).$

###### Proof.

(i) Let $A\subseteq X_{1}$. We will show that
$\alpha(A)^{\blacksquare}=\beta(A^{\blacksquare})$. Let
$a\in\alpha(A)^{\blacksquare}$, $x\in X_{1}$ and $xR_{1}\beta^{-1}(a)$. Then
$\alpha(x)R_{2}a$, as $f$ is a context isomorphism. Therefore
$\alpha(x)\in\alpha(A).$ So $x\in A$, as $\alpha$ is a bijection. This implies
that $\beta^{-1}(a)\in A^{\blacksquare}.$ Hence $a\in\beta(A^{\blacksquare})$.
For the other direction, let $b\in\beta(A^{\blacksquare}).$ Then
$\beta^{-1}(b)\in A^{\blacksquare}$. Let $x^{\prime}\in X_{2}$ and
$x^{\prime}R_{2}b.$ This means $\alpha^{-1}(x^{\prime})R_{1}\beta^{-1}(b)$, as
$f$ is a context isomorphism. So $\alpha^{-1}(x^{\prime})\in A,$ and
$x^{\prime}\in\alpha(A)$, which implies that $b\in\alpha(A)^{\blacksquare}$.

Next let us show that $\beta(A^{\blacklozenge})=\alpha(A)^{\blacklozenge}$.
Let $y\in\beta(A^{\blacklozenge}).$ Then $\beta^{-1}(y)\in A^{\blacklozenge}$,
which implies that $R_{1}^{-1}(\beta^{-1}(y))\cap A\neq\emptyset$. So there
exists $x\in A$ such that $xR_{1}\beta^{-1}(y)$, which gives
$\alpha(x)R_{2}y$, as $f$ is a context isomorphism. This implies that
$R^{-1}_{2}(y)\cap\alpha(A)\neq\emptyset$ and hence
$y\in\alpha(A)^{\blacklozenge}$. On the other hand, let
$y_{0}\in\alpha(A)^{\blacklozenge}.$ Then
$R^{-1}_{2}(y_{0})\cap\alpha(A)\neq\emptyset.$ So there exists
$\alpha(x)\in\alpha(A)$ such that $\alpha(x)R_{2}y_{0}$. As $f$ is a context
isomorphism, $xR_{1}\beta^{-1}(y_{0}).$ This implies that
$R_{1}^{-1}(\beta^{-1}(y_{0}))\cap A\neq\emptyset$. So $\beta^{-1}(y_{0})\in
A^{\blacklozenge}$, whence $y_{0}\in\beta(A^{\blacklozenge}).$
(ii) This can be proved similarly. ∎

######  Definition 20.

Let $\mathbb{K}^{T}_{1}:=((X_{1},\tau),(Y_{1},\rho),R_{1})$ and
$\mathbb{K}^{T}_{2}:=((X_{2},\tau_{1}),(Y_{2},\rho_{1}),R_{2})$ be two CTSCRs
and $f:=(\alpha,\beta)$ be a CTSCR-homeomorphism from $\mathbb{K}^{T}_{1}$ to
$\mathbb{K}^{T}_{2}$. The map
$f_{\alpha\beta}:\mathfrak{R}^{T}(\mathbb{K}^{T}_{2})\rightarrow\mathfrak{R}^{T}(\mathbb{K}^{T}_{1})$
is defined as $f_{\alpha\beta}((A,B)):=(\alpha^{-1}(A),\beta^{-1}(B))$ for all
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{2})$.

######  Theorem 18.

If $f:=(\alpha,\beta)$ is a CTSCR-homeomorphism then $f_{\alpha\beta}$ is a
dBa isomorphism from $\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}_{2})$ to
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}_{1})$. Moreover, the restriction
of $f_{\alpha\beta}$ on $\mathcal{S}^{T}(\mathbb{K}^{T}_{2})$ gives a dBa
isomorphism from $\mathcal{S}^{T}(\mathbb{K}^{T}_{2})$ to
$\mathcal{S}^{T}(\mathbb{K}^{T}_{1})$.

###### Proof.

As $f:=(\alpha,\beta)$ is a CTSCR-isomorphism from $\mathbb{K}^{T}_{1}$ to
$\mathbb{K}^{T}_{2}$, $f$ is a context isomorphism and $\alpha,\beta$ are
continuous functions on $X_{1},X_{2}$ respectively. Let
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{2}).$ Then
$A^{\blacksquare\lozenge}=B^{\lozenge}$. By Proposition 1,
$(\alpha^{-1},\beta^{-1})$ is a context isomorphism. So, using Proposition 17,
$\alpha^{-1}(A)^{\blacksquare\lozenge}=\beta^{-1}(A^{\blacksquare})^{\lozenge}=\alpha^{-1}(A^{\blacksquare\lozenge})=\alpha^{-1}(B^{\lozenge})=\beta^{-1}(B)^{\lozenge}.$
Therefore $(\alpha^{-1}(A),\beta^{-1}(B))$ is an object oriented protoconcept
of the context $\mathbb{K}_{1}$ and
$(\alpha^{-1}(A),\beta^{-1}(B))\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{1})$, as
$\alpha,\beta$ are continuous. Hence $f_{\alpha\beta}$ is a well-defined map.
Let
$x:=(A_{1},B_{1}),y:=(A_{2},B_{2})\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{2})$.
Using Proposition 17 we get,

$\displaystyle f_{\alpha\beta}(x\sqcap y)=$
$\displaystyle(\alpha^{-1}(A_{1}\cup A_{2}),\beta^{-1}((A_{1}\cup
A_{2})^{\blacksquare}))$ $\displaystyle=$ $\displaystyle(\alpha^{-1}(A_{1}\cup
A_{2}),\alpha^{-1}(A_{1}\cup A_{2})^{\blacksquare})$ $\displaystyle=$
$\displaystyle(\alpha^{-1}(A_{1})\cup\alpha^{-1}(A_{2}),(\alpha^{-1}(A_{1})\cup\alpha^{-1}(A_{2}))^{\blacksquare})$
$\displaystyle=$ $\displaystyle f_{\alpha\beta}(x)\sqcap f_{\alpha\beta}(y).$

Dually, $f_{\alpha\beta}(x\sqcup y)=f_{\alpha\beta}(x)\sqcup
f_{\alpha\beta}(y)$. $f_{\alpha\beta}(\lrcorner
x)=(\alpha^{-1}(B_{1}^{c\lozenge}),\beta^{-1}(B_{1}^{c}))=(\beta^{-1}(B_{1}^{c})^{\lozenge},\beta^{-1}(B_{1}^{c}))=\lrcorner
f_{\alpha\beta}(x)$. Similarly, $f_{\alpha\beta}(\neg x)=\neg
f_{\alpha\beta}(x)$. Since $\alpha$ and $\beta$ are bijective,
$f_{\alpha\beta}(\top)=\top$ and $f_{\alpha\beta}(\bot)=\bot$. Therefore
$f_{\alpha\beta}$ is a dBa homomorphism. If
$f_{\alpha\beta}(x)=f_{\alpha\beta}(y)$ then
$\alpha^{-1}(A_{1})=\alpha^{-1}(A_{2})$ and
$\beta^{-1}(B_{1})=\beta^{-1}(B_{2})$. Therefore $A_{1}=A_{2}$ and
$B_{1}=B_{2}$, as $\alpha,\beta$ are bijections, which implies that $x=y.$ So
$f_{\alpha\beta}$ is an injective dBa homomorphism.
Let $(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{1}).$ Then
$A^{\blacksquare\lozenge}=B^{\lozenge}$. Therefore using Proposition 17,
$\alpha(A)^{\blacksquare\lozenge}=\beta(B)^{\lozenge}$. Since $\alpha,\beta$
are homeomorphisms,
$(\alpha(A),\beta(B))\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{2})$. So
$f_{\alpha\beta}((\alpha(A),\beta(B)))=(\alpha^{-1}(\alpha(A)),\beta^{-1}(\beta(B)))=(A,B)$,
as $\alpha,\beta$ are bijective. Hence $f_{\alpha\beta}$ is a dBa isomorphism.
Second part of the theorem follows from Theorem 13. ∎

In the next theorem, we show that the property of being a CTSCR is invariant
under CTS-homeomorphisms.

######  Theorem 19.

Let $\mathbb{K}^{T}_{1}:=((X_{1},\tau_{1}),(Y_{1},\rho_{1}),R_{1})$ and
$\mathbb{K}^{T}_{2}:=((X_{2},\tau_{2}),(Y_{2},\rho_{2}),R_{2})$ be two CTS
such that $\mathbb{K}^{T}_{1}$ is homeomorphic to $\mathbb{K}^{T}_{2}$. Then
$\mathbb{K}^{T}_{1}$ is a CTSCR if and only if $\mathbb{K}^{T}_{2}$ is a
CTSCR.

###### Proof.

Let $f:=(\alpha,\beta)$ from $\mathbb{K}^{T}_{1}$ to $\mathbb{K}^{T}_{2}$ be a
CTS-homeomorphism, and suppose that $\mathbb{K}^{T}_{1}$ is a CTSCR. We show
that $R_{2}$ and $R^{-1}_{2},$ are continuous. Let $O$ be an open set in
$(Y_{2},\rho_{2}).$ By Corollary 4, it is sufficient to show that
$O^{\square}$ and $O^{\lozenge}$ are open in $(X_{2},\tau_{2})$.
$\beta^{-1}(O)$ is open in $(Y_{1},\tau_{1})$, as $\beta$ is continuous and
$O$ is open in $(Y_{2},\rho_{2})$. Now continuity of $R_{1}$ implies that
$\beta^{-1}(O)^{\square}$ and $\beta^{-1}(O)^{\lozenge}$ are open in
$(X_{1},\tau_{1})$. Since $\alpha$ is a homeomorphism from $(X_{1},\tau_{1})$
to $(X_{2},\tau_{2})$, $\alpha(\beta^{-1}(O)^{\square})$ and
$\alpha(\beta^{-1}(O)^{\lozenge})$ are open in $(X_{2},\tau_{2})$. By
Proposition 17(ii),
$\alpha(\beta^{-1}(O)^{\lozenge})=\beta(\beta^{-1}(O))^{\lozenge}=O^{\lozenge}$
and
$\alpha(\beta^{-1}(O)^{\square})=\beta(\beta^{-1}(O))^{\square}=O^{\square}.$
Hence $O^{\square}$ and $O^{\lozenge}$ are open in $(X_{2},\tau_{2})$. Thus
$R_{2}$ is continuous.

The proof of continuity of $R_{2}^{-1}$ is similar, making use of the
continuity of $R_{1}^{-1}$ and the fact that $\beta$ is homeomorphism. ∎

It may be remarked that because of Theorem 19 and Theorem 17, either one of
$\mathbb{K}^{T}_{1},\mathbb{K}^{T}_{2}$ in Theorem 18 could have been taken to
be a CTS and the other as a CTSCR.

Theorem 17 establishes that given a CTSCR, one obtains a fully contextual and
a pure dBa. What about the converse – that is, given a (fully contextual/pure)
dBa D, can one define an “appropriate” CTSCR corresponding to D? We address
this question in the next section, and obtain representation results for dBas
in terms of the corresponding CTSCRs.

##  5 Representation theorem for dBas

Let D be any dBa. In this section, we shall work with the complement of the
standard context
$\mathbb{K}(\textbf{D}):=(\mathcal{F}_{p}(\textbf{D}),\mathcal{I}_{p}(\textbf{D}),\Delta),$
namely
$\mathbb{K}^{c}(\textbf{D}):=(\mathcal{F}_{p}(\textbf{D}),\mathcal{I}_{p}(\textbf{D}),\nabla),$
where $\nabla$ is the complement of $\Delta$. So for
$F\in\mathcal{F}_{p}(\textbf{D})$ and $I\in\mathcal{I}_{p}(\textbf{D})$,
$F\nabla I$ if and only in $F\cap I=\emptyset$. By Proposition 7 (cf. Section
2.3),
$\mathbb{K}^{c}(\textbf{D})=(\mathcal{F}_{pr}(\textbf{D}),\mathcal{I}_{pr}(\textbf{D}),\nabla)$.
The context $\mathbb{K}^{c}(\textbf{D})$ will henceforth be denoted by
$\mathbb{K}_{pr}(\textbf{D})$. We shall define a CTS
$\mathbb{K}_{pr}^{T}(\textbf{D})$ based on the context
$\mathbb{K}_{pr}(\textbf{D})$. Recall the sets $F_{x},I_{x}$ given in Notation
5.

######  Proposition 18.

For any $x\in D$, the following hold.

  1. 1.

(a) $F_{x}^{\blacksquare}=I_{\neg x}$, (b)
$F_{x}^{\blacklozenge}=I_{\lrcorner(x\sqcap x)}$.

  2. 2.

(a) $I_{x}^{\square}=F_{\lrcorner x}$, (b) $I_{x}^{\lozenge}=F_{\neg(x\sqcup
x)}$.

###### Proof.

(i)(a). Let $I\in I_{\neg x}$ and $F\cap I=\emptyset$, for some
$F\in\mathcal{F}_{pr}(\textbf{D})$. As $\neg x\in I$, we get $\neg x\notin F$.
So $x\in F$, $F$ being a primary filter. $F\nabla I$ then implies that $F\in
F_{x}$. Therefore $I_{\neg x}\subseteq F_{x}^{\blacksquare}$.
On the other hand, let $I\in F_{x}^{\blacksquare}$, and if possible, suppose
$\neg x\notin I$. Then $I\cap F(\\{\neg x\\})=\emptyset,$ where $F(\\{\neg
x\\})$ is the filter generated by $\neg x$, as otherwise $\neg x\in I$.
Therefore by Theorem 3, there exists a primary filter $F$ containing
$F(\\{\neg x\\})$ and such that $F\cap I=\emptyset$. This implies that $\neg
x\in F$, that is $x\notin F$ – which contradicts the assumption that $I\in
F_{x}^{\blacksquare}.$ Hence $F_{x}^{\blacksquare}=I_{\neg x}.$
(i)(b). Let $I\in I_{\lrcorner(x\sqcap x)}.$ Then $x\sqcap x\notin I$. Let
$F_{0}$ be the filter generated by $x\sqcap x$. $F_{0}\cap I=\emptyset$, as
otherwise $x\sqcap x\in I$. Therefore by Theorem 3, there is a primary filter
$F$ containing $F_{0}$ such that $F\cap I=\emptyset$ and $x\in F$. Hence
$I_{\lrcorner(x\sqcap x)}\subseteq F_{x}^{\blacklozenge}$.
Now suppose $I\in F_{x}^{\blacklozenge}$. Then there exists $F_{1}\in F_{x}$
such that $I\cap F_{1}=\emptyset$. Since $x\in F_{1}$, $x\sqcap x\in F_{1}$ so
$x\sqcap x\notin I$ and hence $\lrcorner(x\sqcap x)\in I$.
The proofs of (ii)(a) and (b) are similar to the above. ∎

######  Definition 21.

A topology $\mathcal{J}$ on $\mathcal{I}_{pr}(\textbf{D})$ is defined by
taking $\mathcal{B}:=\\{I_{x}:x\in D\\}$ as a subbase for the closed sets of
$\mathcal{J}.$ Similarly, a topology $\mathcal{T}$ on
$\mathcal{F}_{pr}(\textbf{D})$ is defined by taking
$\mathcal{B}_{0}:=\\{F_{x}:x\in D\\}$ as a subbase for the closed sets of
$\mathcal{T}.$

######  Note 2.

For each $x\in D$, $I_{x}$ is clopen in
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$ and $F_{x}$ is clopen in
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$. Moreover, for each open set $O$
in $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$, $O^{c}=\cap_{j\in
J}\cup_{a\in D_{j}}F_{a}$, where $D_{j}$ is some finite subset of $D$ for each
$j$ belonging to some index set $J$. Therefore $O=\cup_{j\in J}\cap_{a\in
D_{j}}F^{c}_{a}=\cup_{j\in J}\cap_{a\in D_{j}}F_{\neg a}$.
Similarly, for each open set $O$ in
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$, $O=\cup_{j\in J}\cap_{a\in
D_{j}}I_{\lrcorner a}$.

Let us recall

######  Theorem 20 (Alexander’s Subbase Lemma [29]).

Let $(X,\tau)$ be a topological space and $S_{0}$ a subbase of closed sets of
$(X,\tau)$. If every family of closed sets in $S_{0}$ with finite intersection
property has a non-empty intersection, then $(X,\tau)$ is compact.

The following can now be established.

######  Theorem 21.

$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ and
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$ are compact and totally
disconnected topological spaces. Hence both are Hausdorff spaces.

###### Proof.

We prove for $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$; the other case
follows dually.
To show that $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ is compact, we
consider the subbase $\mathcal{B}_{0}$ for $\mathcal{T}$ given in Definition
21 and a family $\mathfrak{C}\subseteq\mathcal{B}_{0}$ having the finite
intersection property. We show that $\cap\mathfrak{C}\neq\emptyset$ and use
Theorem 20.
Let $S:=\\{a\in D:F_{a}\in\mathfrak{C}\\}$ and $F(S)$ be the filter generated
by $S$. We claim that $F(S)\neq D$. Indeed, if not, $\bot\in F(S)$ – which
implies that $a_{1}\sqcap\ldots\sqcap a_{n}\sqsubseteq\bot$ for some
$a_{1},\ldots,a_{n}$ in $S.$ Therefore $F_{a_{1}\sqcap\ldots\sqcap
a_{n}}=\emptyset$, as $F_{\bot}=\emptyset.$ So
$\cap^{n}_{i=1}F_{a_{i}}=F_{\sqcap^{n}_{i=1}a_{i}}=\emptyset$, which is a
contradiction.
As $F(S)\neq D$, by Theorem 14, there exists
$F\in\mathcal{F}_{pr}(\textbf{D})$ such that $S\subseteq F.$ Hence
$F\in\cap\mathfrak{C}$, that is $\cap\mathfrak{C}\neq\emptyset$.
Now let $x\in D.$ $F_{x}$ is clopen in $\mathcal{F}_{pr}(\textbf{D})$. Let
$F_{1},F_{2}\in\mathcal{F}_{pr}(\textbf{D})$ and $F_{1}\neq F_{2}.$ Then
either $F_{1}\nsubseteq F_{2}$ or $F_{2}\nsubseteq F_{1}$. Without loss of
generality, suppose $F_{1}\nsubseteq F_{2}$. So there is $x\in F_{1}$ but
$x\notin F_{2}$, implying that $F_{1}\in F_{x}$ and $F_{2}\notin F_{x}.$ Hence
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ is totally disconnected. ∎

Corresponding to the dBa D, we next consider a CTS
$\mathbb{K}_{pr}^{T}(\textbf{D}):=((\mathcal{F}_{pr}(\textbf{D}),\mathcal{T}),(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J}),\nabla)$
that is based on the context $\mathbb{K}_{pr}(\textbf{D})$ (mentioned at the
beginning of this section) and topologies $\mathcal{T}$ and $\mathcal{J}$ as
given in Definition 21. It will be shown that
$\mathbb{K}^{T}_{pr}(\textbf{D})$ is a CTSCR (Theorem 23 below). To prove
this, we shall use Rado’s Selection Principle.

######  Theorem 22 (Rado’s Selection Principle [28]).

Let $I$ be an arbitrary index set and $S$ be any set. Let
$\mathcal{U}:=\\{A_{i}:~{}i\in I\\}$ be a family of non-empty finite subsets
of a set $S$ and $\mathfrak{I}$ be the collection of all non-empty finite
subsets of $I$. Further, for each $J\in\mathfrak{I}$, let there be a function
$f_{J}:J\rightarrow\cup_{i\in J}A_{i}~{}\mbox{such that}~{}f_{J}(i)\in
A_{i}~{}\mbox{for all}~{}i\in J.$ Then there exists a function
$f:I\rightarrow\cup_{i\in I}A_{i}~{}\mbox{such that}~{}f(i)\in
A_{i}~{}\mbox{for all}~{}i\in I$, and for every $J\in\mathfrak{I}$ there is
some $K\in\mathfrak{I}$ with $J\subseteq K$ such that $f_{K}(j)=f(j)$ for all
$j\in J$.

Let us fix some notations for the dBa D.

######  Notation 11.

$\mathcal{U}_{0}:=\\{D_{i}\\}_{i\in J}$ denotes the family of all non-empty
finite subsets $D_{i}$ of $D$, indexed over a set $J$. $\mathfrak{J}_{0}$
denotes the set of all non-empty finite subsets of $J$. For each $i\in J$,
$D_{i\sqcap}:=\\{a\sqcap a~{}:~{}a\in D_{i}\\}$ and $D_{i\sqcup}:=\\{a\sqcup
a~{}:~{}a\in D_{i}\\}$. Note that $D_{i\sqcap}\subseteq D_{\sqcap}$ and
$D_{i\sqcup}\subseteq D_{\sqcup}$.
For each $E\in\mathfrak{J}_{0}$,
$\mathfrak{F}_{E}:=\\{f:E\rightarrow\cup_{j\in J}D_{i\sqcap}~{}\mbox{such
that}~{}f(j)\in D_{j\sqcap}\\}$ and
$\mathfrak{G}_{E}:=\\{g:E\rightarrow\cup_{j\in J}D_{i\sqcup}~{}\mbox{such
that}~{}g(j)\in D_{j\sqcup}\\}.$
$\sqcap f(E):=\sqcap_{j\in E}f(j)$ for each $f\in\mathfrak{F}_{E}$, and
$\sqcup g(E):=\sqcup_{j\in E}g(j)$ for each $g\in\mathfrak{G}_{E}$.
Note that $\sqcap f(E)\in D_{\sqcap}$ and $\sqcup g(E)\in D_{\sqcup}$ for each
$f\in\mathfrak{F}_{E}$ and $g\in\mathfrak{G}_{E}$ respectively.
$\mathcal{D}_{E}:=\\{\sqcap f(E)~{}:~{}f\in\mathfrak{F}_{E}\\}$,
$\mathcal{B}_{E}:=\\{\sqcup g(E)~{}:~{}g\in\mathfrak{G}_{E}\\}$.

######  Observation 7.

  1. 1.

$\mathfrak{F}_{E}$ is non-empty and finite, as both $E$ and $\cup_{j\in
E}D_{i\sqcap}$ are non-empty and finite. Similarly, $\mathfrak{G}_{E}$ is non-
empty and finite.

  2. 2.

$\mathcal{D}_{E}(\mathcal{B}_{E})$ is a non-empty and finite subset of
$D_{\sqcap}(D_{\sqcup})$, as $\mathfrak{F}_{E}(\mathfrak{G}_{E})$ is non-empty
and finite.

Let us recall Notation 10 and introduce

######  Notation 12.

For each $E\in\mathfrak{J}_{0}$,
$a_{E}:=\vee\mathcal{D}_{E}=\vee_{f\in\mathfrak{F}_{E}}\sqcap f(E)$ and
$b_{E}:=\wedge\mathcal{B}_{E}=\wedge_{g\in\mathfrak{G}_{E}}\sqcup g(E)$.

Recall Proposition 2 stating that
$\textbf{D}_{\sqcap}:=(D_{\sqcap},\sqcap,\vee,\neg,\bot,\neg\bot)$ and
$\textbf{D}_{\sqcup}:=(D_{\sqcup},\sqcup,\wedge,\lrcorner,\top,\lrcorner\top)$
are Boolean algebras, so that $\vee$ gives the least upper bound operator in
$\textbf{D}_{\sqcap}$ and $\wedge$ gives the greatest lower bound operator in
$\textbf{D}_{\sqcup}$. Therefore, $\sqcap f(E)\sqsubseteq a_{E}$ for any
$f\in\mathfrak{F}_{E}$ and $b_{E}\sqsubseteq\sqcup g(E)$ for all
$g\in\mathfrak{G}_{E}$.

######  Theorem 23.

$\mathbb{K}_{pr}^{T}(\textbf{D})$ is a CTSCR.

###### Proof.

We prove that $\nabla^{-1}$ is (i) upper semi-continuous and (ii) lower semi-
continuous. Proofs similar to those of (i) and (ii) will imply that $\nabla$
has these two properties as well.
(i) Using Theorem 16 (II), it is sufficient to show that for each open set $O$
in $\mathcal{T}$, $O^{\blacksquare}$ is open in $\mathcal{J}$. Suppose $O$ is
an open set in $\mathcal{T}$. Then from Theorem 4(vi) it follows that
$O^{\blacksquare}_{\nabla^{-1}}=O^{c\prime}_{\Delta^{-1}}$. $O^{c}$ is closed
in $\mathcal{T}$, therefore $O^{c}=\cap_{j\in J}\cup_{a\in D_{j}}F_{a}$. To
show $O^{c\prime}_{\Delta^{-1}}$ is open, we will show that
$O^{c\prime}_{\Delta^{-1}}$ is the union of some open sets $I_{a}$, for which
we need to find elements $a$ in D. Let
$B:=\\{a_{E}~{}:~{}E\in\mathfrak{J}_{0}\\}$. We show that
$O_{\Delta^{-1}}^{c\prime}=\cup_{a\in B}I_{a}$.

Let $I\in\cup_{a\in B}I_{a}.$ Then there exists $a_{E_{0}}\in B$ such that
$a_{E_{0}}\in I$. Let $F\in O^{c}$ – that is if and only if $F\in\cup_{a\in
D_{j}}F_{a}$ for all $j\in J$, which is equivalent to $F\cap
D_{j}\neq\emptyset$ for all $j\in J.$ Since $F$ is a filter, $F\cap
D_{j}\neq\emptyset$ if and only if $F\cap D_{j\sqcap}\neq\emptyset$ for all
$j\in J.$ So $F\in O^{c}$ implies that $F\cap D_{j\sqcap}\neq\emptyset$ for
all $j\in E_{0}$. As $F\cap D_{j\sqcap}$ is finite, we choose and fix
$b_{j}\in F\cap D_{j\sqcap}$ for each $j\in E_{0}$ and define a function $f$
from $E_{0}$ to $\cup_{j\in E_{0}}D_{j\sqcap}$ by $f(j):=b_{j}$ for all $j\in
E_{0}.$ Then $f\in\mathfrak{F}_{E_{0}}$. Since $F$ is a filter in D, $\sqcap
f(E_{0})\in F$. As pointed out before the theorem, $\sqcap f(E_{0})\sqsubseteq
a_{E_{0}}$. So $a_{E_{0}}\in F$, as $F$ is a filter in D. Therefore $F\cap
I\neq\emptyset$, which implies that $I\in O_{\Delta^{-1}}^{c\prime}$. Thus
$\cup_{a\in B}I_{a}\subseteq O_{\Delta^{-1}}^{c\prime}$.

The other direction, that is $O_{\Delta^{-1}}^{c\prime}\subseteq\cup_{a\in
B}I_{a}$, is proved by contraposition. Let $I$ be a primary ideal of D such
that $a\notin I$ for all $a\in B$. By Observation 7,
$\mathfrak{F}_{E}\neq\emptyset$ for each $E\in\mathfrak{J}_{0}$. Now we show
that there exists a function $f_{E}\in\mathfrak{F}_{E}$ such that $\sqcap
f_{E}(E)\notin I$.

Indeed, if possible suppose that $\sqcap f(E)\in I$ for all
$f\in\mathfrak{F}_{E}$. Then $\sqcup_{f\in\mathfrak{F}_{E}}\sqcap f(E)\in I$,
as $I$ is an ideal and $\mathfrak{F}_{E}$ finite. Now by Proposition 14,
$a_{E}:=\vee\mathcal{D}_{E}=\vee_{f\in\mathfrak{F}_{E}}\sqcap
f(E)\sqsubseteq\sqcup_{f\in\mathfrak{F}_{E}}\sqcap f(E)$. So $a_{E}\in I$,
which is a contradiction as $a_{E}\in B$.

Therefore by Rado’s Selection Principle, there exists a function $f_{0}$ from
$J$ to $\cup_{j\in J}D_{j\sqcap}$ such that $f_{0}(j)\in D_{j\sqcap}$, and for
each $E\in\mathfrak{J}_{0}$ there exists $K\in\mathfrak{J}_{0}$ with
$E\subseteq K$ such that $f_{0}(j)=f_{K}(j)$ for all $j\in E$. Let $F$ be the
filter in D generated by $\\{f_{0}(j):j\in J\\}$. We claim that $I\cap
F=\emptyset$. If possible, suppose there exists $x\in D$ such that $x\in I\cap
F$. Then there is $E\in\mathfrak{J}_{0}$ such that $\sqcap f_{0}(E)\sqsubseteq
x$, as $x\in F$. So there exists $K\in\mathfrak{J}_{0}$ such that $E\subseteq
K$ and for all $j\in E$, $f_{K}(j)=f_{0}(j)$, which implies that $\sqcap
f_{0}(E)=\sqcap f_{K}(E)\sqsupseteq\sqcap f_{K}(K)$. Then $\sqcap
f_{K}(K)\sqsubseteq x$. This implies $\sqcap f_{K}(K)\in I$, which is a
contradiction. Therefore $I\cap F=\emptyset$ and by the prime ideal theorem
for dBas (Theorem 3), there exists $F_{0}\in\mathcal{F}_{pr}(\textbf{D})$ such
that $I\cap F_{0}=\emptyset$ and $F\subseteq F_{0}$. Then $F_{0}\cap
D_{j\sqcap}\neq\emptyset$ for each $j\in J$, as $\\{f_{0}(j):j\in
J\\}\subseteq F\subseteq F_{0}$. This implies that $F_{0}\cap
D_{j}\neq\emptyset$ for each $j\in J$. So $F_{0}\in O^{c}$ and hence $I\notin
O^{c\prime}_{\Delta^{-1}}$. So $I\in O^{c\prime}_{\Delta^{-1}}$ implies that
$I\in\cup_{a\in B_{J}}I_{a}$.

Thus $O^{c\prime}_{\Delta^{-1}}=\cup_{a\in B}I_{a}$, and
$O^{\blacksquare}_{\nabla^{-1}}=O^{c\prime}_{\Delta^{-1}}=\cup_{a\in B}I_{a}$,
which is open as $I_{a}$ is open for all $a\in D$. Hence $\nabla^{-1}$ is
upper semi-continuous.

(ii) Let $O$ be an open set in $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$.
Then by Note 2, $O=\cup_{j\in J}\cap_{a\in D_{j}}F_{\neg a}=\cup_{j\in
J}F_{\neg(\vee_{a\in D_{j}}a)},$ and $O^{\blacklozenge}=(\cup_{j\in
J}F_{\neg(\vee_{a\in D_{j}}a)})^{\blacklozenge}=\cup_{j\in
J}F^{\blacklozenge}_{\neg(\vee_{a\in D_{j}}a)}=\cup_{j\in
J}I_{\lrcorner(\neg(\vee_{a\in D_{j}}a)\sqcap\neg(\vee_{a\in
D_{j}}a))}=\cup_{j\in J}I_{\lrcorner(\neg(\vee_{a\in D_{j}}a))}$,
which is an open set as $I_{a}$ is open for all $a\in D$. Hence by Theorem
16(I), $\nabla^{-1}$ is lower semi-continuous. ∎

######  Corollary 7.

If D is Boolean then $\nabla$ is a homeomorphism from
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ to
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$.

###### Proof.

Continuity of $\nabla$ and $\nabla^{-1}$ follows from Theorem 23. We show that
$\nabla$ is a bijection from $\mathcal{F}_{pr}(\textbf{D})$ to
$\mathcal{I}_{pr}(\textbf{D})$. Let $F\in\mathcal{F}_{pr}(\textbf{D})$, and
$I_{1},I_{2}\in\mathcal{I}_{pr}(\textbf{D})$ be such that $F\nabla
I_{1},F\nabla I_{2}$. Then $F\cap I_{1}=\emptyset$, $F\cap I_{2}=\emptyset$,
which implies that $I_{1},I_{2}\subseteq F^{c}$. Since D is a Boolean algebra,
the notions of primary filter and ideal coincide with those of prime filter
and ideal respectively. $F^{c}$ is then a prime ideal as $F$ is a prime
filter, and $I_{1}=F^{c}=I_{2}$ as prime ideals are maximal. So $\nabla$ is a
function from $\mathcal{F}_{pr}(\textbf{D})$ to
$\mathcal{I}_{pr}(\textbf{D})$. Similarly, one can show that $\nabla^{-1}$ is
a function from $\mathcal{I}_{pr}(\textbf{D})$ to
$\mathcal{F}_{pr}(\textbf{D})$. Clearly, $\nabla^{-1}\circ\nabla$ is the
identity map on $\mathcal{F}_{pr}(\textbf{D})$. So $\nabla$ is a bijection
from $\mathcal{F}_{pr}(\textbf{D})$ to $\mathcal{I}_{pr}(\textbf{D})$. ∎

How are the CTSCRs corresponding to two quasi-isomorphic dBas related? To see
that, let us introduce the following functions.

######  Definition 22.

Let M and D be two dBas and $h:\textbf{M}\rightarrow\textbf{D}$ a dBa
homomorphism. The function
$\alpha_{h}:\mathcal{F}_{pr}(\textbf{D})\rightarrow\mathcal{F}_{pr}(\textbf{M})$
is given by $\alpha_{h}(F):=h^{-1}(F)$ for any
$F\in\mathcal{F}_{pr}(\textbf{D}),$ and the function
$\beta_{h}:\mathcal{I}_{pr}(\textbf{D})\rightarrow\mathcal{I}_{pr}(\textbf{M})$
is given by $\beta_{h}(I):=h^{-1}(I)$ for any
$I\in\mathcal{I}_{pr}(\textbf{D})$.

By Proposition 15(ii and iii) (cf. Section 3), $\alpha_{h}$ and $\beta_{h}$
are well-defined functions. Moreover,

######  Proposition 19.

$\alpha_{h}^{-1}(F_{x})=F_{h(x)}$ and $\beta_{h}^{-1}(I_{x})=I_{h(x)}$, for
all $x\in D$.

###### Proof.

Let $x\in D$. Then
$\alpha_{h}^{-1}(F_{x})=\\{F\in\mathcal{F}_{pr}(\textbf{D}):x\in
h^{-1}(F)\\}=\\{F\in\mathcal{F}_{pr}(\textbf{D}):h(x)\in F\\}=F_{h(x)}\\\
\mbox{and}~{}\beta_{h}^{-1}(I_{x})=\\{I\in\mathcal{I}_{pr}(\textbf{D}):x\in
h^{-1}(I)\\}=\\{I\in\mathcal{I}_{pr}(\textbf{D}):h(x)\in I\\}=I_{h(x)}$. ∎

We show in Theorem 25 below that if dBas M and D are quasi-isomorphic, the
CTSCRs
$\mathbb{K}^{T}_{pr}(\textbf{M}):=((\mathcal{F}_{pr}(\textbf{M}),\mathcal{T}),(\mathcal{I}_{pr}(\textbf{M}),\mathcal{J}),\nabla_{1})$
and
$\mathbb{K}^{T}_{pr}(\textbf{D}):=((\mathcal{F}_{pr}(\textbf{D}),\mathcal{T}),(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J}),\nabla_{2})$
corresponding to M and D are homeomorphic. In order to prove the result, we
use the following.

######  Theorem 24.

[25] Let $(X,\tau),(Y,\rho)$ be two topological spaces and $f:X\rightarrow Y$
be a bijective continuous function. If $(X,\tau)$ is compact and $(Y,\rho)$ is
Hausdorff, then $f$ is a homeomorphism.

######  Theorem 25.

Let $h:\textbf{M}\rightarrow\textbf{D}$ be a dBa quasi-isomorphism. Then
$(\alpha_{h},\beta_{h}):\mathbb{K}^{T}_{pr}(\textbf{D})\rightarrow\mathbb{K}^{T}_{pr}(\textbf{M})$
is a CTSCR-homeomorphism.

###### Proof.

Let $F\in\mathcal{F}_{pr}(\textbf{D})$ and $I\in\mathcal{I}_{pr}(\textbf{D})$
such that $F\cap I=\emptyset$. If possible, let $h^{-1}(F)\cap
h^{-1}(I)\neq\emptyset$. Then there exists $x\in h^{-1}(F)\cap h^{-1}(I).$
Hence $h(x)\in F\cap I$, which is a contradiction. So $h^{-1}(F)\cap
h^{-1}(I)=\emptyset$. Now let $\alpha_{h}(F)\cap\beta_{h}(I)=h^{-1}(F)\cap
h^{-1}(I)=\emptyset$ and if possible, let $F\cap I\neq\emptyset.$ Then there
exists $x\in F\cap I$ and $y_{1}\in h^{-1}(F),y_{2}\in h^{-1}(I)$ such that
$h(y_{1})=h(y_{2})=x$. Therefore $y_{2}\in h^{-1}(F)$, implying that $y_{2}\in
h^{-1}(F)\cap h^{-1}(I)$, which is a contradiction. Hence $F\cap I=\emptyset$.
So $F\nabla_{1}I$ if and only if $\alpha_{h}(F)\nabla_{2}\beta_{h}(I)$, which
implies that $(\alpha_{h},\beta_{h})$ is a context homomorphism. Now let us
define a correspondence $f_{h}$ from $\mathcal{F}_{pr}(\textbf{M})$ to
$\mathcal{F}_{pr}(\textbf{D})$ by $f_{h}(F):=h(F)$ for all
$F\in\mathcal{F}_{pr}(\textbf{M}).$ By Proposition 15(v) it follows that
$f_{h}$ is a well-defined map. Similarly, $g_{h}$ from
$\mathcal{I}_{pr}(\textbf{M})$ to $\mathcal{I}_{pr}(\textbf{D})$ defined by
$g_{h}(I):=h(I)$ is a well-defined map. As $h$ is surjective,
$f_{h}\circ\alpha_{h}=Id_{\mathcal{F}_{pr}(\textbf{D})}$, the identity map on
$\mathcal{F}_{pr}(\textbf{D})$ and
$g_{h}\circ\beta_{h}=Id_{\mathcal{I}_{pr}(\textbf{D})}$, the identity map on
$\mathcal{I}_{pr}(\textbf{D})$. Therefore $f_{h}=\alpha^{-1}_{h}$ and
$g_{h}=\beta^{-1}_{h}$, that is $\alpha_{h},\beta_{h}$ are bijections. Hence
$(\alpha_{h},\beta_{h})$ is a context isomorphism.

Next we show that $(\alpha_{h},\beta_{h})$ is a CTSCR-homeomorphism, for which
$\alpha_{h}$ and $\beta_{h}$ both must be shown to be continuous. Let $O$ be
an open set in $(\mathcal{F}_{pr}(\textbf{M}),\mathcal{T}).$ By Note 2,
$O=\cup_{j\in J}\cap_{x\in M_{j}}F_{\neg x}$, where $M_{j}$ is a finite subset
of $M$ for all $j\in J.$ So $\alpha_{h}^{-1}(O)=\alpha_{h}^{-1}(\cup_{j\in
J}\cap_{x\in M_{j}}F_{\neg x})=\cup_{j\in J}\cap_{x\in
M_{j}}\alpha_{h}^{-1}(F_{\neg x})=\cup_{j\in J}\cap_{x\in M_{j}}F_{\neg h(x)}$
by Proposition 19. Hence $\alpha_{h}$ is continuous. By Theorems 24 and 21 it
follows that $\alpha_{h}$ is a homeomorphism. Similarly we can show that
$\beta_{h}$ is a homeomorphism. Hence $(\alpha_{h},\beta_{h})$ is a CTSCR-
homeomorphism. ∎

######  Corollary 8.

If $h$ is a dBa isomorphism from M to $\textbf{D},$ $(\alpha_{h},\beta_{h})$
is a CTSCR-homeomorphism from $\mathbb{K}^{T}_{pr}(\textbf{D})$ to
$\mathbb{K}^{T}_{pr}(\textbf{M})$.

###### Proof.

Follows from Theorem 25, as any dBa isomorphism is a dBa quasi-isomorphism. ∎

Let us note the following.

######  Proposition 20.

For all $x\in D$, $(F_{\neg
x},I_{x})\in\mathfrak{R}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$. Moreover, if D
is a pure dBa, $(F_{\neg
x},I_{x})\in\mathfrak{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$ for all $x\in
D$.

###### Proof.

$F_{x}$ is clopen in $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ and $I_{x}$
is clopen in $(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$ for each $x\in D$
(cf. Note 2). Now $F_{\neg x}^{\blacksquare\lozenge}=F_{\neg(x\sqcup
x)}=I_{x}^{\lozenge}$ by Proposition 18. Therefore $(F_{\neg x},I_{x})$ is an
object oriented protoconcept. So $(F_{\neg
x},I_{x})\in\mathfrak{R}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$.

In case $D$ is a pure dBa, either $x=x\sqcap x$ or $x=x\sqcup x$ for all $x\in
D$. If $x=x\sqcap x$, $(F_{\neg x},I_{x})=(F_{\neg x},I_{x\sqcap x}).$ Then it
follows from Proposition 18(i(a)) that $(F_{\neg x},I_{x})$ is an object
oriented semiconcept, as $\neg\neg x=x\sqcap x.$ On the other hand, if
$x=x\sqcup x$, $(F_{\neg x},I_{x})=(F_{\neg(x\sqcup x)},I_{x}).$ From
Proposition 18(ii(b)) it follows that $(F_{\neg x},I_{x})$ is an object
oriented semiconcept. So $(F_{\neg
x},I_{x})\in\mathfrak{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$. ∎

We now give a characterization of the set of clopen object oriented
semiconcepts of $\mathbb{K}^{T}_{pr}(\textbf{D})$. This will be used to obtain
the representation theorems for fully contextual as well as pure dBas.

######  Theorem 26 (Characterization theorem).

Let D be a dBa. Then
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$ if and only if
$(A,B)=(F_{\neg x},I_{x})$ for some $x\in D_{p}$. Moreover, if
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{\sqcap}$ then
$x\in D_{\sqcap}$ and if
$(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{\sqcup}$ then
$x\in D_{\sqcup}$.

###### Proof.

Let us recall that
$\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}=\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{\sqcap}\cup\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{\sqcup}$.
Then the following cases arise.
(i) Let $(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{\sqcap}$.
Then $(A,B)=(A,B)\sqcap(A,B)=(A,A^{\blacksquare})$. Since $A$ is closed,
$A=\cap_{j\in J}\cup_{a\in D_{j}}F_{a}$, where $D_{j}^{\prime}s$ are finite
subsets of $D$ for each $j\in J$. As $A$ is open, $A^{c}=\cup_{j\in
J}\cap_{a\in D_{j}}F^{c}_{a}$ is a closed subset of
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$. Therefore $A^{c}$ is compact in
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$, as
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ is compact. So there exists a
finite subset $E$ of $J$ such that $A^{c}=\cup_{j\in E}\cap_{a\in
D_{j}}F^{c}_{a}=\cup_{j\in E}\cap_{a\in D_{j}}F_{\neg a}=\cup_{j\in
E}F_{\sqcap_{a\in D_{j}}\neg a}=\cup_{j\in E}F_{\neg(\vee_{a\in D_{j}}a)}$ by
Lemma 1. This means $A=\cap_{j\in E}F_{\neg\neg(\vee_{a\in
D_{j}}a)}=\cap_{j\in E}F_{\vee_{a\in D_{j}}a}$, as $\vee_{a\in D_{j}}a$ in
$D_{\sqcap}$ for each $j\in E$ by Note 1. This implies that $A=F_{\sqcap_{j\in
E}(\vee_{a\in D_{j}}a)}=F_{x}$, where $x=\sqcap_{j\in E}(\vee_{a\in D_{j}}a)$.
Therefore $(A,A^{\blacksquare})=(F_{x},F^{\blacksquare}_{x})=(F_{x},I_{\neg
x})=(F_{x\sqcap x},I_{\neg x})=(F_{\neg\neg x},I_{\neg x})$, by Lemma 1(iv),
Proposition 18, and 5(iii). So $(A,B)=(F_{\neg\neg x},I_{\neg x})$, where
$\neg x\in D_{\sqcap}\subseteq D_{p}$ by Proposition 5(i).
(ii) Let $(A,B)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{\sqcup}$.
Then $(A,B)=(A,B)\sqcup(A,B)=(B^{\lozenge},B)$. Similar to the proof of (i),
we can show that $(B^{\lozenge},B)=((I_{b})^{\lozenge},I_{b})=(F_{\neg(b\sqcup
b)},I_{b})$ for some $b\in D$. Using Lemma 1(iii), $(A,B)=(F_{\neg(b\sqcup
b)},I_{b\sqcup b})$, where $b\sqcup b\in D_{\sqcup}\subseteq D_{p}$ by
Proposition 5(iv). ∎

######  Corollary 9.

Let D be a dBa. Then $F\nabla I$ if and only if for all
$(X,Y)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p},I\in Y$ implies
that $F\in X$.

###### Proof.

Let $F\nabla I$ and
$(X,Y)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$ such that
$I\in Y$. Then $F\cap I=\emptyset$ and by Theorem 26, $(X,Y)=(F_{x},I_{\neg
x})$ for some $x\in D_{p}$. $\neg x\in I$ and $F\cap I=\emptyset$ imply that
$\neg x\notin F$, and so $x\in F$. Hence $F\in X=F_{x}$. For the converse, let
$F\in\mathcal{F}_{pr}(\textbf{D})$, and $I\in\mathcal{I}_{pr}(\textbf{D})$ be
such that for all
$(X,Y)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$, $I\in Y$
implies $F\in X$. If possible, let $F\cap I\neq\emptyset$. Then there exists
$a\in F\cap I$. Therefore $I\in I_{a\sqcup a}$ and $a\sqcup a\in F$, that is
$\neg(a\sqcup a)\notin F$. So $F\notin F_{\neg(a\sqcup a)}$, which is a
contradiction, as $(F_{\neg(a\sqcup a)},I_{a\sqcup
a})\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$. Therefore $F\cap
I=\emptyset$, which implies $F\nabla I.$ ∎

As mentioned in Theorem 8 (cf. Section 2.5), any dBa D can be quasi-embedded
into the algebra $\underline{\mathfrak{R}}(\mathbb{K}_{pr}(\textbf{D}))$ of
object oriented protoconcepts, through the map
$h:D\rightarrow\mathfrak{R}(\mathbb{K}_{pr}(\textbf{D}))$ given by
$h(x):=(F_{\neg x},I_{x})$ for all $x\in D$. Proposition 20 gives that for all
$x\in D$, $h(x)\in\mathfrak{R}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$, which is
a subset of $\mathfrak{R}(\mathbb{K}_{pr}(\textbf{D}))$. The next
representation theorem establishes that, in fact, the dBa D can be quasi-
embedded into the subalgebra
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$ of
$\underline{\mathfrak{R}}(\mathbb{K}_{pr}(\textbf{D}))$, using the map $h$.

######  Theorem 27 (Representation theorem for dBas and contextual dBas).

  1. 1.

Any dBa D is quasi-embedded into
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$, where
$h:D\rightarrow\mathfrak{R}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$ defined by
$h(x):=(F_{\neg x},I_{x})$ for all $x\in D$, gives the required quasi-
injective dBa homomorphism. Moreover, $\textbf{D}_{p}$ is isomorphic to
$\mathcal{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$.

  2. 2.

Any contextual dBa D is embedded into
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$, the above map
$h$ giving the required injective dBa homomorphism from D into
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$.

###### Proof.

(i) From Proposition 20 it follows that $h$ is a well-defined map. Let $x,y\in
D$. Then
$h(x\sqcap y)=(F_{\neg(x\sqcap y)},I_{x\sqcap y})=(F_{\neg
x},I_{x})\sqcap(F_{\neg y},I_{y})=h(x)\sqcap h(y)$, and
$h(x\sqcup y)=(F_{\neg(x\sqcup y)},I_{x\sqcup y})=(F_{\neg
x},I_{x})\sqcup(F_{\neg y},I_{y})=h(x)\sqcup h(y)$.
Similarly, one can show that
$h(\lrcorner x)=\lrcorner h(x),h(\neg x)=\neg
h(x),h(\top)=(\emptyset,\emptyset)$ and
$h(\bot)=(\mathcal{F}_{pr}(\textbf{D}),\mathcal{I}_{pr}(\textbf{D}))$.
Therefore $h$ is a dBa homomorphism.

Now we show that $h$ is quasi-injective. Let $x,y\in D$ and $h(x)\sqsubseteq
h(y)$. If possible, suppose $x\not\sqsubseteq y$. Then by Proposition 2(iii),
either $x_{\sqcap}\not\sqsubseteq_{\sqcap}y_{\sqcap}$ or
$x_{\sqcup}\not\sqsubseteq_{\sqcup}y_{\sqcup}$. If
$x_{\sqcap}\not\sqsubseteq_{\sqcap}y_{\sqcap}$ then by the prime ideal theorem
of Boolean algebras, there exists a prime filter $F_{0}$ in
$\textbf{D}_{\sqcap}$ such that $x\sqcap x\in F_{0}$ and $y\sqcap y\notin
F_{0}$. Therefore by Lemma 2(ii) and Proposition 7, it follows that there
exists a primary filter $F$ such that $F_{0}=F\cap D_{\sqcap}$. So $x\sqcap
x\in F$ and $\neg y\in F$ (as $y\sqcap y\notin F$). Thus $F\in F_{\neg y}$ but
$F\notin F_{\neg x}$, which implies $F_{\neg y}\not\subseteq F_{\neg x}$. Now
if $x_{\sqcup}\not\sqsubseteq_{\sqcup}y_{\sqcup}$ then dually we can show that
$I_{y}\not\subseteq I_{x}$. Therefore in both cases, $h(x)\not\sqsubseteq
h(y)$, which is a contradiction. So $x\sqsubseteq y$. Conversely, let
$x\sqsubseteq y.$ Then $I_{y}\subseteq I_{x}$. Further
$x_{\sqcap}\sqsubseteq_{\sqcap}y_{\sqcap}$ and
$x_{\sqcup}\sqsubseteq_{\sqcup}y_{\sqcup}$. Therefore by Proposition 5(ii),
$\neg y=\neg y_{\sqcap}\sqsubseteq_{\sqcap}\neg x_{\sqcap}=\neg x$. So
$F_{\neg y}\subseteq F_{\neg x}.$ Hence $h(x)\sqsubseteq h(y)$, using
Proposition 12.

From Theorem 26 it follows that $h|_{D_{p}}$ is a surjective dBa homomorphism
from $\textbf{D}_{p}$ onto
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$. The
restriction of quasi order $\sqsubseteq$ on $D_{p}$ becomes a partial order by
Proposition 3. So $h|_{D_{p}}$ is a dBa isomorphism from $\textbf{D}_{p}$ to
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$.
Proposition 11 applied to the case of
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$ and
$\mathcal{S}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))$ gives
$\underline{\mathfrak{R}}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}=\mathcal{S}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))$.

(ii) If D is a contextual dBa then the quasi-order $\sqsubseteq$ becomes a
partial order, whence $h$ becomes an injective dBa homomorphism. ∎

Recall the special case when a dBa D is finite, as remarked upon in Section
2.5. We now show that in this case, the topological spaces
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ and
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$ become discrete. So, effectively,
the topologies do not play any role in the representation theorem above and
the result coincides with Corollary 3.

######  Proposition 21.

For a finite dBa D, $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ and
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$ are discrete topological spaces.

###### Proof.

(i) Consider $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$. We show that any
subset $A$ of $\mathcal{F}_{pr}(\textbf{D})$ is open in
$(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$. Let
$A:=\\{F_{1},\ldots,F_{n}\\}$, where
$F_{i}\in\mathcal{F}_{pr}(\textbf{D}),i=1,\ldots,n$. Then $F_{i}=\\{x\in
D~{}:~{}z\sqsubseteq x~{}\mbox{for some }~{}z\in F_{0i}\\}$, where
$F_{0i}=\uparrow a_{i}$ is a principal filter of the Boolean algebra
$\textbf{D}_{\sqcap}$ generated by an atom $a_{i}$ of $\textbf{D}_{\sqcap}$.
We claim that $A=F_{\vee_{i=1}^{n}a_{i}}$. Indeed, let $F_{k}\in A$,
$k\in\\{1,\ldots,n\\}$. As $a_{k}\sqsubseteq a_{k}$, $a_{k}\in F_{k}$. By
Proposition 2 and Note 1, it follows that
$a_{k}\sqsubseteq\vee_{i=1}^{n}a_{i}$. As $F_{k}$ is a filter,
$\vee_{i=1}^{n}a_{i}\in F_{k}$, whence $F_{k}\in F_{\vee_{i=1}^{n}a_{i}}$.

Now let $F\in F_{\vee_{i=1}^{n}a_{i}}$. Since $F$ is a primary filter,
$F=\\{x\in D~{}:~{}z\sqsubseteq x~{}\mbox{for some}~{}z\in F_{0}\\}$, where
$F_{0}=\uparrow a$ is a principal filter of the Boolean algebra
$\textbf{D}_{\sqcap}$ generated by the atom $a$. As $\vee_{i=1}^{n}a_{i}\in
F$, $a\sqsubseteq\vee_{i=1}^{n}a_{i}$. So by Proposition 2 and Note 1,
$a\sqsubseteq_{\sqcap}\vee_{i=1}^{n}a_{i}$, which implies that $a=a_{k}$ for
some $k\in\\{1,\ldots,n\\}$, as $a$ is an atom of $\textbf{D}_{\sqcap}$.
Therefore $F\in A$. Hence $A=F_{\vee_{i=1}^{n}a_{i}}$.
(ii) For the case of $(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$, similar to
the proof in (i), one can show that any subset $B$ of
$\mathcal{I}_{pr}(\textbf{D})$ has the form $B=I_{\wedge_{i=1}^{n}b_{i}}$ for
some coatoms $b_{i}$ of the Boolean algebra $\textbf{D}_{\sqcup}$. ∎

We next prove the isomorphism theorem for fully contextual dBas. For that, the
following result will be used.

######  Lemma 3.

For any dBa D and $a\in D$, $F_{\neg a}=F_{\neg(a\sqcap a)}$.

###### Proof.

Let $F\in F_{\neg a}$. Then $\neg a\in F$. By Proposition 4(v) it follows that
$a\sqcap a\sqsubseteq a$ and Proposition 5(ii) gives $\neg
a\sqsubseteq\neg(a\sqcap a)$. Therefore $\neg(a\sqcap a)\in F$, as $F$ is a
filter. So $F_{\neg a}\subseteq F_{\neg(a\sqcap a)}$. Conversely, let
$\neg(a\sqcap a)\in F$. As $F$ is a primary filter, $a\sqcap a\notin F$, which
implies that $a\notin F$ (otherwise $a\sqcap a\in F$, as $F$ is a filter).
Therefore $\neg a\in F$, as $F$ is a primary filter. This gives
$F_{\neg(a\sqcap a)}\subseteq F_{\neg a}$. So $F_{\neg(a\sqcap a)}=F_{\neg
a}$. ∎

######  Theorem 28 (Representation theorem for fully contextual dBas).

Any fully contextual dBa D is isomorphic to
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$, the map
$h:D\rightarrow\mathfrak{R}^{T}_{pr}(\mathbb{K}_{pr}^{T}(\textbf{D}))$ defined
by $h(x):=(F_{\neg x},I_{x})$.

###### Proof.

From Theorem 27 and the fact that a fully contextual dBa D is a contextual
dBa, it follows that $h$ is a dBa embedding. To complete the proof it remains
to show that $h$ is surjective. For that, let
$(X,Y)\in\mathfrak{R}^{T}_{pr}(\mathbb{K}_{pr}^{T}(\textbf{D}))$. Then $X$ is
clopen in $(\mathcal{F}_{pr}(\textbf{D}),\mathcal{T})$ and $Y$ is clopen in
$(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J})$. Further,
$X^{\blacksquare\lozenge}=Y^{\lozenge}$, which is equivalent to
$X^{\blacksquare}=Y^{\lozenge\blacksquare}$, by Observation 3. Now we consider
the object oriented protoconcepts
$(X,X^{\blacksquare}),(Y^{\lozenge},Y)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$.
Then the following equations hold.
$(X,X^{\blacksquare})_{\sqcup}=(X,X^{\blacksquare})\sqcup(X,X^{\blacksquare})=(X^{\blacksquare\lozenge},X^{\blacksquare})=(Y^{\lozenge},Y^{\lozenge\blacksquare})=(Y^{\lozenge},Y)\sqcap(Y^{\lozenge},Y)=(Y^{\lozenge},Y)_{\sqcap}$.
As
$(X,X^{\blacksquare}),(Y^{\lozenge},Y)\in\mathfrak{R}^{T}(\mathbb{K}^{T}_{pr}(\textbf{D}))_{p}$,
$(X,X^{\blacksquare})=(F_{\neg x},I_{x})$ and $(Y^{\lozenge},Y)=(F_{\neg
b},I_{b})$ for some $x\in D_{\sqcap}$ and $b\in D_{\sqcup}$ by Theorem 26.

From the above it follows that $(F_{x},I_{\neg x})_{\sqcup}=(F_{\neg(b\sqcup
b)},I_{b})_{\sqcap}$, which is equivalent to $(I_{\neg x}^{\lozenge},I_{\neg
x})=(F_{\neg(b\sqcup b)},F_{\neg(b\sqcup b)}^{\blacksquare})$. Therefore
$(F_{\neg(\neg x\sqcup\neg x)},I_{\neg x})=(F_{\neg(b\sqcup
b)},I_{\neg\neg(b\sqcup b)})$, which is equivalent to $(F_{\neg(\neg
x\sqcup\neg x)},\\\ I_{\neg x\sqcup\neg x})=(F_{\neg\neg\neg(b\sqcup
b)},I_{\neg\neg(b\sqcup b)})$, as $I_{\neg x\sqcup\neg x}=I_{\neg x}$ by Lemma
1(iii) and $\neg\neg\neg(b\sqcup b)=\neg(b\sqcup b)$ by Proposition 5(ix).
Therefore $h(\neg x\sqcup\neg x)=h(\neg\neg(b\sqcup b))$. As $h$ is injective,
$\neg x\sqcup\neg x=\neg\neg(b\sqcup b)=(b\sqcup b)\sqcap(b\sqcup b)$ by
Proposition 5(iii). $\neg x\sqcap\neg x=\neg x\in D_{\sqcap}$ and $(b\sqcup
b)\sqcup(b\sqcup b)=(b\sqcup b)\in D_{\sqcup}$ by Proposition 5(i and iv).
Since D is a fully contextual dBa, there exists a unique $c\in D$ such that
$c\sqcap c=\neg x$ and $c\sqcup c=b\sqcup b$. Therefore $X=F_{x}=F_{x\sqcap
x}=F_{\neg\neg x}$ by Lemma 1(iv) and Proposition 5(iii), which implies that
$X=F_{\neg(c\sqcap c)}=F_{\neg c}$ by Lemma 3, and $Y=I_{b}=I_{b\sqcup
b}=I_{c\sqcup c}=I_{c}$, by Lemma 1(iii). Therefore $h(c)=(F_{\neg
c},I_{c})=(X,Y)$, which implies that $h$ is surjective. Hence $h$ is a dBa
isomorphism. ∎

######  Corollary 10.

If D is a fully contextual dBa then $\mathbb{K}^{T}_{pr}(\textbf{D})$ is
homeomorphic to
$\mathbb{K}^{T}_{pr}(\mathfrak{R}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D})))$.

###### Proof.

Follows directly from Theorems 28 and 25. ∎

Now we obtain the representation theorem for the class of pure dBas.

######  Theorem 29 (Representation theorem for pure dBas).

Any pure dBa D is isomorphic to
$\mathcal{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$.

###### Proof.

Since D is pure, $\textbf{D}_{p}=\textbf{D}$ by Proposition 6. From Theorem 27
it follows that D is isomorphic to
$\mathcal{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D}))$. ∎

######  Corollary 11.

If D is a pure dBa then $\mathbb{K}^{T}_{pr}(\textbf{D})$ is homeomorphic to
$\mathbb{K}^{T}_{pr}(\mathcal{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{D})))$.

###### Proof.

By Theorems 29 and 25. ∎

######  Corollary 12 (Representation theorem for Boolean algebras).

Any Boolean algebra B is isomorphic to
$\mathcal{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{B}))$.

###### Proof.

Follows from Theorem 29, as any Boolean algebra is a pure dBa (by Theorem 12).
∎

In [19] it is shown that every Boolean algebra B is isomorphic to
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{B}))$. Note that
this also follows from Theorem 28 as a Boolean algebra B is fully contextual
(Theorem 12). Furthermore, this is in consonance with Corollary 12, as
$\underline{\mathfrak{R}}^{T}(\mathbb{K}_{pr}^{T}(\textbf{B}))=\mathcal{S}^{T}(\mathbb{K}_{pr}^{T}(\textbf{B}))$.
In fact, the set of all clopen object oriented concepts of
$\mathbb{K}^{T}_{pr}(\textbf{B})$ coincides with both the set of clopen object
oriented protoconcepts and that of clopen object oriented semiconcepts.

We note here that, using the relations between $\blacksquare,\lozenge$ and ′
operators given in Theorem 4(vi), one can rewrite the representation results
for dBas obtained in this section, in terms of the algebra of protoconcepts
and that of semiconcepts (cf. Notation 3).

##  6  Duality results for dBas

Pure dBas as objects and dBa isomorphisms as morphisms constitute a category,
denoted PDBA. Fully contextual dBas and dBa isomorphisms also form a category,
denoted FCDBA. On the other hand, abstraction of properties of the CTSCR
$\mathbb{K}^{T}_{pr}(\textbf{D})$ corresponding to any dBa D, leads us to the
definition of Stone contexts. Stone contexts and CTSCR-homeomorphisms are
observed to constitute a category, denoted Scxt. Moreover, relations between
the collections of morphisms of these categories have already been obtained.
By Theorem 13, for fully contextual dBas D and M, any map $f:D\rightarrow M$
is a dBa isomorphism if and only if $f|_{D_{p}}$ is a dBa isomorphism between
the pure subalgebras $\textbf{D}_{p}$ and $\textbf{M}_{p}$ of D and M
respectively. By Corollary 8, corresponding to any dBa isomorphism from dBa D
to dBa M, there is a CTSCR- homeomorphism from
$\mathbb{K}^{T}_{pr}(\textbf{M})$ to $\mathbb{K}^{T}_{pr}(\textbf{D})$. A
natural question then is to relate the categories
$\textbf{FCDBA},\textbf{PDBA}$ and Scxt. It is shown in this section that
FCDBA and PDBA are equivalent, whereas PDBA and Scxt are dually equivalent.

We divide the section into two subsections. In Section 6.1 the goal is to
define Stone contexts and give examples; Section 6.2 presents the
relationships among the categories $\textbf{FCDBA},\textbf{PDBA}$ and Scxt.

######  Notation 13.

For objects A, B of a locally small category [2] $\mathfrak{C}$, the set of
morphisms from A to B is denoted by
$Hom_{\mathfrak{C}}(\textbf{A},\textbf{B})$. The categories we work on in this
paper are all locally small.

Let us briefly recall here, some properties of functors. Suppose $F$ is a
functor from category $\mathfrak{C}$ to category $\mathfrak{E}$. $F$ is an
equivalence [2] if and only if it is faithful, full, and essential surjective.
$F$ is faithful and full if and only if the restriction of $F$ on
$Hom_{\mathfrak{C}}(\textbf{A},\textbf{B})$ is a bijection from
$Hom_{\mathfrak{C}}(\textbf{A},\textbf{B})$ to
$Hom_{\mathfrak{E}}(F(\textbf{A}),F(\textbf{B}))$. To show $F$ is essential
surjective, for each object X of the category $\mathfrak{E}$ one finds an
object A in $\mathfrak{C}$ such that $F(\textbf{A})$ is isomorphic to X.

###  6.1 Stone contexts

On abstraction of properties of the CTSCR $\mathbb{K}_{pr}^{T}(\textbf{D})$
corresponding to any dBa D, one obtains the definition of a Stone context
(Definition 23 below). Let us note that a totally disconnected space is a
topological space $(X,\tau)$ in which for any two $x,y\in X$ such that $x\neq
y$, there is a clopen set $U$ in $(X,\tau)$ such that $x\in U$ and $y\notin
U.$ We may remark here that, usually, a space $(G,\tau)$ is called totally
disconnected if and only if every quasi-component consists of a single point.
But for compact topological spaces, these two definitions coincide. Recall
that a topological space $(X,\tau)$ is called a Stone space if and only if it
is compact and totally disconnected [7].

######  Definition 23.

A CTSCR $\mathbb{K}^{T}:=((G,\tau),(M,\rho),R)$ is called a Stone context if
the following hold:

  1. 1.

$(G,\rho)$ and $(M,\tau)$ are Stone spaces,

  2. 2.

$gRm,$ if for all $(A,B)\in\mathfrak{S}^{T}(\mathbb{K}^{T}),m\in B$ implies
$g\in A$.

######  Theorem 30.

For any dBa D,
$\mathbb{K}^{T}_{pr}(\textbf{D}):=((\mathcal{F}_{pr}(\textbf{D}),\mathcal{T}),(\mathcal{I}_{pr}(\textbf{D}),\mathcal{J}),\nabla)$
is a Stone context.

###### Proof.

Follows from Theorems 23 and 21, and Corollary 9. ∎

Apart from the example of Stone context provided by
$\mathbb{K}^{T}_{pr}(\textbf{D}),$ we have the following.

######  Example 4.

Any context $\mathbb{K}:=(X,Y,R)$, where X,Y are finite and $R\subseteq
X\times Y$ is an arbitrary relation, can be trivially looked upon as a Stone
context with finite discrete topology.

######  Example 5.

$\mathbb{K}_{+}^{T}$ given in Example 3 is a Stone context, when
$(X,\tau_{1})$ and $(Y,\tau_{2})$ are non-empty Stone spaces.

###### Proof.

It is already established in Example 3 that $\mathbb{K}_{+}^{T}$ is a CTSCR.
We just verify condition (b) of Definition 23. Let $g\in X$ and $m\in Y$ such
that for all $(A,B)\in\mathfrak{S}^{T}(\mathbb{K}_{+}^{T}),$ $m\in B$ implies
that $g\in A$. If possible, assume that $g\cancel{R}m$. Then $m\notin C$ and
so $m\in C^{c}$. Now we take $B=C^{c}$. From the proof of Example 3, it
follows that $(\emptyset,C^{c})$ is an object oriented semiconcept. Since
$\emptyset$ and C are clopen in $\tau_{1},\tau_{2}$ respectively,
$(\emptyset,C^{c})\in\mathfrak{S}^{T}(\mathbb{K}_{+}^{T})$. Therefore we get a
clopen object oriented semiconcept $(\emptyset,C^{c})$ such that $m\in C^{c}$,
but $g\notin\emptyset$. Thus we have a contradiction, and $gRm$ must hold.
Hence $\mathbb{K}_{+}^{T}$ is a Stone context. ∎

A familiar example of a Stone space is the Cantor set [35]. One can construct
a Stone context using the Cantor set.

######  Example 6.

Let $X=Y=C$, where $C$ is the Cantor set and $\tau$ be the subspace topology
on $C$ induced by the usual topology on $\mathbb{R}$. Let
$\mathbb{K}^{T}:=((X,\tau),(Y,\tau),R),$ where $R(x)=[0,x]\cap C$ for all
$x\in C$. Then $\mathbb{K}^{T}$ is a Stone context.

###### Proof.

$(X,\tau)$ and $(Y,\tau)$ both are Stone spaces. We show that $\mathbb{K}^{T}$
is a CTSCR, and check condition (b) of Definition 23.
To show that $R$ is continuous, let $O$ be an open set in $(Y,\tau)$. We
verify that $O^{\lozenge},O^{\square}$ are open in $(X,\tau)$. For any subset
$A\subseteq C$, $A\subseteq A^{\lozenge}$, as for all $x\in A$, $([0,x]\cap
C)\cap A\neq\emptyset$. Let $z\in O^{\lozenge}.$ Then $([0,z]\cap C)\cap
O\neq\emptyset$. Therefore there exists $x_{0}\in([0,z]\cap C)\cap O.$ Now if
$z=x_{0}$ then $z\in O\subseteq O^{\lozenge}.$ If $x_{0}<z$ then
$n=\mbox{inf}(O)<z$, where $\mbox{inf}(O)$ is the infimum of $O$. We choose
$\epsilon=z-n.$ Then $z\in(z-(z-n),z+(z-n))\cap C=(n,2z-n)\cap
C\neq\emptyset$. Let $x\in(n,2z-n)\cap C.$ $([0,x]\cap C)\cap O\neq\emptyset$,
as $n<x$. So $(n,2z-n)\cap C\subseteq O^{\lozenge}$. So in both cases, $z$ is
an interior point of $O^{\lozenge}$. Thus $O^{\lozenge}$ is open in
$(X,\tau)$. For the case of $O^{\square}$, let $z\in O^{\square}.$ This means
$[0,z]\cap C\subseteq O$. So $z\in O=O^{*}\cap C$, where $O^{*}$ is an open
set in $\mathbb{R}$. Therefore there exists an open interval $(a,b)$ in
$\mathbb{R}$ such that $z\in(a,b)\cap C\subseteq O$. Let $x\in(a,b)\cap C$.