# On Geometric Implications

Amirhossein Akbar Tabatabai
Institute of Mathematics, Czech Academy of Sciences

###### Abstract

It is a well-known fact that although the poset of open sets of a topological
space is a Heyting algebra, its Heyting implication is not necessarily stable
under the inverse image of continuous functions and hence is not a geometric
concept. This leaves us wondering if there is any stable family of
implications that can be safely called geometric. In this paper, we will first
recall the abstract notion of implication as a binary modality introduced in
[1]. Then, we will use a weaker version of categorical fibrations to define
the geometricity of a category of pairs of spaces and implications over a
given category of spaces. We will identify the greatest geometric category
over the subcategories of open-irreducible (closed-irreducible) maps as a
generalization of the usual injective open (closed) maps. Using this
identification, we will then characterize all geometric categories over a
given category $\mathcal{S}$, provided that $\mathcal{S}$ has some basic
closure properties. Specially, we will show that there is no non-trivial
geometric category over the full category of spaces. Finally, as the
implications we identified are also interesting in their own right, we will
spend some time to investigate their algebraic properties. We will first use a
Yoneda-type argument to provide a representation theorem, making the
implications a part of an adjunction-style pair. Then, we will use this result
to provide a Kripke-style representation for any arbitrary implication.

Keywords: modal algebras, implications, geometricity, frame representation.

##  1 Introduction

It is well-known that the poset of open sets of a topological space is a
(complete) Heyting algebra, an observation that provides the topological
interpretation of the intuitionistic propositional logic $\mathsf{IPC}$. As
the interpretation turns out to be complete, one may be tempted to consider
$\mathsf{IPC}$ as the logic of spaces in the same way that the classical
propositional logic is the logic of the discrete spaces or simply the
unstructured sets. Despite the beauty of such a philosophical temptation, the
Heyting implication, although present in any of these locales, is not
preserved under the inverse image of continuous functions and hence can not be
considered a truly geometric concept. (To see more about locales, see [16]).
To solve the issue, one may eliminate the Heyting implication from the
language and restrict its expressive power to the so-called coherent fragment.
It is also possible to be more faithful to the nature of space (and hence less
to the elementary nature of the language) to allow the infinitary disjunctions
and achieve the so-called geometric logic [11, 18]. In many contexts [19],
these logics are the natural logical systems to consider and although they
seem to be weak at first glance, they prove their sophistication through their
natural role in the real practice. For instance, geometric theories play a
crucial role in topos theory, where they characterize all Grothendieck topoi
as the mathematical universes freely constructed from the free model of
geometric theories [15, 12, 13]. Even the finitary coherent fragment is more
powerful than it appears as any classical theory has a coherent conservative
extension [5].

Having said that, the implication is not what one wants to ignore permanently.
Philosophically speaking, implication is the machinery to internalize the
meta-relation of the “entailment order between the propositions $A$ and $B$”
into a “proposition $A\to B$” to empower the language to talk about its own
entailment behavior. The logical realm is full of different instances of
implications from the more philosophically motivated conditionals [3] and the
weak implications [20, 21, 17] addressing the impredicativity problem of the
Heyting implication to the more mathematically motivated implications such as
the ones appeared in provability logic [21] and preservability logic [8, 9,
14].

Opening the horizon to the alternative implications, one may wonder if there
is any sort of geometric implication, powerful enough to internalize some
parts of the structures, on the one hand, and be geometric, on the other. To
address such a problem formally, we must first be precise about what we mean
by an implication. Reading the internalization process algebraically,
implications are some binary operations over the posets where they internalize
the order of the poset, mapping the predicate $a\leq b$ into the element $a\to
b$. Naturally, there are many structures and properties to internalize. For
instance, the fact that the order is reflexive, i.e., $a\leq a$ internalizes
to $a\to a=1$, its transitivity internalizes to $(a\to b)\wedge(b\to
c)\leq(a\to c)$ and the existence of the binary meets to $a\to(b\wedge
c)=(a\to b)\wedge(a\to c)$. To provide a definition for a general notion of
implication, we must choose the minimum level of internalization to enforce,
and we think that the natural minimum property of an order to internalize is
simply the fact that it is an order, i.e., that it is reflexive and
transitive, see [1].

######  Definition 1.1.

Let $\mathcal{A}=(A,\leq,\wedge,\vee,1,0)$ be a bounded distributive lattice.
A binary operator $\to$ over $\mathcal{A}$, decreasing in its first argument
and increasing in its second is called an _implication_ over $\mathcal{A}$ if:

$(i)$

(_internal reflexivity_) $a\to a=1$, for any $a\in\mathcal{A}$,

$(ii)$

(_internal transitivity_) $(a\to b)\wedge(b\to c)\leq a\to c$, for any
$a,b,c\in\mathcal{A}$.

An implication is called _meet internalizing_ if $a\to(b\wedge c)=(a\to
b)\wedge(a\to c)$ and _join internalizing_ if $(a\vee b)\to c=(a\to
c)\wedge(b\to c)$, for any $a,b,c\in\mathcal{A}$. For any implication, $\neg
a$ is an abbreviation for $a\to 0$. If $\to$ is an implication over
$\mathcal{A}$, the pair $(\mathcal{A},\to)$ is called a _strong algebra_. If
$\mathcal{A}=\mathcal{O}(X)$, for some space $X$, then the pair $(X,\to)$ is
called a _strong space_. By a _strong algebra map_
$f:(\mathcal{A},\to_{\mathcal{A}})\to(\mathcal{B},\to_{\mathcal{B}})$, we mean
a bounded lattice map preserving the implication, i.e.,
$f(a\to_{\mathcal{A}}b)=f(a)\to_{\mathcal{B}}f(b)$, for any
$a,b\in\mathcal{A}$. A _strong space map_ is a continuous map between spaces
such that its inverse image preserves the implication.

######  Remark 1.2.

In [1], it is shown that implications can be equivalently defined as the
binary operations over $\mathcal{A}$ satisfying the conditions:

$(i^{\prime})$

If $a\leq b$ then $a\to b=1$, for any $a,b\in\mathcal{A}$

$(ii)$

$(a\to b)\wedge(b\to c)\leq a\to c$, for any $a,b,c\in\mathcal{A}$.

To see a more detailed discussion to motivate the aforementioned definition,
the reader may consult [1]. However, it is illuminating to think of an
implication as a special case of a general setting in which a category
internalizes its hom structure, i.e., its identity and its composition. The
general formalization for such a generalized function space is introduced in
[7], where it is called an arrow. The categorical formalization of the arrows
that act as the generalized internal hom functors can be found in [10]. In
this broader story, our implications are nothing but arrows enriched over the
category $\\{0\leq 1\\}$ rather than $\mathbf{Set}$ and hence they are just
the propositional shadows of the more structured arrows.

######  Example 1.3.

Over any bounded distributive lattice $\mathcal{A}$, there is a _trivial
implication_ defined by $a\to_{t}b=1$, for any $a,b\in\mathcal{A}$. The
Boolean and the Heyting implications are also implications. To construct a new
implication from the old, assume that $(\mathcal{B},\to_{\mathcal{B}})$ is a
strong algebra, $f:\mathcal{A}\to\mathcal{B}$ is an order-preserving map and
$g:\mathcal{B}\to\mathcal{A}$ is a finite meet preserving map. Then, it is
easy to check that the operator
$a\to_{\mathcal{A}}b=g(f(a)\to_{\mathcal{B}}f(b))$ is an implication over
$\mathcal{A}$. It is also possible to show that any implication is
constructible from the Heyting implication in this way, expanding the base
lattice to a locale, see [1].

Having a definition for implication, it is now reasonable to search for the
geometric implications, i.e., the family of implications over the locales of
opens of spaces stable under the inverse image of _all_ continuous functions.
We will show that there is only one such family, namely the one with the
trivial implications. To prove that surprising result, we employ a weaker
version of the categorical fibrations to develop the relative notion of
geometricity of a category $\mathcal{C}$ of strong spaces over a category
$\mathcal{S}$ of spaces. Here, geometricity simply means that the implications
of the strong spaces in $\mathcal{C}$ are stable under the inverse image of
the maps in $\mathcal{S}$. We will then continue by identifying the greatest
geometric categories over the subcategories of the open- and closed-
irreducible maps. These two families of maps can be considered as the
generalizations of the injective open (closed) maps. The implications stable
under the open-irreducible maps are the ones for which $c\wedge a\leq b$
implies $c\leq a\to b$. These implications behave similarly to a well-known
family of implications called the basic implication introduced in [21] in
provability logic and later in [17] for philosophical reasons. For the closed-
irreducible maps, the implications are the ones for which $a\leq b\vee c$
implies $(a\to b)\vee c=1$. We will show that having these two properties
forces the implications to behave similarly to the Boolean implications as
they satisfy the equation $a\to b=\neg a\vee b$. Using these implications and
their relationship with the geometricity for the open- and closed-irreducible
maps, we will then identify all the geometric categories over a given category
$\mathcal{S}$, provided that $\mathcal{S}$ has some basic closure properties.

Completing the characterization of the geometric categories, as the
implications we identified are also interesting in their own right, we will
spend the last section to provide a representation theorem for them. We will
first use a Yoneda-type argument, making the implications a part of an
adjunction-style pair. Then, we will use this result to represent an arbitrary
implication as the implication of a topological version of a combination of an
intuitionistic Kripke and a neighbourhood frame.

##  2 Preliminaries

In this section, we will recall some basic notions and their corresponding
theorems we need throughout the paper. Let $\mathcal{P}=(P,\leq)$ be a poset.
A subset $S\subseteq P$ is called an _upset_ if for any $x,y\in P$, if $x\in
S$ and $x\leq y$ then $y\in S$. The _downsets_ are defined dually. The set of
all upsets of $(P,\leq)$ is denoted by $U(P,\leq)$. For any $S\subseteq P$,
the greatest lower bound of $S$ (resp. the least upper bound of $S$), if it
exists, is called the _meet_ (resp. _join_) of the elements of $S$ and is
denoted by $\bigwedge S$ (resp. $\bigvee S$). If $S=\\{a,b\\}$, the meet
$\bigwedge S$ and the join $\bigvee S$ are denoted by $a\wedge b$ and $a\vee
b$, respectively. Moreover, $\bigwedge\varnothing$ and $\bigvee\varnothing$,
i.e., the greatest and the least elements of $P$, if exist, are denoted by $1$
and $0$, respectively. A poset is called a _bounded lattice_ , if for any
finite subset $S\subseteq P$, both $\bigwedge S$ and $\bigvee S$ exist and it
is called _complete_ if for any set $S\subseteq P$, the meet $\bigwedge S$
exists. A bounded lattice is called _distributive_ , if $a\wedge(b\vee
c)=(a\wedge b)\vee(a\wedge c)$, for any $a,b,c\in P$. It is called a _locale_
, if for any $S\subseteq P$, the join $\bigvee S$ exists and
$a\wedge\bigvee_{b\in S}b=\bigvee_{b\in S}(a\wedge b)$, for any $a\in P$ and
$S\subseteq P$. By the _Heyting implication_ over a bounded lattice
$\mathcal{A}$, we mean the binary operation $\Rightarrow$ over $\mathcal{A}$
such that $a\wedge b\leq c$ iff $a\leq b\Rightarrow c$, for any
$a,b,c\in\mathcal{A}$. A bounded lattice $\mathcal{H}$ is called a _Heyting
algebra_ if it has the Heyting implication. A bounded lattice $\mathcal{B}$ is
called a _Boolean algebra_ if all the elements of $\mathcal{B}$ have a
complement, i.e., for any $a\in\mathcal{B}$, there is $b\in\mathcal{B}$ such
that $a\vee b=1$ and $a\wedge b=0$. A subset of a bounded lattice is called a
_filter_ , if it is an upset and closed under all finite meets. A filter $F$
is called _prime_ if $0\notin F$ and $a\vee b\in F$ implies either $a\in F$ or
$b\in F$. The set of all prime filters of a lattice $\mathcal{A}$ is denoted
by $\mathcal{F}_{p}(\mathcal{A})$. A subset of $\mathcal{A}$ is called an
_ideal_ , if it is a downset and closed under all finite joins. The following
theorem is a useful tool when working with bounded distributive lattices:

######  Theorem 2.1.

[4, 6](Prime filter theorem) Let $\mathcal{A}$ be a bounded distributive
lattice, $F$ be a filter and $I$ be an ideal such that $F\cap I=\varnothing$.
Then, there exists a prime filter $P$ such that $F\subseteq P$ and $P\cap
I=\varnothing$.

Let $(P,\leq_{P})$ and $(Q,\leq_{Q})$ be two posets and $f:P\to Q$ be a
function. It is called an _order-preserving map_ , if it preserves the order,
meaning $f(a)\leq_{Q}f(b)$, for any $a\leq_{P}b$. An order-preserving map is
called an _order embedding_ or simply an embedding, if for any $a,b\in P$, the
inequality $f(a)\leq_{Q}f(b)$ implies $a\leq_{P}b$. An order-preserving map
between two bounded lattices (locales) is called a _bounded lattice map_
(_locale map_), if it preserves all finite meets and finite joins (arbitrary
joins). For two order-preserving maps $f:P\to Q$ and $g:Q\to P$, the pair
$(f,g)$ is called an _adjunction_ , denoted by _$f\dashv g$_ , if
$f(a)\leq_{Q}b$ is equivalent to $a\leq_{P}g(b)$, for any $a\in P$ and $b\in
Q$. If $f\dashv g$, the map $f$ is called the _left adjoint_ of $g$ and $g$ is
called the _right adjoint_ of $f$.

######  Theorem 2.2.

[2] (Adjoint functor theorem for posets) Let $(P,\leq_{P})$ be a complete
poset and $(Q,\leq_{Q})$ be a poset. Then, an order-preserving map
$f:(P,\leq_{P})\to(Q,\leq_{Q})$ has a right (left) adjoint iff it preserves
all joins (meets).

Let $X$ be a topological space. We denote the locale of its open subsets by
$\mathcal{O}(X)$. A topological space is called $T_{0}$, if for any two
different points $x,y\in X$, there is an open set which contains one of these
points and not the other. It is called _Hausdorff_ if for any two distinct
points $x,y\in X$, there are opens $U,V\in\mathcal{O}(X)$ such that $x\in U$,
$y\in V$ and $U\cap V=\varnothing$. A pair $(X,\leq)$ of a topological space
and a partial order is called a _Priestley space_ if $X$ is compact and for
any $x,y\in X$, if $x\nleq y$, there exists a clopen upset $U$ such that $x\in
U$ and $y\notin U$. For any bounded distributive lattice $\mathcal{A}$, the
pair $(\mathcal{F}_{p}(\mathcal{A}),\subseteq)$ is a Priestley space, where
$\mathcal{F}_{p}(\mathcal{A})$ is the set of all prime filters of
$\mathcal{A}$ and the topology on $\mathcal{F}_{p}(\mathcal{A})$ is defined by
the basis of the opens in the form $\\{P\in\mathcal{F}_{p}(\mathcal{A})\mid
a\in P\;\text{and}\;b\notin P\\}$, for any $a,b\in\mathcal{A}$. Denoting
$\\{P\in\mathcal{F}_{p}(\mathcal{A})\mid a\in P\\}$ by $i(a)$, it is known
that any clopen upset in this Priestley space equals to $i(a)$, for some
$a\in\mathcal{A}$ and any clopen set is in form
$\bigcup_{r=1}^{n}[i(a_{r})\cap i(b_{r})^{c}]$, for some finite sets
$\\{a_{1},\ldots,a_{n}\\},\\{b_{1},\ldots,b_{n}\\}\subseteq\mathcal{A}$. For a
comprehensive explanation, see [4].

##  3 Open, Closed and Weakly Boolean Implications

In this section, we first introduce the three families of open, closed, and
weakly Boolean implications, some of their natural examples and a method to
construct the new ones from the old. Then, in Subsection 3.1, we provide a
characterization for the weakly Boolean implications defined on the locale of
opens of a topological space needed in the next section. Finally, in
Subsection 3.2, we introduce two families of continuous maps as the
generalized versions of the injective open and closed maps. These families
provide the real motivation to consider the above-mentioned families of
implications as the classes of the open and closed implications are the
greatest classes of implications that are stable under the inverse image of
the open- and closed-irreducible maps, respectively. In other words, the
conditions we put on any of these two families of implications are necessary
if we want them to be stable under the corresponding classes of continuous
maps.

######  Definition 3.1.

An implication $\to$ over a bounded distributive lattice $\mathcal{A}$ is
called _open_ if $a\wedge b\leq c$ implies $a\leq b\to c$, for any
$a,b,c\in\mathcal{A}$. It is called _closed_ if $a\leq b\vee c$ implies $(a\to
b)\vee c=1$, for any $a,b,c\in\mathcal{A}$. An implication is called a _weakly
Boolean_ implication (WBI, for short), if it is both open and closed. A strong
algebra $(\mathcal{A},\to)$ is called open, closed, or weakly Boolean, if its
implication is.

######  Remark 3.2.

As mentioned before, the real motivation to investigate the three families of
implications introduced in Definition 3.1 is geometric and will be covered in
Subsection 3.2. However, it is also worth providing a logical motivation for
them in this remark. For that purpose, first, consider the following sequent-
style rule for the classical implication in the usual calculus $\mathbf{LK}$
for classical logic:

$\Gamma,A\Rightarrow B,\Delta$ $\Gamma\Rightarrow A\to B,\Delta$
---

The condition for the open implications is half of the adjunction property of
Heyting implications and is reminiscent of the above rule, except that in the
rule $\Delta$ is considered as empty. The condition for the closed
implications is also reminiscent of the above rule. However, this time the
restriction changes to the emptiness of $\Gamma$. It is easy to see that an
implication is weakly Boolean iff it admits the full rule, i.e., if $c\wedge
a\leq b\vee d$ implies $c\leq(a\to b)\vee d$, for any $a,b,c,d\in\mathcal{A}$.

It is practically helpful to simplify the definition of the open and closed
implications from an implication between two inequalities to just one
inequality. It is also theoretically important as it shows that the each of
these families form a variety. Here is the simplification.

######  Lemma 3.3.

Let $(\mathcal{A},\to)$ be a strong algebra. Then:

$(i)$

$\to$ is open iff $a\leq b\to a\wedge b$, for any $a,b\in\mathcal{A}$.
Specially, $a\leq b\to a$ and $a\wedge\neg 1=a\wedge\neg a$, for any
$a,b\in\mathcal{A}$.

$(ii)$

$\to$ is closed iff $(a\vee b\to a)\vee b=1$, for any $a,b\in\mathcal{A}$.
Specially, $b\vee\neg b=1$, for any $b\in\mathcal{A}$.

$(iii)$

If $\to$ is closed, then $c\wedge a\leq b$ implies $c\leq\neg a\vee b$, for
any $a,b,c\in\mathcal{A}$.

###### Proof.

For $(i)$, if an implication is open, as $a\wedge b\leq a\wedge b$, we have
$a\leq b\to a\wedge b$. Conversely, if $a\wedge b\leq c$, we have $a\leq
b\to(a\wedge b)\leq b\to c$. For the special cases, notice that by $a\wedge
b\leq a$, we reach $a\leq b\to a$. To prove $a\wedge\neg a=a\wedge\neg 1$,
since $a\leq 1$, we reach $\neg 1\leq\neg a$ which itself implies $a\wedge\neg
1\leq a\wedge\neg a$. For the converse, note that as $a\leq 1\to a$, we have
$a\wedge\neg a\leq(1\to a)\wedge(a\to 0)\leq 1\to 0=\neg 1$. Hence,
$a\wedge\neg a\leq a\wedge\neg 1$. For $(ii)$, its first part is similar to
that of $(i)$. For the special case, by setting $a=0$, we have $b\vee\neg
b=1$. For $(iii)$, if $c\wedge a\leq b$, then $c\leq(c\vee\neg
a)\wedge(a\vee\neg a)=(c\wedge a)\vee\neg a\leq\neg a\vee b$. ∎

######  Example 3.4.

The trivial implication $\to_{t}$ defined as the constant function $1$ is both
open and closed. The Heyting implication is clearly open, while it is not
necessarily closed. In fact, it is closed iff it is Boolean. To have a closed
implication that is not open, we refer the reader to Example 3.8, where a
general machinery to construct the open and closed implications are provided.
Here, however, we provide an example of an implication that is neither open
nor closed. Let $\mathcal{A}$ be $\mathcal{O}(\mathbb{R})$ and $\Rightarrow$
be its Heyting implication. Now, putting $f(U)=U+1=\\{x+1\mid x\in U\\}$ and
$g=id_{\mathcal{A}}$ in Example 1.3, the operation $U\to
V=[(U+1)\Rightarrow(V+1)]$ is an implication. However, it is not open as
$[\mathbb{R}\to(0,1)]=[\mathbb{R}\Rightarrow(1,2)]=(1,2)\nsupseteq(0,1)$ and
it is not closed as
$(-\infty,0)\cup[(-\infty,0)\to\varnothing]=(-\infty,0)\cup[(-\infty,1)\Rightarrow\varnothing]=(-\infty,0)\cup(1,\infty)\neq\mathbb{R}$.
Finally, to provide a family of implications that are both open and closed,
over any Boolean algebra $\mathcal{B}$ define $a\to b=\bar{a}\vee b\vee m$,
where $\bar{a}$ is the complement of $a$ and $m\in\mathcal{B}$ is a fixed
element. It is easy to see that $\to$ is a WBI.

In the following definition, we provide a combination of an intuitionistic
Kripke frame and a neighbourhood frame, serving as an order-theoretic and
hence concrete machinery to construct different families of implications.
Later, in Section 5, we will see that these frames are powerful enough to
represent all possible implications.

######  Definition 3.5.

A _Kripke-Neighbourhood frame_ (KN-frame, for short) is a tuple
$\mathcal{K}=(K,\leq,R,B,N)$ of a poset $(K,\leq)$, a binary relation $R$ on
$K$, a set $B\subseteq P(X)$ and a map $N:X\to P(U(K,\leq,B))$, where
$U(K,\leq,B)$ is the set of all upsets in $B$, such that:

  * $\bullet$

$R$ is compatible with the order, i.e., if $x\leq y$ and $(y,z)\in R$, then
$(x,z)\in R$,

  * $\bullet$

for any $x\in X$ and any upsets $U,V\in B$, if $U\subseteq V$ and $U\in N(x)$
then $V\in N(x)$,

  * $\bullet$

$B$ is closed under finite union (including $K$ as the nullary union),
complement and the operation $\lozenge_{R}$ defined by
$\lozenge_{R}(U)=\\{x\in X\mid\exists y\in U,(x,y)\in R\\}$,

  * $\bullet$

$j(U)=\\{x\in X\mid U\in N(x)\\}$ is in $B$, for any upset $U\in B$.

A KN-frame is called _full_ if $B=P(X)$. It is called _standard_ if $B=P(X)$
and $N(k)=\\{U\in U(K,\leq)\mid k\in U\\}$. We denote a standard KN-frame by
$(K,\leq,R)$ as its only non-trivial ingredients. A KN-frame is called _open_
when for any $x,y\in K$ and any upsets $U,V\in B$, if $x\in U$, $(x,y)\in R$
and $V\in N(y)$, then $U\cap V\in N(y)$. It is called _closed_ when for any
$x,y\in K$ and any upsets $U,V\in B$, if $x\notin U$, $(x,y)\in R$ and $U\cup
V\in N(y)$, then $V\in N(y)$.

######  Example 3.6.

Let $\mathcal{K}=(K,\leq,R,B,N)$ be a KN-frame. Then, the bounded distributive
lattice $U(K,\leq,B)$ of the upsets in $B$ is closed under the operation
$U\to_{\mathcal{K}}V=\\{x\in K\mid\forall y\in K\,[(x,y)\in
R\;\text{and}\;U\in N(y)\;\text{then}\;V\in N(y)]\\}$ and the pair
$\mathfrak{A}(\mathcal{K})=(U(K,\leq,B),\to_{\mathcal{K}})$ is a strong
algebra. Moreover, if $\mathcal{K}$ is open (closed), then so is
$\mathfrak{A}(\mathcal{K})$. To prove the closure of $U(K,\leq,B)$ under the
operation $\to_{\mathcal{K}}$, let $U$ and $V$ be two upsets in $B$. Then,
notice that $(U\to_{\mathcal{K}}V)^{c}=\Diamond_{R}(j(U)\cap j(V)^{c})$. As
$B$ is closed under complement, finite intersection and $\Diamond_{R}$ and $j$
maps the upsets in $B$ to the elements of $B$, we can conclude that
$(U\to_{\mathcal{K}}V)^{c}$ and hence $U\to_{\mathcal{K}}V$ is in $B$. Also,
using the compatibility of the order with $R$, it is easy to see that
$U\to_{\mathcal{K}}V$ is an upset. Hence, $U\to_{\mathcal{K}}V\in
U(K,\leq,B)$.
To prove that $\to_{\mathcal{K}}$ is an implication, the only non-trivial part
is to prove that if $U\subseteq V$, then $U\to_{\mathcal{K}}V=K$, for any
upsets $U,V\in B$. Let $x\in K$ be an arbitrary element and assume $(x,y)\in
R$ and $U\in N(y)$. As $U\subseteq V$ and $N(y)$ is upward closed for the
upsets in $B$, we reach $V\in N(y)$. Therefore, $x\in U\to_{\mathcal{K}}V$.
Finally, for the open and closed conditions, if $\mathcal{K}$ is open, using
Lemma 3.3, it is enough to prove that $U\subseteq V\to_{\mathcal{K}}U\cap V$,
for any upsets $U,V\in B$. Let $x\in U$, $(x,y)\in R$ and $V\in N(y)$. Then,
by the openness of the KN-frame, we know $U\cap V\in N(y)$, which completes
the proof. A similar argument works for the closed case.

######  Remark 3.7.

First, notice that a KN-frame is a combination of an intuitionistic Kripke
frame with an independent monotone neighbourhood function restricted to the
upsets of a given Boolean algebra of the subsets of $K$. The presence of the
neighbourhood function is crucial as in the standard KN-frames, the definition
of the implication simplifies to $U\to_{\mathcal{K}}V=\\{x\in K\mid\forall
y\in K\,[(x,y)\in R\;\text{and}\;y\in U\;\text{then}\;y\in V]\\}$ which is
always meet- and join-internalizing. Therefore, without the neighbourhood
structure, the KN-frames are not capable of representing all implications.
Secondly, note that starting from a KN-frame, it is always possible to drop
the Boolean algebra $B$ to reach a full KN-frame and hence a greater strong
algebra. More precisely, let $\mathcal{K}=(K,\leq,R,B,N)$ be a KN-frame and
define $\mathcal{K}^{f}$ as $(K,\leq,R,P(K),N^{f})$, where $N^{f}(x)=\\{U\in
U(K,\leq)\mid\exists V\in N(x)\,V\subseteq U\\}$. It is easy to see that
$\mathcal{K}^{f}$ is a full KN-frame. Moreover, as $N(x)$ is upward closed for
the upsets in $B$, it is clear that $U\in N(x)$ iff $U\in N^{f}(x)$, for any
$U\in U(K,\leq,B)$. Therefore, the strong algebra $\mathfrak{A}(\mathcal{K})$
is a subalgebra of the strong algebra $\mathfrak{A}(\mathcal{K}^{f})$. Notice
that the passage from $\mathcal{K}$ to $\mathcal{K}^{f}$ does not necessarily
preserve the openness or the closedness of the original KN-frame.

######  Example 3.8.

For any standard KN-frame $\mathcal{K}=(K,\leq,R)$, if $R\subseteq\,\leq$, the
implication $\to_{\mathcal{K}}$ is open and if $R^{op}\subseteq\,\leq$, it is
closed, where by $R^{op}$, we mean $\\{(k,l)\in K^{2}\mid(l,k)\in R\\}$. For
the first claim, using Lemma 3.3, it is enough to show $U\subseteq
V\to_{\mathcal{K}}U\cap V$, for any $U,V\in U(K,\leq)$. For that purpose,
assume $k\in U$. Then, for any $l\in V$, if $(k,l)\in R$, as
$R\subseteq\,\leq$, we have $k\leq l$ and since $U$ is an upset, we have $l\in
U$. Hence, $l\in U\cap V$. Therefore, $k\in V\to_{\mathcal{K}}(U\cap V)$. For
the second claim, again using Lemma 3.3, it is enough to show $[(U\cup
V)\to_{\mathcal{K}}V]\cup U=K$, for any $U,V\in U(K,\leq)$. Suppose $k\notin
U$. Then, for any $l\in U\cup V$, if $(k,l)\in R$, as $R^{op}\subseteq\,\leq$,
we have $l\leq k$. Hence, as $U$ is an upset and $k\notin U$, we have $l\notin
U$. Hence, $l\in V$. Therefore, $k\in(U\cup V)\to_{\mathcal{K}}V$.
Employing these two families of implications, it is easy to provide a closed
implication that is not open. Set $K=\\{k,l\\}$, $k\leq l$ and
$R=\\{(l,k),(k,k)\\}$ and consider $\mathcal{K}=(K,\leq,R)$. It is clear that
$R$ is compatible with the order and $R^{op}\subseteq\,\leq$. Hence,
$\to_{\mathcal{K}}$ is closed. To show that it is not open, we show
$K\to_{\mathcal{K}}\\{l\\}=\varnothing$ and as $\\{l\\}\nsubseteq
K\to_{\mathcal{K}}\\{l\\}$, the implication cannot be open. For
$K\to_{\mathcal{K}}\\{l\\}=\varnothing$, if either $k$ or $l$ is in
$K\to_{\mathcal{K}}\\{l\\}$, as $(k,k),(l,k)\in R$, we must have $k\in\\{l\\}$
which is impossible.

######  Remark 3.9.

It is not hard to see that being closed (as opposed to being open) is a very
demanding condition restricting the form of the closed implications and hence
the WBI’s in a very serious way (see Theorem 3.12). For instance, it is easy
to see that the condition $R^{op}\subseteq\,\leq$ in Example 3.8 together with
the compatibility condition of $R$ with respect to $\leq$, restricts the
relation $R$ only to the ones in the from $\\{(f(k),k)\in K^{2}\mid k\in
L\\}\cup\\{(k,k)\in K^{2}\mid k\in L\\}$, where $f:L\to K$ is an injective
function, $L$ is a subset of the minimal elements of $K$ and $f(k)\geq k$, for
any $k\in L$:

${{k_{0}}}$${{k_{1}}}$${{k_{2}}}$${\cdots}$${{f(k_{0})}}$${{f(k_{1})}}$${{f(k_{2})}}$$\scriptstyle{R}$$\scriptstyle{R}$$\scriptstyle{R}$

For the WBI’s, even the function $f$ collapses to the identity and the only
remaining data will be the set $L$. We do not prove these claims as they will
not be used in the present paper. They are mentioned here only to convey the
feeling that the study of these two families is not as justified as one might
expect. However, we spend some time studying these two families as we need
their behavior to prove the rarity of geometric implications in the next
section.

So far, we have seen some concrete examples of the open and closed
implications. The following theorem modifies the method of Example 1.3 to
construct the new open and closed implications from the old.

######  Theorem 3.10.

Let $(\mathcal{B},\to_{\mathcal{B}})$ be a strong algebra,
$f:\mathcal{A}\to\mathcal{B}$ be an order-preserving and
$g:\mathcal{B}\to\mathcal{A}$ be a finite meet preserving map. Then,
$a\to_{\mathcal{A}}b=g(f(a)\to_{\mathcal{B}}f(b))$ is:

$(i)$

an open implication over $\mathcal{A}$, if $\to_{\mathcal{B}}$ is open, $f$
preserves all binary meets and $gf(a)\geq a$, for any $a\in\mathcal{A}$.

$(ii)$

a closed implication over $\mathcal{A}$, if $\to_{\mathcal{B}}$ is closed, $f$
preserves all binary joins and $c\vee f(a)=1$ implies $g(c)\vee a=1$, for any
$a\in\mathcal{A}$ and $c\in\mathcal{B}$.

###### Proof.

To check whether $\to_{\mathcal{A}}$ is open or closed, we use the criterion
of Lemma 3.3. For $(i)$, as $\to_{\mathcal{B}}$ is open, $f$ preserves the
binary meets and $gf(a)\geq a$, we have $b\to_{\mathcal{A}}a\wedge
b=g(f(b)\to_{\mathcal{B}}f(a\wedge b))=g(f(b)\to_{\mathcal{B}}f(a)\wedge
f(b))\geq g(f(a))\geq a$. For $(ii)$, as $\to_{\mathcal{B}}$ is closed and $f$
preserves the binary joins, we have $[f(a\vee b)\to_{\mathcal{B}}f(b)]\vee
f(a)=[f(a)\vee f(b)\to_{\mathcal{B}}f(b)]\vee f(a)=1$. Hence, by the property,
we have $g(f(a\vee b)\to_{\mathcal{B}}f(b))\vee a=1$ which implies $[(a\vee
b)\to_{\mathcal{A}}b]\vee a=1$. ∎

######  Remark 3.11.

In practice, part $(i)$ in Lemma 3.10 is useful in situations where $f\dashv
g$ and $f$ is binary meet preserving and part $(ii)$ is used when $g\dashv f$,
$g$ is finite meet preserving and $f$ is binary join preserving.

The following theorem provides a complete characterization of the WBI’s. We
will see that their general form is not far from the one provided in Example
3.4.

######  Theorem 3.12.

Let $\mathcal{A}$ be a bounded distributive lattice. Then, for any weakly
Boolean implication $\to$ over $\mathcal{A}$, the interval $[\neg
1,1]=\\{x\in\mathcal{A}\mid\neg 1\leq x\leq 1\\}$ with its induced order is a
Boolean algebra. Moreover, $a\to b=\neg a\vee b$ and $\neg a$ is the
complement of $a\vee\neg 1$ in $[\neg 1,1]$. Conversely, if for some
$m\in\mathcal{A}$, the interval $[m,1]$ with its induced order is a Boolean
algebra, and $n(a)$ is the complement of $a\vee m$ in $[m,1]$, then
$a\to_{m}b=n(a)\vee b$ is a WBI over $\mathcal{A}$. Note that $\neg_{m}a=n(a)$
and $\neg_{m}1=n(1)=m$.

###### Proof.

For the first part, by Lemma 3.3, we have $(a\vee\neg 1)\wedge\neg
a=(a\wedge\neg a)\vee(\neg 1\wedge\neg a)=(a\wedge\neg 1)\vee\neg 1=\neg 1$
and $(a\vee\neg 1)\vee\neg a=1$, for any $a\in\mathcal{A}$. Therefore, $\neg
a$ is the complement of $a\vee\neg 1$ over $[\neg 1,1]$. Moreover, for any
$a\geq\neg 1$, as $a\vee\neg 1=a$, the element $\neg a$ is the complement of
$a$ which implies that $[\neg 1,1]$ is a Boolean algebra. To show $a\to b=\neg
a\vee b$, note that $\neg a=(a\to 0)\leq(a\to b)$. As the implication is open,
by Lemma 3.3, we have $b\leq a\to b$. Hence, $\neg a\vee b\leq a\to b$. For
the converse, as the implication is also closed, by Lemma 3.3, $b\vee\neg
b=1$. Hence, $a\to b=(a\to b)\wedge(b\vee\neg b)=((a\to b)\wedge b)\vee((a\to
b)\wedge\neg b)\leq b\vee(a\to 0)=\neg a\vee b$.
Conversely, let $[m,1]$ be a Boolean algebra and $n(a)$ be the complement of
$a\vee m$ in $[m,1]$. First, we prove that $a\vee n(a)=1$ and $b\wedge
n(b)\leq n(a)$, for any $a,b\in\mathcal{A}$. For the former, as $n(a)$ is the
complement of $a\vee m$, we have $n(a)\vee(a\vee m)=1$ and as $n(a)\in[m,1]$,
we reach $m\leq n(a)$. Hence, $n(a)\vee a=1$. For the latter, as $n(b)$ is the
complement of $b\vee m$ over $[m,1]$, we have $(b\vee m)\wedge n(b)=m$ which
implies $b\wedge n(b)\leq m$. Again as $m\leq n(a)$, we reach $b\wedge
n(b)\leq n(a)$. Now, to show that $a\to_{m}b=n(a)\vee b$ is an implication, we
must check the properties in Remark 1.2. For $(i^{\prime})$, if $a\leq b$,
then $1=n(a)\vee a\leq n(a)\vee b=a\to_{m}b$. For $(ii)$, using the
distributivity, we have $(a\to_{m}b)\wedge(b\to_{m}c)=(n(a)\vee
b)\wedge(n(b)\vee c)\leq n(a)\vee(b\wedge n(b))\vee c\leq n(a)\vee n(a)\vee
c=a\to_{m}c$. Finally, to show that $\to_{m}$ is both open and closed, as
$a\to_{m}(a\wedge b)=n(a)\vee(a\wedge b)=(n(a)\vee a)\wedge(n(a)\vee
b)=(n(a)\vee b)\geq b$, the implication is clearly open. For closedness, note
that $((a\vee b)\to_{m}a)\vee b=n(a\vee b)\vee a\vee b=1$. ∎

######  Remark 3.13.

Note that in a Boolean algebra, the complement of $a\vee m$ in $[m,1]$ is
$\bar{a}\vee m$, where $\bar{a}$ is the complement of $a$. Therefore, by
Theorem 3.12, it is clear that the implications $a\to_{m}b=\bar{a}\vee b\vee
m$ are the only WBI’s over a Boolean algebra. Having made this observation,
one wonders whether the presence of a WBI forces the ground lattice to be a
Boolean algebra itself. However, as the trivial implication is weakly Boolean
and it is definable over any bounded distributive lattice, it is clear that
the ground lattice structure of a weakly Boolean strong algebra can be quite
general and not necessarily Boolean.

###  3.1 Weakly Boolean Spaces

In this subsection, we will use Theorem 3.12 to provide a characterization for
the WBI’s over the locales of opens of the topological spaces. The
characterization will be useful in the next section. First, let us recall some
basic notions from topology that we need below. A topological space $X$ is
called _discrete_ if all of its subsets are open. It is called _indiscrete_ if
its only opens are $\varnothing$ and $X$. It is called _locally indiscrete_ if
for any $x\in X$, there exists an open $U\subseteq X$ such that $x\in U$ and
the induced topology on $U$ is indiscrete. A space is locally indiscrete iff
all of its closed subsets are open. More generally:

######  Lemma 3.14.

Let $X$ be a topological space and $M\subseteq X$ be an open subset. Then, the
following are equivalent:

$(i)$

For any closed $K\subseteq X$, the union $K\cup M$ is open.

$(ii)$

The subspace $X-M$ is locally indiscrete.

###### Proof.

To prove $(ii)$ from $(i)$, let $L$ be a closed subset of $X-M$. We have to
show that $L$ is also open. As $L$ is closed in $X-M$, there is a closed
subset $K$ of $X$ such that $L=K\cap(X-M)$. By $(i)$, the set $K\cup M$ is
open. As $L=(K\cup M)\cap(X-M)$, the set $L$ is open in $X-M$. Conversely, to
prove $(i)$ from $(ii)$, assume that $K$ is a closed subset in $X$. Hence,
$K\cap(X-M)$ is closed and hence open in $X-M$, by $(ii)$. This implies the
existence of an open $U$ in $X$ such that $K\cap(X-M)=U\cap(X-M)$. Hence,
$K\cup M=U\cup M$. Finally, as both $U$ and $M$ are open in $X$, the subset
$K\cup M$ is open in $X$. ∎

By definition, it is clear that any discrete space is locally indiscrete. The
converse also holds for $T_{0}$ spaces.

######  Lemma 3.15.

Any locally indiscrete $T_{0}$ space is discrete.

###### Proof.

Let $x\in X$ be a point. As $X$ is $T_{0}$, for any $y\neq x$, there is an
open $U_{y}$ such that either $x\in U_{y}$ and $y\notin U_{y}$ or $x\notin
U_{y}$ and $y\in U_{y}$. As any open is closed in the space, w.l.o.g, we can
assume the first case. As closed and open subsets are identical, opens are
closed under arbitrary intersections. Hence, set $U=\bigcap_{y\neq x}U_{y}$.
It is easy to see that $U=\\{x\\}$. Hence, the singletons and consequently all
subsets are open and hence the space is discrete. ∎

We are now ready to provide a characterization for all the WBI’s over the
locales of topological spaces, as promised.

######  Corollary 3.16.

Let $X$ be a topological space and $M\subseteq X$ be an open subset such that
$X-M$ is locally indiscrete. Then, the binary map $U\to V=U^{c}\cup V\cup M$
is a WBI over $\mathcal{O}(X)$. Conversely, any WBI over $\mathcal{O}(X)$ is
in the form $U\to V=U^{c}\cup V\cup M$, where $M=\neg X$ is an open subset
such that $X-M$ is locally indiscrete.

###### Proof.

For the first part, by Lemma 3.14, $\mathcal{O}(X)$ is closed under the
operation $U\mapsto U^{c}\cup M$ as $X-M$ is locally indiscrete and hence the
subset $U^{c}\cup M$ is open, for any open $U$. This proves that $\to$ is
well-defined over $\mathcal{O}(X)$. Now, note that the interval $[M,X]$ in
$\mathcal{O}(X)$ is a Boolean algebra, simply because $U^{c}\cup M$ is the
complement of $U\supseteq M$. Therefore, by Theorem 3.12, the operation $U\to
V=n(U)\cup V$ is a WBI, where $n(U)$ is the complement of $U\cup M$ in
$[M,X]$. As $n(U)=U\cup M^{c}$, we know that $U\to V=U^{c}\cup V\cup M$ is a
WBI. Conversely, if $\to$ is a WBI over $\mathcal{O}(X)$, then by Theorem 3.12
again, the interval $[\neg X,X]$ is a Boolean algebra and $\neg U$ is the
complement of $U\cup\neg X$ in $[\neg X,X]$. Set $M=\neg X$ and note that $M$
is open, as $\neg X\in\mathcal{O}(X)$. As the complement of $U\cup M$ in
$[M,X]$ is $U^{c}\cup M$, we have $U\to V=U^{c}\cup V\cup M$. Moreover, as
$\neg U=U^{c}\cup M$ is open, for any open $U$, the space $X-M$ is locally
indiscrete, by Lemma 3.14. ∎

######  Remark 3.17.

For any WBI over $\mathcal{O}(X)$, its unique $M=\neg X$ is called its _core_.
Note that if $M=\varnothing$, then the space $X$ is locally indiscrete and the
trivial and the Boolean implications are the WBI’s with the cores $X$ and
$\varnothing$, respectively. The strong space $(X,\to)$ is called a _weakly
Boolean space_ , (WBS, for short), if $\to$ is a WBI over $\mathcal{O}(X)$.
Note that if a space is locally indiscrete, then by Corollary 3.16, for any
open $N$, there is a WBI with the core $N$, because $X-N$ is locally
indiscrete. Also notice that a continuous map $f:X\to Y$ induces a strong map
$f:(X,\to_{X})\to(Y,\to_{Y})$ between two WBS’s, iff $f^{-1}(M_{Y})=M_{X}$.

######  Example 3.18.

Let $X$ be a topological space and $N\subseteq X$ be a closed discrete
subspace. Then, by Corollary 3.16, $U\to V=U^{c}\cup V\cup(X-N)$ is a WBI, as
$X-N$ is open and $N$ is locally indiscrete. Corollary 3.16 also shows that if
$X$ is $T_{0}$, then these are the only WBI’s over $X$, as in a $T_{0}$ space,
the only locally indiscrete subspaces are the discrete ones. Moreover, for the
later reference, note that if $X$ is a Hausdorff space, then $U\to V=U^{c}\cup
V\cup(X-N)$ is a WBI, for any finite $N\subseteq X$.

###  3.2 Open-irreducible and Closed-irreducible Maps

In this subsection, we will first introduce the two families of open- and
closed-irreducible maps as the generalizations of the injective open and
closed continuous functions, respectively. Then, we will prove that an
implication $\to$ over $\mathcal{O}(X)$ is open (closed) iff the inverse image
of any open-irreducible (closed-irreducible) map into $X$ transforms $\to$ to
another implication.

######  Definition 3.19.

A topological space $X$ is _open-irreducible (closed-irreducible)_ , if $A\cap
B=\varnothing$ implies $A=\varnothing$ or $B=\varnothing$, for any open
(closed) subsets $A,B\subseteq X$.

Intuitively, the open-irreducible (closed-irreducible) spaces are roughly the
spaces in which the open (closed) subsets are so big that any two non-empty
open (closed) subsets intersect.

######  Example 3.20.

Any singleton space is both open- and closed-irreducible. Any space that is
the closure of a single point is open-irreducible because if
$X=\overline{\\{x\\}}$ and $U$ and $V$ are two nonempty opens, then $x\in U$
as otherwise, $x\in U^{c}$ and since $U^{c}$ is closed, we have
$U^{c}=\overline{\\{x\\}}$ which implies $U=\varnothing$. Similarly, $x\in V$.
Therefore, $U\cap V\neq\varnothing$. For closed-irreducible spaces, it is easy
to see that $X$ is closed-irreducible iff
$\overline{\\{x\\}}\cap\overline{\\{y\\}}$ is non-empty, for any $x,y\in X$.
For instance, if $(P,\leq)$ is a poset, where any two elements have an upper
bound, then $P$ with the topology of the upsets as the closed subsets is
closed-irreducible. The reason is that $\overline{\\{x\\}}$ is $\\{z\in P\mid
z\geq x\\}$. Hence, if $w$ is the upper bound of $x$ and $y$, then
$w\in\overline{\\{x\\}}\cap\overline{\\{y\\}}$.

For our purpose, we must look into the relative version of the open-
irreducible (closed-irreducible) space, where a space is replaced by a
continuous map.

######  Definition 3.21.

A continuous map $f:X\to Y$ is called _open-irreducible (closed-irreducible)_
if it is open (closed) and $f[A]\cap f[B]=f[A\cap B]$, for any open (closed)
subsets $A,B\subseteq X$. As the identity function is open-irreducible
(closed-irreducible) and the composition of any two open-irreducible (closed-
irreducible) maps is also open-irreducible (closed-irreducible), considering
all topological spaces together with the open-irreducible (resp. closed-
irreducible) maps forms a category that we denote by $\mathbf{OI}$ (resp.
$\mathbf{CI}$).

######  Example 3.22.

The unique map $!:X\to\\{*\\}$ is open-irreducible (closed-irreducible) iff
$X$ is open-irreducible (closed-irreducible). It is easy to prove that an open
(a closed) map is open-irreducible (closed-irreducible) iff the fiber
$f^{-1}(y)$ is open-irreducible (closed-irreducible) as a subspace of $X$, for
any $y\in Y$. As a special case, all injective open (closed) maps are
trivially open-irreducible (closed-irreducible), as their fibers are
singletons or empty.

Note that if $f:X\to Y$ is an open map, then the map
$f_{!}:\mathcal{O}(X)\to\mathcal{O}(Y)$ defined by $f_{!}(U)=f[U]$ is the left
adjoint of $f^{-1}$, as $U\subseteq f^{-1}(V)$ iff $f[U]\subseteq V$. The map
$f_{!}$ is well-defined as $f$ is open. Having this left adjoint, we can say
that for any open map $f$, it is open-irreducible iff the left adjoint of
$f^{-1}$ preserves the binary meets. Similarly, if $f:X\to Y$ is a closed map,
then the map $f_{*}:\mathcal{O}(X)\to\mathcal{O}(Y)$ defined by
$f_{*}(U)=f[U^{c}]^{c}$ is the right adjoint of $f^{-1}$, as
$f^{-1}(V)\subseteq U$ iff $V\subseteq f[U^{c}]^{c}$. Note that $f_{*}$ is
well-defined as $f$ is closed. Again, using this right adjoint, we can say
that for any closed map $f$, it is closed-irreducible iff the right adjoint of
$f^{-1}$ preserves the binary joins.

In the following theorem, we will characterize the open (closed) implications
based on their preservability under the inverse image of the open-irreducible
(closed-irreducible) maps.

######  Theorem 3.23.

For a strong space $(Y,\to_{Y})$, the followings are equivalent:

$(i)$

For any topological space $X$ and any open-irreducible map $f:X\to Y$, there
is an open implication $\to_{X}$ over $\mathcal{O}(X)$ such that
$f^{-1}(U\to_{Y}V)=f^{-1}(U)\to_{X}f^{-1}(V)$, for any $U,V\in\mathcal{O}(Y)$.

$(ii)$

For any topological space $X$ and any open-irreducible map $f:X\to Y$, there
is an implication $\to_{X}$ over $\mathcal{O}(X)$ such that we have
$f^{-1}(U\to_{Y}V)=f^{-1}(U)\to_{X}f^{-1}(V)$, for any $U,V\in\mathcal{O}(Y)$.

$(iii)$

For any topological space $X$ and any open embedding $f:X\to Y$, there is an
implication $\to_{X}$ over $\mathcal{O}(X)$ such that
$f^{-1}(U\to_{Y}V)=f^{-1}(U)\to_{X}f^{-1}(V)$, for any $U,V\in\mathcal{O}(Y)$.

$(iv)$

The implication $\to_{Y}$ is open localizable, i.e., for any open subset
$Z\subseteq Y$ and any $U_{1},V_{1},U_{2},V_{2}\in\mathcal{O}(Y)$, if
$U_{1}\cap Z=U_{2}\cap Z$ and $V_{1}\cap Z=V_{2}\cap Z$, then
$(U_{1}\to_{Y}V_{1})\cap Z=(U_{2}\to_{Y}V_{2})\cap Z$.

$(v)$

The implication $\to_{Y}$ is open.

The same also holds, replacing open with closed, everywhere in the theorem.

###### Proof.

The parts $(i)$ to $(ii)$ and $(ii)$ to $(iii)$ are trivial. For $(iii)$ to
$(iv)$, as $Z$ is open in $Y$, the inclusion map $j:Z\to Y$ is an open
embedding. Therefore, we can apply $(iii)$ to $j$. Then, as
$j^{-1}(U_{1})=U_{1}\cap Z=U_{2}\cap Z=j^{-1}(U_{2})$ and
$j^{-1}(V_{1})=V_{1}\cap Z=V_{2}\cap Z=j^{-1}(V_{2})$, we have
$j^{-1}(U_{1})\to_{Z}j^{-1}(V_{1})=j^{-1}(U_{2})\to_{Z}j^{-1}(V_{2})$.
Therefore, $j^{-1}(U_{1}\to_{Y}V_{1})=j^{-1}(U_{2}\to_{Y}V_{2})$ and hence,
$(U_{1}\to_{Y}V_{1})\cap Z=(U_{2}\to_{Y}V_{2})\cap Z$.
For $(iv)$ to $(v)$, using Lemma 3.3, it is enough to prove $V\subseteq
U\to_{Y}U\cap V$, for any $U,V\in\mathcal{O}(Y)$. Set $Z=V$ in $(iv)$. As
$Z\cap(U\cap V)=Z\cap U$, we have $(U\to_{Y}U\cap V)\cap Z=(U\to_{Y}U)\cap Z$,
by $(iv)$. As $\to_{Y}$ is an implication, we have $U\to_{Y}U=Y$ and hence
$V=Z\subseteq(U\to_{Y}U\cap V)$.
Finally, to prove $(i)$ from $(v)$, define
$U^{\prime}\to_{X}V^{\prime}=f^{-1}(f[U^{\prime}]\to_{Y}f[V^{\prime}])$, for
any $U^{\prime},V^{\prime}\in\mathcal{O}(X)$. First, note that as $f$ is open,
$\to_{X}$ is well-defined over $\mathcal{O}(X)$. Secondly, by Example 1.3,
$\to_{X}$ is clearly an implication. Thirdly, recall that $f_{!}(U)=f[U]$ is
the left adjoint of $f^{-1}$ and preserves all binary meets. Hence, as
$\to_{Y}$ is open, by Theorem 3.10, the implication $\to_{X}$ is open.
Finally, to prove $f^{-1}(U)\to_{X}f^{-1}(V)=f^{-1}(U\to_{Y}V)$, for any
$U,V\in\mathcal{O}(Y)$, we first prove that $f^{-1}(U_{1})=f^{-1}(U_{2})$ and
$f^{-1}(V_{1})=f^{-1}(V_{2})$ imply
$f^{-1}(U_{1}\to_{Y}V_{1})=f^{-1}(U_{2}\to_{Y}V_{2})$, for any opens
$U_{1},V_{1},U_{2},V_{2}\in\mathcal{O}(Y)$. First, notice that by the
assumptions $f^{-1}(U_{1})=f^{-1}(U_{2})$ and $f^{-1}(V_{1})=f^{-1}(V_{2})$,
we reach $U_{1}\cap f[X]=U_{2}\cap f[X]$ and $V_{1}\cap f[X]=V_{2}\cap f[X]$.
As $f[X]$ is open, $\to_{Y}$ is an open implication, and $U_{2}\cap
f[X]\subseteq U_{1}$ and $V_{1}\cap f[X]\subseteq V_{2}$, we have
$f[X]\subseteq(U_{2}\to_{Y}U_{1})\cap(V_{1}\to_{Y}V_{2})$. Therefore,
$f[X]\cap(U_{1}\to_{Y}V_{1})\subseteq(U_{2}\to_{Y}U_{1})\cap(U_{1}\to_{Y}V_{1})\cap(V_{1}\to_{Y}V_{2})\subseteq
U_{2}\to_{Y}V_{2}$. By symmetry, we also have
$f[X]\cap(U_{2}\to_{Y}V_{2})\subseteq U_{1}\to_{Y}V_{1}$. Therefore,
$f[X]\cap(U_{1}\to_{Y}V_{1})=f[X]\cap(U_{2}\to_{Y}V_{2})$ which implies
$f^{-1}(U_{1}\to_{Y}V_{1})=f^{-1}(U_{2}\to_{Y}V_{2})$.
Now, as $f_{!}\dashv f^{-1}$, we have $f^{-1}f_{!}f^{-1}(U)=f^{-1}(U)$ and
$f^{-1}f_{!}f^{-1}(V)=f^{-1}(V)$, for any $U,V\in\mathcal{O}(Y)$. Therefore,
using what we just proved, we have
$f^{-1}(U)\to_{X}f^{-1}(V)=f^{-1}(f_{!}f^{-1}(U)\to_{Y}f_{!}f^{-1}(V))=f^{-1}(U\to_{Y}V)$.
For the closed case, the only non-trivial cases are $(iv)$ to $(v)$ and $(v)$
to $(i)$. For the first case, using Lemma 3.3, it is enough to show $(U\cup
V\to_{Y}V)\cup U=Y$ or equivalently $U^{c}\subseteq U\cup V\to_{Y}V$, for any
$U,V\in\mathcal{O}(Y)$. Set $Z=U^{c}$ in $(iv)$ and note that $Z$ is closed in
$Y$. As $(U\cup V)\cap Z=V\cap Z$, we have $(U\cup V\to_{Y}V)\cap
Z=(V\to_{Y}V)\cap Z$, by $(iv)$. As $\to_{Y}$ is an implication, we have
$V\to_{Y}V=Y$ and hence $U^{c}=Z\subseteq(U\cup V\to_{Y}V)$. Therefore,
$U^{c}\subseteq U\cup V\to_{Y}V$.
To prove $(i)$ from $(v)$, define
$U^{\prime}\to_{X}V^{\prime}=f^{-1}(f_{*}(U^{\prime})\to_{Y}f_{*}(V^{\prime}))$,
for any $U^{\prime},V^{\prime}\in\mathcal{O}(X)$, where
$f_{*}(W)=(f[W^{c}])^{c}$, for any $W\in\mathcal{O}(X)$. By Example 1.3,
$\to_{X}$ is clearly an implication. As $f^{-1}\dashv f_{*}$, the implication
$\to_{Y}$ is closed and $f_{*}$ preserves all binary joins, by Theorem 3.10,
$\to_{X}$ is also closed. To prove
$f^{-1}(U\to_{Y}V)=f^{-1}(U)\to_{X}f^{-1}(V)$, we first prove that
$f^{-1}(U_{1})=f^{-1}(U_{2})$ and $f^{-1}(V_{1})=f^{-1}(V_{2})$ imply
$f^{-1}(U_{1}\to_{Y}V_{1})=f^{-1}(U_{2}\to_{Y}V_{2})$, for any
$U_{1},V_{1},U_{2},V_{2}\in\mathcal{O}(Y)$. As before, by
$f^{-1}(U_{1})=f^{-1}(U_{2})$ and $f^{-1}(V_{1})=f^{-1}(V_{2})$, we reach
$U_{1}\cap f[X]=U_{2}\cap f[X]$ and $V_{1}\cap f[X]=V_{2}\cap f[X]$.
Therefore, $U_{2}\subseteq f[X]^{c}\cup U_{1}$ and $V_{1}\subseteq
f[X]^{c}\cup V_{2}$. As $f[X]^{c}$ is open and the implication $\to_{Y}$ is
closed, we have
$f[X]^{c}\cup(U_{2}\to_{Y}U_{1})=f[X]^{c}\cup(V_{1}\to_{Y}V_{2})=Y$. Hence,
$U_{1}\to_{Y}V_{1}\subseteq f[X]^{c}\cup(U_{1}\to_{Y}V_{1})$ which is itself a
subset of

$(f[X]^{c}\cup(U_{1}\to_{Y}V_{1}))\cap(f[X]^{c}\cup(U_{2}\to_{Y}U_{1}))\cap(f[X]^{c}\cup(V_{1}\to_{Y}V_{2})])=$
$f[X]^{c}\cup[(U_{2}\to_{Y}U_{1})\cap(U_{1}\to_{Y}V_{1})\cap(V_{1}\to_{Y}V_{2})]\subseteq
f[X]^{c}\cup(U_{2}\to_{Y}V_{2}).$

Thus, $U_{1}\to_{Y}V_{1}\subseteq f[X]^{c}\cup(U_{2}\to_{Y}V_{2})$. By
symmetry, $U_{2}\to_{Y}V_{2}\subseteq f[X]^{c}\cup(U_{1}\to_{Y}V_{1})$.
Therefore, $f[X]\cap(U_{1}\to_{Y}V_{1})=f[X]\cap(U_{2}\to_{Y}V_{2})$ which
implies $f^{-1}(U_{1}\to_{Y}V_{1})=f^{-1}(U_{2}\to_{Y}V_{2})$.
Finally, as $f^{-1}\dashv f_{*}$, we have the equalities
$f^{-1}f_{*}f^{-1}(U)=f^{-1}(U)$ and $f^{-1}f_{*}f^{-1}(V)=f^{-1}(V)$, for any
$U,V\in\mathcal{O}(Y)$. Therefore,

$f^{-1}(U)\to_{X}f^{-1}(V)=f^{-1}(f_{*}f^{-1}(U)\to_{Y}f_{*}f^{-1}(V))$

which is equal to $f^{-1}(U\to_{Y}V)$, for any $U,V\in\mathcal{O}(Y)$. ∎

##  4 Geometric Implications

In this section, we will introduce a natural notion of geometricity for a
category of strong spaces over a category of spaces to formalize the informal
concept of geometricity of a family of implications over a family of
continuous maps. Then, we will provide a characterization theorem for the
geometric categories over the categories of spaces satisfying some basic
closure properties. Specifically, we will see that the only geometric category
over $\mathbf{Top}$ is the category of the strong spaces with the trivial
implication. We first need the following lemma.

######  Lemma 4.1.

$(i)$

Let $(\mathcal{A},\to_{\mathcal{A}})$ be a strong algebra, $\mathcal{B}$ be a
bounded distributive lattice and $f:\mathcal{A}\to\mathcal{B}$ be a bounded
lattice map with a left inverse, i.e, a monotone function
$g:\mathcal{B}\to\mathcal{A}$ such that $gf=id_{\mathcal{A}}$. Then, there is
an implication $\to_{\mathcal{B}}$ over $\mathcal{B}$ such that
$f:(\mathcal{A},\to_{\mathcal{A}})\to(\mathcal{B},\to_{\mathcal{B}})$ is a
strong algebra map.

$(ii)$

Let $(\mathcal{A},\to_{\mathcal{A}})$ be a weakly Boolean strong algebra,
$\mathcal{B}$ be a bounded distributive lattice and
$f:\mathcal{A}\to\mathcal{B}$ be a surjective bounded distributive lattice
morphism. Then, there is a unique weakly Boolean implication
$\to_{\mathcal{B}}$ over $\mathcal{B}$ such that
$f:(\mathcal{A},\to_{\mathcal{A}})\to(\mathcal{B},\to_{\mathcal{B}})$ is a
strong algebra map. If $\mathcal{B}$ is Boolean, it holds even without the
surjectivity condition.

###### Proof.

For $(i)$, define $c\to_{\mathcal{B}}d=f(g(c)\to_{\mathcal{A}}g(d))$, for any
$c,d\in\mathcal{B}$, where $g:\mathcal{B}\to\mathcal{A}$ is the left inverse
of $f$. By Example 1.3, the operator $\to_{\mathcal{B}}$ is an implication. To
prove that $f$ is a strong algebra map, note that
$f(a)\to_{\mathcal{B}}f(b)=f(gf(a)\to_{\mathcal{A}}gf(b))$. As
$gf=id_{\mathcal{A}}$, we reach the conclusion.
For $(ii)$, as $\to_{\mathcal{A}}$ is weakly Boolean, by Theorem 3.12, the
interval $[\neg_{\mathcal{A}}1,1]$ is a Boolean algebra,
$a\to_{\mathcal{A}}b=\neg_{\mathcal{A}}a\vee b$ and $\neg_{\mathcal{A}}a$ is
the complement of $a\vee\neg_{\mathcal{A}}1$ in $[\neg_{\mathcal{A}}1,1]$. Set
$n=f(\neg_{\mathcal{A}}1)$. Note that $f(\neg_{\mathcal{A}}a)$ is the
complement of $f(a)\vee n$ in $[n,1_{\mathcal{B}}]$, for any
$a\in\mathcal{A}$, because $f(a)\vee n\vee
f(\neg_{\mathcal{A}}a)=f(a\vee\neg_{\mathcal{A}}a\vee\neg_{\mathcal{A}}1)=f(1_{\mathcal{A}})=1_{\mathcal{B}}$
and $(f(a)\vee n)\wedge
f(\neg_{\mathcal{A}}a)=f(a\vee\neg_{\mathcal{A}}1)\wedge
f(\neg_{\mathcal{A}}a)=f((a\vee\neg_{\mathcal{A}}1)\wedge\neg_{\mathcal{A}}a)=f(\neg_{\mathcal{A}}1)=n$.
Now, to show that the interval $[n,1_{\mathcal{B}}]$ in $\mathcal{B}$ is a
Boolean algebra, we use either the surjectivity of $f$ or the Booleanness of
$\mathcal{B}$. If $f$ is surjective, then every $c\in\mathcal{B}$ is equal to
$f(a)$, for some $a\in\mathcal{A}$. Hence, if $c=f(a)\geq n$, as
$f(\neg_{\mathcal{A}}a)$ is the complement of $f(a)\vee n=f(a)$ in
$[n,1_{\mathcal{B}}]$, then any $c\geq n$ has a complement in
$[n,1_{\mathcal{B}}]$. If $\mathcal{B}$ is Boolean, then the complement of
$c\geq n$ in $[n,1_{\mathcal{B}}]$ is $\bar{c}\vee n$, where $\bar{c}$ is the
complement of $c$ in $\mathcal{B}$. Now, by Theorem 3.12, the operation
$c\to_{\mathcal{B}}d=n(c)\vee d$ is a WBI over $\mathcal{B}$, where $n(c)$ is
the complement of $c\vee n$ in $[n,1_{\mathcal{B}}]$. Note that we proved
$n(f(a))=f(\neg_{\mathcal{A}}a)$. Hence, as $\neg_{\mathcal{B}}f(a)=n(f(a))$,
we reach $f(a\to_{\mathcal{A}}b)=f(\neg_{\mathcal{A}}a\vee
b)=f(\neg_{\mathcal{A}}a)\vee b=\neg_{\mathcal{B}}f(a)\vee
f(b)=f(a)\to_{\mathcal{B}}f(b)$. For uniqueness, let $\to$ be another WBI over
$\mathcal{B}$ such that $f(a\to_{\mathcal{A}}b)=f(a)\to f(b)$, for any
$a,b\in\mathcal{A}$. Then, $\neg 1_{\mathcal{B}}=f(\neg_{\mathcal{A}}1)=n$. As
$\neg c$ and $\neg_{\mathcal{B}}c$ are both the complements of $c\vee n$ in
$[n,1_{\mathcal{B}}]$, they must be equal. As a WBI is determined uniquely by
its negation, we have $\to_{\mathcal{B}}=\,\to$. ∎

As geometricity is the stability of a family of implications under the inverse
image of a family of continuous functions, to formalize this notion, we need
to be precise about two ingredients: the continuous maps we use and the family
of implications we choose. For the former, it is reasonable to start with a
subcategory $\mathcal{S}$ of $\mathbf{Top}$ to have a relative version of
geometricity. For the latter, as any implication must be over a space in this
case, a natural formalization of a family of implications is some sort of
fibration that to each space $X$ in $\mathcal{S}$ assigns a fiber of strong
spaces over $X$. Having these two ingredients fixed, the geometricity simply
means the stability of the fibres under the inverse image of the maps in
$\mathcal{S}$. In other words, it states that for any map $f:X\to Y$ in
$\mathcal{S}$, the inverse image map $f^{-1}$ must map a fiber over $Y$ into
the fiber over $X$. The following is the formalization of this idea.

######  Definition 4.2.

Let $\mathcal{S}$ be a (not necessarily full) subcategory of $\mathbf{Top}$. A
category $\mathcal{C}$ of strong spaces is called _geometric over
$\mathcal{S}$_, if the forgetful functor $U:\mathcal{C}\to\mathbf{Top}$
mapping $(X,\to_{X})$ to $X$ has the following conditions:

$(i)$

$U$ maps $\mathcal{C}$ to $\mathcal{S}$ and it is surjective on the objects of
$\mathcal{S}$, and

$(ii)$

for any object $(Y,\to_{Y})$ in $\mathcal{C}$, any object $X$ in $\mathcal{S}$
and any map $f:X\to Y=U(Y,\to_{Y})$ in $\mathcal{S}$, there exists an object
$(X,\to_{X})$ in $\mathcal{C}$ such that $f$ induces a strong space map
$f:(X,\to_{X})\to(Y,\to_{Y})$ in $\mathcal{C}$:

${{\mathcal{C}}}$${{(X,\to_{X})}}$${{(Y,\to_{Y})}}$${{\mathcal{S}}}$${X}$${{Y=U(Y,\to_{Y})}}$$\scriptstyle{U}$$\scriptstyle{f}$$\scriptstyle{f}$

Let us explain the connection between the definition and the previous
discussion we had. Using the functor $U$, the category $\mathcal{C}$ is
nothing but a way to provide a fiber of strong spaces or equivalently a fiber
of implications over any space in $\mathcal{S}$. The condition $(i)$, then,
demands that the fibers and the maps between them are all lying over
$\mathcal{S}$ and none of the fibers are empty. The condition $(ii)$ is the
geometricity condition that states that for any map $f:X\to Y$, its inverse
image can pull back any implication over $Y$ to an implication over $X$. In
other words, the fibers are stable under the inverse image of the maps of
$\mathcal{S}$.
Note that if $\mathcal{C}$ is geometric over $\mathcal{S}$, then the maps of
$\mathcal{C}$ and $\mathcal{S}$ are the same. One direction is easy as $U$
maps $\mathcal{C}$ into $\mathcal{S}$. For the other direction, notice that
for any map $f:X\to Y$, as the fiber over $Y$ is non-empty, there exists an
object $(Y,\to_{Y})$ in $\mathcal{C}$. Now, use $(ii)$ to show that $f$
actually lives in $\mathcal{C}$. This implies that to show the equality of two
geometric categories over a fixed category, it is enough to show that their
objects are the same.

######  Example 4.3.

For any category $\mathcal{S}$ of spaces, let $\mathcal{S}_{t}$ be the
category of strong spaces $(X,\to_{t})$, where $X$ is in $\mathcal{S}$ and
$\to_{t}$ is the trivial implication together with all the maps of
$\mathcal{S}$ as the morphisms. It is well defined as the inverse image of any
map $f:X\to Y$ in $\mathcal{S}$ preserves the implication, since
$f^{-1}(Y)=X$. It is clear that $\mathcal{S}_{t}$ is a geometric category over
$\mathcal{S}$. This category is called the _trivial_ geometric category over
$\mathcal{S}$.

If $\mathcal{S}$ only consists of locally indiscrete spaces, then there are
three other degenerate geometric categories over $\mathcal{S}$. The first is
the category $\mathcal{S}_{b}$ of strong spaces $(X,\to_{b})$, where $X$ is in
$\mathcal{S}$ and $\to_{b}$ is the Boolean implication together with the maps
of $\mathcal{S}$ as the morphisms. This category is well defined, since the
locally indiscreteness of $X$ implies the Booleanness of $\mathcal{O}(X)$ and
the inverse images always preserve all the Boolean operators. It is easy to
see that $\mathcal{S}_{b}$ is actually geometric over $\mathcal{S}$. The
second example is the union of $\mathcal{S}_{b}$ and $\mathcal{S}_{t}$ that we
denote by $\mathcal{S}_{bt}$. This category is also clearly geometric over
$\mathcal{S}$. The third example is $\mathcal{S}_{a}$, the subcategory of
strong spaces $(X,\to)$, where $X$ is in $\mathcal{S}$ and $\to$ is a WBI,
together with the strong space morphisms that $U$ maps into $\mathcal{S}$. The
fibers are clearly non-empty, because of the existence of the trivial
implication over any space. It is geometric by part $(ii)$ of Lemma 4.1.

To have a non-degenerate example, note that the category of all strong spaces
$(X,\to)$, where $\to$ is open (closed) together with the strong maps that are
open-irreducible (closed-irreducible) is geometric over the category
$\mathbf{OI}$ (resp. $\mathbf{CI}$), by Theorem 3.23. Note that Theorem 3.23
actually proves that these categories are the greatest geometric categories
over $\mathbf{OI}$ and $\mathbf{CI}$, respectively, because if the inverse
image of any open-irreducible (closed-irreducible) map into $X$ maps an
implication $\to$ over $\mathcal{O}(X)$ to another implication, then $\to$ is
open (resp. closed). Finally, it is worth mentioning that the category of all
strong spaces with the strong space morphisms that are also surjective is
geometric over the category of all spaces with all surjections,
$\mathbf{Srj}$. To prove, note that if $f:X\to Y$ is a continuous surjection,
then $f^{-1}$ has a left inverse that is also monotone. Now, use part $(i)$ of
Lemma 4.1.

To provide a characterization for all geometric categories over a given
category $\mathcal{S}$, it is reasonable to assume some closure properties for
$\mathcal{S}$.

######  Definition 4.4.

A subcategory $\mathcal{S}$ of $\mathbf{Top}$ is called _local_ if it has at
least one non-empty object and it is closed under all embeddings, i.e., for
any space $X$ in $\mathcal{S}$ and any embedding $f:Y\to X$, both $Y$ and $f$
belongs to $\mathcal{S}$. A space $X$ is called _full_ in $\mathcal{S}$ if
$\mathcal{S}$ has $X$ as an object and all maps $f:Y\to X$ as maps, for any
object $Y$ in $\mathcal{S}$.

First, note that a local subcategory $\mathcal{S}$ of $\mathbf{Top}$ has all
singleton spaces as its objects. Because, as $\mathcal{S}$ has a non-empty
space $X$, it is possible to pick $x\in X$ and set $f:\\{*\\}\to X$ as a
function mapping $*$ to $x$. As $f$ is an embedding and $\mathcal{S}$ is
local, the space $\\{*\\}$ is in $\mathcal{S}$. Secondly, notice that although
the singletons are all present in any local subcategory $\mathcal{S}$ of
$\mathbf{Top}$, it does not necessarily mean that $\mathcal{S}$ has a terminal
object. To have a counter-example, it is enough to set $\mathcal{S}$ as the
subcategory of all embeddings in $\mathbf{Top}$. $\mathcal{S}$ is trivially
local. However, if $T$ is its terminal object, then it must be a singleton, as
otherwise, there are either two or none embeddings from $\\{*\\}$ to $T$.
However, if $T$ is a singleton, then there is no embedding from the discrete
space $\\{*,\dagger\\}$ with two points into $T$ which is a contradiction.
Using an argument similar to what we had above, it is not hard to see that a
terminal object in a local subcategory of $\mathbf{Top}$, if exists, must be a
singleton space. Moreover, a singleton space is a terminal object iff it is
full in the subcategory.

######  Theorem 4.5.

Let $\mathcal{S}$ be a local subcategory of $\mathbf{Top}$ with a terminal
object:

$(i)$

If $\mathcal{S}$ has at least one non-locally-indiscrete space, then the only
geometric category over $\mathcal{S}$ is $\mathcal{S}_{t}$.

$(ii)$

If $\mathcal{S}$ only consists of locally-indiscrete spaces, includes a non-
indiscrete space and a full discrete space with two points, then the only
geometric categories over $\mathcal{S}$ are the four distinct categories
$\mathcal{S}_{t}$, $\mathcal{S}_{b}$, $\mathcal{S}_{bt}$ and
$\mathcal{S}_{a}$.

$(iii)$

If $\mathcal{S}$ only consists of indiscrete spaces, then the only geometric
categories over $\mathcal{S}$ are the three distinct categories
$\mathcal{S}_{t}$, $\mathcal{S}_{b}$, and $\mathcal{S}_{bt}$.

###### Proof.

We first show that the locality of $\mathcal{S}$ implies that the strong
spaces in $\mathcal{C}$ are all WBS. Let $(Y,\to_{Y})$ be an object in
$\mathcal{C}$. As $\mathcal{S}$ is closed under all embeddings, by
geometricity of $\mathcal{C}$ over $\mathcal{S}$, for any embedding $f:X\to
Y$, there is a strong space $(X,\to_{X})$ such that $f$ induces a strong space
map from $(X,\to_{X})$ to $(Y,\to_{Y})$. Therefore, by Theorem 3.23, $\to_{Y}$
is both open and closed and hence a WBI. Now, note that by Corollary 3.16,
$\to_{Y}$ has a core. Denote this core by $M_{Y}$. Then, the geometricity
implies that for any $f:X\to Y$ in $\mathcal{S}$, there exists a strong space
$(X,\to_{X})$ in $\mathcal{C}$ such that $f$ induces the strong space map
$f:(X,\to_{X})\to(Y,\to_{Y})$. Therefore, by Remark 3.17, we can conclude that
for any $f:X\to Y$ in $\mathcal{S}$, there is a WBI over $X$ in $\mathcal{C}$
with the core $M_{X}=f^{-1}(M_{Y})$. We will use this property frequently in
the proof.
To prove $(i)$, it is enough to show that $M_{Y}=Y$, for any $(Y,\to_{Y})$ in
$\mathcal{C}$. Because, by Remark 3.17, any strong space over $Y$ must have
the trivial implication and as the fiber is non-empty, it also has the trivial
strong space. Thus, the objects of $\mathcal{C}$ are the same as the objects
of $\mathcal{S}_{t}$ and hence $\mathcal{C}=\mathcal{S}_{t}$. Now, let $X$ be
a non-locally-indiscrete space in $\mathcal{S}$ and for the sake of
contradiction, assume $M_{Y}\neq Y$, for some $(Y,\to_{Y})$ in $\mathcal{C}$.
Set $y\in Y-M_{Y}$. As $\mathcal{S}$ is local and has a terminal object, the
singleton space $\\{*\\}$ is a full object in $\mathcal{S}$. As $\mathcal{S}$
is closed under the embeddings, $\mathcal{S}$ has $Y-M_{Y}$ as an object and
the embeddings $i:Y-M_{Y}\to Y$ and $k:\\{*\\}\to Y-M_{Y}$ as the morphisms,
where $i$ is the inclusion and $k$ maps $*$ to $y$. Consider the map
$f=ik!:X\to Y$, where $!:X\to\\{*\\}$ is the constant function. As $\\{*\\}$
is full, $!$ and hence $f$ lives in $\mathcal{S}$. By the geometricity of
$\mathcal{C}$ over $\mathcal{S}$, there is an implication $\to_{X}$ over $X$
such that $f^{-1}(M_{Y})=M_{X}$. As $f$ is a constant function mapping every
element of $X$ to $y\notin M_{Y}$, we have $f^{-1}(M_{Y})=\varnothing$. Hence,
$M_{X}=\varnothing$. Therefore, by Remark 3.17, $X$ is locally indiscrete
which is a contradiction. Therefore, $M_{Y}=Y$.
For $(ii)$, it is easy to see that the four geometric categories over
$\mathcal{S}$ are different. Let $X$ be a non-indiscrete space in
$\mathcal{S}$. As it is not indiscrete, there is a non-trivial open
$\varnothing\subset M\subset X$. As $X$ is locally indiscrete, by Remark 3.17,
there is a WBI over $X$ with the core $M$. This WBI is clearly different from
the Boolean and the trivial ones, as its core, i.e., $M$ is different from the
cores of the other two that are $\varnothing$ and $X$. Hence, the fibers over
$X$ in the four categories $\mathcal{S}_{t}$, $\mathcal{S}_{b}$,
$\mathcal{S}_{bt}$ and $\mathcal{S}_{a}$ are different. To prove that these
four categories are the only geometric categories over $\mathcal{S}$, we will
consider the following two cases. Either, $M_{Y}=\varnothing$ or $M_{Y}=Y$,
for any $(Y,\to_{Y})$ in $\mathcal{C}$ or there exists a strong space
$(Y,\to_{Y})$ in $\mathcal{C}$ such that $\varnothing\subset M_{Y}\subset Y$.
In the first case, we show that $\mathcal{C}$ is uniquely determined by the
fiber over $\\{*\\}$ and it is one of $\mathcal{S}_{t}$, $\mathcal{S}_{b}$ or
$\mathcal{S}_{bt}$. In the second case, we show that
$\mathcal{C}=\mathcal{S}_{a}$. For the first case, as $\\{*\\}$ has only two
subsets, there are only two possible implications over $\\{*\\}$. If both are
in $\mathcal{C}$, then as the constant function $!:Y\to\\{*\\}$ lives in
$\mathcal{S}$, by the fullness of $\\{*\\}$, the geometricity dictates the
existence of two strong spaces $(Y,\to_{Y})$ and $(Y,\to^{\prime}_{Y})$ in
$\mathcal{C}$ such that $!^{-1}(\\{*\\})=M_{Y}$ and
$!^{-1}(\varnothing)=M^{\prime}_{Y}$. Therefore, $M_{Y}=Y$ and
$M^{\prime}_{Y}=\varnothing$. Hence, over $Y$ the implications with both cores
appear. Hence, on objects, $\mathcal{C}$ and $\mathcal{S}_{bt}$ are similar
which implies $\mathcal{C}=\mathcal{S}_{bt}$. For the other case, if the fiber
over $\\{*\\}$ has just one implication, denote its core by $M_{*}$. Now, pick
an arbitrary non-empty strong space $(Y,\to_{Y})$ in $\mathcal{C}$ with the
core $M_{Y}$. Note that it is either $\varnothing$ or $Y$, by the assumption.
As $Y$ is non-empty, there is a map $i:\\{*\\}\to Y$. This is an embedding and
hence it is in $\mathcal{S}$. Therefore, by geometricity,
$i^{-1}(M_{Y})=M_{*}$. Hence, if $M_{*}=\varnothing$, we have
$M_{Y}=\varnothing$ and if $M_{*}=\\{*\\}$, we reach $M_{Y}=Y$. Therefore, by
the same reasoning as before, either $\mathcal{C}=\mathcal{S}_{b}$ or
$\mathcal{C}=\mathcal{S}_{t}$.
For the second case, assume the existence of a strong space $(Y,\to_{Y})$ in
$\mathcal{C}$ such that $\varnothing\subset M_{Y}\subset Y$. For any $Z$ in
$\mathcal{S}$ and any open $N\subseteq Z$, define $\to_{N}$ as the WBI with
the core $N$. As $Y$ is locally indiscrete, this is possible by Remark 3.17.
We show that $(Z,\to_{N})$ is in $\mathcal{C}$ and as $N$ is arbitrary, we
conclude $\mathcal{C}=\mathcal{S}_{a}$. Let $\\{*,\dagger\\}$ be the full
discrete space with two elements in $\mathcal{S}$. Set $y\in M_{Y}$ and $z\in
Y-M_{Y}$ and let $f:Z\to\\{*,\dagger\\}$ be the function mapping $N$ to $*$
and $Z-N$ to $\dagger$. As $N$ is open and $Z$ is locally indiscrete, $N$ is
also clopen. Hence, $f$ is continuous. As $\mathcal{S}$ has all maps into
$\\{*,\dagger\\}$, the map $f$ is in $\mathcal{S}$. Now, consider the map
$i:\\{*,\dagger\\}\to Y$ mapping $*$ to $y$ and $\dagger$ to $z$. As
$\\{*,\dagger\\}$ is discrete, the map $i$ is continuous. As $M_{Y}$ is open
and $Y$ is locally indiscrete, $M_{Y}$ is clopen. Hence, it is easy to see
that $i$ is an embedding. Therefore, $i$ is in $\mathcal{S}$. Consider the map
$g:Z\to Y$ defined by $g=if$ and notice that $g$ is also in $\mathcal{S}$. As
$\mathcal{C}$ is geometric over $\mathcal{S}$, there must be a strong space
$(Z,\to_{Z})$ with the core $M_{Z}$ such that $g^{-1}(M_{Y})=M_{Z}$. By
construction, we have $g^{-1}(M_{Y})=f^{-1}(i^{-1}(M_{Y}))=f^{-1}(\\{*\\})=N$.
Therefore, $M_{Z}=N$. As any WBI is uniquely determined by its core, we have
$\to_{Z}=\to_{N}$. Hence, $(Z,\to_{N})$ is in $\mathcal{C}$.
For $(iii)$, note that the trivial and the Boolean implications are different
over $\mathcal{O}(\\{*\\})$, as their cores, i.e., $\\{*\\}$ and
$\\{*\\}^{c}=\varnothing$ are different. Hence, the three categories
$\mathcal{S}_{t}$, $\mathcal{S}_{b}$ and $\mathcal{S}_{bt}$ are different, as
their fibers over $\\{*\\}$ are different. To prove that these are the only
geometric categories over $\mathcal{S}$, assume that $(X,\to_{X})$ is an
arbitrary object in $\mathcal{C}$. As $X$ is indiscrete and has only two
opens, the core is either $\varnothing$ or $X$. The rest is similar to the
first case in $(ii)$. ∎

######  Corollary 4.6.

Let $\mathcal{S}$ be a full subcategory of $\mathbf{Top}$ that is also local.
Then:

$(i)$

If $\mathcal{S}$ has at least one non-locally-indiscrete space, then the only
geometric category over $\mathcal{S}$ is $\mathcal{S}_{t}$.

$(ii)$

If $\mathcal{S}$ only consists of locally-indiscrete spaces and includes a
non-indiscrete space, then the only geometric categories over $\mathcal{S}$
are the four distinct categories $\mathcal{S}_{t}$, $\mathcal{S}_{b}$,
$\mathcal{S}_{bt}$ and $\mathcal{S}_{a}$.

$(iii)$

If $\mathcal{S}$ only consists of indiscrete spaces, then the only geometric
categories over $\mathcal{S}$ are the three distinct categories
$\mathcal{S}_{t}$, $\mathcal{S}_{b}$, and $\mathcal{S}_{bt}$.

###### Proof.

As $\mathcal{S}$ is local, it has the singleton object and as $\mathcal{S}$ is
a full subcategory, the singleton is a full object and hence it is the
terminal object of $\mathcal{S}$. Having made that observation, for $(i)$ and
$(iii)$, it is enough to use Theorem 4.5. For $(ii)$, let $Y$ be a non-
indiscrete yet locally indiscrete space in $\mathcal{S}$. Hence, it has a non-
trivial open $\varnothing\subset U\subset Y$ and as $Y$ is locally indiscrete,
$U$ is clopen. Pick two elements $x\in U$ and $y\in Y-U$. The map
$g:\\{*,\dagger\\}\to X$ mapping $*$ to $x$ and $\dagger$ to $y$ is continuous
as $\\{*,\dagger\\}$ is discrete. It is an embedding, as $U$ is clopen. Hence,
as $\mathcal{S}$ is local, it must have the discrete space $\\{*,\dagger\\}$.
Finally, as $\mathcal{S}$ is a full subcategory, the space $\\{*,\dagger\\}$
is a full object. Now, use Theorem 4.5 to complete the proof. ∎

######  Corollary 4.7.

$\mathbf{Top}_{t}$ is the only geometric category over $\mathbf{Top}$.

The following example shows that the existence of the terminal object in
Theorem 4.5 is crucial, as without the condition, it is possible to have
infinitely many geometric categories over some local subcategories of
$\mathbf{Top}$.

######  Theorem 4.8.

Let $\mathcal{S}$ be a category of Hausdorff spaces with injective continuous
maps such that the size of the objects is not bounded by any finite number.
Then, there are infinitely many geometric categories over $\mathcal{S}$.

###### Proof.

For any natural number $n\geq 0$, define the category $\mathcal{C}_{n}$ as
follows. For objects, consider all strong spaces $(X,\to_{M})$, where $X$ is
in $\mathcal{S}$, $M\subseteq X$ is a subset such that $X-M$ has at most $n$
elements and $\to_{M}$ is the WBI with the core $M$. Note that as $X-M$ is
finite and $X$ is Hausdorff, $X-M$ is closed and discrete. Therefore, the
implication is well-defined by Example 3.18. For the morphisms, consider all
maps $f:(X,\to_{M})\to(Y,\to_{N})$, where $f:X\to Y$ is in $\mathcal{S}$ and
$f^{-1}(N)=M$. By Remark 3.17, the maps are strong space maps and hence well
defined. We prove that $\mathcal{C}_{n}$ is geometric over $\mathcal{S}$. It
is clear that $U$ maps $\mathcal{C}_{n}$ into $\mathcal{S}$. The fibers are
non-empty as the trivial space $(X,\to_{t})$ is in $\mathcal{C}_{n}$. For
geometricity, for any $(Y,\to_{N})$ and any map $f:X\to Y$ in $\mathcal{S}$,
as $Y-N$ has at most $n$ elements and $f$ is injective, the set
$f^{-1}(Y-N)=X-f^{-1}(N)$ has at most $n$ elements, as well. Hence, $f$ lifts
to the strong space map $f:(X,\to_{f^{-1}(N)})\to(Y,\to_{N})$. Finally, it is
clear that $\mathcal{C}_{n}$ is a subcategory of $\mathcal{C}_{n+1}$. We show
that this inclusion is proper. To prove it, note that the size of the objects
in $\mathcal{S}$ is not bounded by a finite number. Hence, for any $n$, there
exists a space $X$ with at least $n+1$ elements. Choose $M$ such that $X-M$
has exactly $n+1$ elements. Then, $(X,\to_{M})$ appears in $\mathcal{C}_{n+1}$
but not in $\mathcal{C}_{n}$. ∎

##  5 A Kripke-style Representation Theorem

In the previous sections, we have introduced three families of implications
and their corresponding geometric characterizations. These implications are
also interesting in their own right for their algebraic and logical aspects.
Algebraically, they are different internalizations of the order of the base
lattice, while logically, as observed in Remark 3.2, they are reflecting
different versions of the right implication rule. Among the three, the closed
and the weakly Boolean implications are rather too restricted in their form
and hence too close to the well-understood Boolean implication. However, the
open implications may come in many different forms, appearing in many
different logical disciplines. For instance, the open meet-internalizing and
join-internalizing implications are well-studied in provability logics [21]
and philosophical discussions around the impredicativity of the intuitionistic
implication [17]. Inspired by these aspects, in this section, we intend to
provide a representation theorem for all implications, in general, and the
open implications, in particular.

To provide this representation theorem, we first need to use a Yoneda-type
technique augmented by an ideal completion to embed any strong algebra in a
strong algebra over a locale, where the implication becomes a part of an
adjunction-like situation. The machinery is explained in detail in [1].
However, for the sake of completeness and to address the new case of the open
implications, we will repeat the main construction here and refer the reader
to [1], for a more detailed explanation.

######  Lemma 5.1.

For any strong algebra $(\mathcal{A},\to_{\mathcal{A}})$, there exist a locale
$\mathcal{H}$, an implication $\to_{\mathcal{H}}$ over $\mathcal{H}$, an
embedding
$i:(\mathcal{A},\to_{\mathcal{A}})\to(\mathcal{H},\to_{\mathcal{H}})$, a join
preserving operator $\nabla:\mathcal{H}\to\mathcal{H}$ and an order-preserving
map $F:\mathcal{H}\to\mathcal{H}$ such that $\nabla x\wedge F(y)\leq F(z)$ iff
$x\leq y\to_{\mathcal{H}}z$, for any $x,y,z\in\mathcal{H}$. If
$\to_{\mathcal{A}}$ is open, then $\to_{\mathcal{H}}$ can be chosen as open.

###### Proof.

Let $\mathcal{B}$ be the set of all functions
$f:\prod_{n=0}^{\infty}\mathcal{A}\to\mathcal{A}$ with the pointwise ordering
$\leq_{\mathcal{B}}$. It is clear that $(\mathcal{B},\leq_{\mathcal{B}})$ is a
bounded distributive lattice. It is also possible to define an implication
$\to_{\mathcal{B}}$ over $\mathcal{B}$ in a pointwise fashion. Define
$s:\mathcal{B}\to\mathcal{B}$ as a shift function on the inputs, i.e.,
$[s(f)](\\{x_{i}\\}_{i=0}^{\infty})=f(\\{x_{i}\\}_{i=1}^{\infty})$. Notice
that $s$ respects the order $\leq_{\mathcal{B}}$, the meets and the joins as
they are defined pointwise. Set $\mathcal{H}$ as the locale of the ideals of
$\mathcal{B}$ and define $i:\mathcal{A}\to\mathcal{H}$ as
$i(a)=\\{f\in\mathcal{B}\mid f\leq c_{a}\\}$, where $c_{a}$ is the constant
function mapping every input into $a$. It is clear that $i$ is an embedding
respecting all finite joins and meets. Define $\nabla
I=\\{f\in\mathcal{B}\mid\exists g\in I\;[f\leq s(g)]\\}$ and
$F(I)=\\{f\in\mathcal{B}\mid\exists g\in
I\,[f\leq(x_{0}\to_{\mathcal{B}}s(g))]\\}$. Note that both $\nabla$ and $F$
are order-preserving. It is easy to see that $\nabla$ is also join preserving.
As $\mathcal{H}$ is complete, $\nabla$ has a right adjoint $\Delta$. Define
$I\to_{\mathcal{H}}J=\Delta(F(I)\Rightarrow F(J))$, where $\Rightarrow$ is the
Heyting implication of $\mathcal{H}$. The operation $\to_{\mathcal{H}}$ is an
implication, by Example 1.3. By the adjunction $\nabla\dashv\Delta$, it is
clear that $\nabla K\cap F(I)\subseteq F(J)$ iff $K\subseteq
I\to_{\mathcal{H}}J$, for any ideal $I,J,K$ of $\mathcal{B}$. The only
remaining thing to prove is the identity
$i(a\to_{\mathcal{A}}b)=i(a)\to_{\mathcal{H}}i(b)$, for any
$a,b\in\mathcal{A}$. To prove, it is enough to show that $\nabla I\cap
F(i(a))\subseteq F(i(b))$ iff $I\subseteq i(a\to_{\mathcal{A}}b)$, for any
ideal $I$ of $\mathcal{B}$. For the forward direction, if $f\in I$, then
$s(f)\in\nabla I$ and as $c_{a}\in i(a)$, we have
$s(f)\wedge(x_{0}\to_{\mathcal{B}}s(c_{a}))\in\nabla I\cap F(i(a))$.
Therefore, $s(f)\wedge(x_{0}\to_{\mathcal{B}}s(c_{a}))\in F(i(b))$ which
implies
$s(f)\wedge(x_{0}\to_{\mathcal{B}}s(c_{a}))\leq(x_{0}\to_{\mathcal{B}}s(c_{b}))$.
This means, $f(\\{x_{i}\\}_{i=1}^{\infty})\wedge(x_{0}\to_{\mathcal{A}}a)\leq
x_{0}\to_{\mathcal{A}}b$, for any $\\{x_{i}\\}_{i=0}^{\infty}$. Putting
$x_{0}=a$, we reach $f(\\{x_{i}\\}_{i=1}^{\infty})\leq a\to_{\mathcal{A}}b$,
for any $\\{x_{i}\\}_{i=0}^{\infty}$. Thus, as $\\{x_{i}\\}_{i=0}^{\infty}$ is
arbitrary, we have $f(\\{x_{i}\\}_{i=0}^{\infty})\leq a\to_{\mathcal{A}}b$,
for any $\\{x_{i}\\}_{i=0}^{\infty}$. Therefore, $f\leq
c_{a\to_{\mathcal{A}}b}$ which implies $f\in i(a\to_{\mathcal{A}}b)$. For the
backward direction, if $f\in\nabla I\cap F(i(a))$, then there exist $g\in I$
and $h\in i(a)$ such that $f\leq s(g)\wedge(x_{0}\to_{\mathcal{B}}s(h))$. As
$g\in I$, we have $g\leq c_{a\to_{\mathcal{A}}b}$. Hence, $f\leq
c_{a\to_{\mathcal{A}}b}\wedge(x_{0}\to_{\mathcal{B}}c_{a})$ which means
$f(\\{x_{i}\\}_{i=0}^{\infty})\leq(a\to_{\mathcal{A}}b)\wedge(x_{0}\to_{\mathcal{A}}a)$,
for any $\\{x_{i}\\}_{i=0}^{\infty}$. Therefore,
$f(\\{x_{i}\\}_{i=0}^{\infty})\leq x_{0}\to_{\mathcal{A}}b$ which implies
$f\leq x_{0}\to_{\mathcal{B}}s(c_{b})$. Hence, $f\in F(i(b))$.
Finally, for the preservation of the openness, if $\to_{\mathcal{A}}$ is open,
using Lemma 3.3, it is enough to prove $I\subseteq J\to_{\mathcal{H}}I\cap J$,
for any $I,J\in\mathcal{H}$. Using the adjunction-style condition for
$\to_{\mathcal{H}}$, we must show that $\nabla I\cap F(J)\subseteq F(I\cap
J)$, for any $I,J\in\mathcal{H}$. To prove it, let $f\in\nabla I\cap F(J)$.
Hence, there are $g\in I$ and $h\in J$ such that $f\leq s(g)$ and $f\leq
x_{0}\to_{\mathcal{B}}s(h)$, i.e., $f(\\{x_{i}\\}_{i=0}^{\infty})\leq
g(\\{x_{i}\\}_{i=1}^{\infty})$ and $f(\\{x_{i}\\}_{i=0}^{\infty})\leq
x_{0}\to_{\mathcal{A}}h(\\{x_{i}\\}_{i=1}^{\infty})$, for any
$\\{x_{i}\\}_{i=0}^{\infty}$. We want to show that

$f(\\{x_{i}\\}_{i=0}^{\infty})\leq
x_{0}\to_{\mathcal{A}}[g(\\{x_{i}\\}_{i=1}^{\infty})\wedge
h(\\{x_{i}\\}_{i=1}^{\infty})].$

As $\to_{\mathcal{A}}$ is open, $g(\\{x_{i}\\}_{i=1}^{\infty})\leq
h(\\{x_{i}\\}_{i=1}^{\infty})\to_{\mathcal{A}}g(\\{x_{i}\\}_{i=1}^{\infty})\wedge
h(\\{x_{i}\\}_{i=1}^{\infty})$. Therefore, $f(\\{x_{i}\\}_{i=0}^{\infty})$ is
less than or equal to

$(x_{0}\to_{\mathcal{A}}h(\\{x_{i}\\}_{i=1}^{\infty}))\wedge(h(\\{x_{i}\\}_{i=1}^{\infty})\to_{\mathcal{A}}[g(\\{x_{i}\\}_{i=1}^{\infty})\wedge
h(\\{x_{i}\\}_{i=1}^{\infty})]).$

Hence, $f(\\{x_{i}\\}_{i=0}^{\infty})\leq
x_{0}\to_{\mathcal{A}}[g(\\{x_{i}\\}_{i=1}^{\infty})\wedge
h(\\{x_{i}\\}_{i=1}^{\infty})]$. Now, set $k=g\wedge h$. Therefore, $k\in
I\cap J$ and $f\leq x_{0}\to_{\mathcal{B}}s(k)$. Hence, $f\in F(I\cap J)$. ∎

In the following, we define a topological version of the KN-frames introduced
before to represent all (open) strong algebras. Apart from providing a more
specific representation, the topology helps to represent the open strong
algebras by the open KN-frames. Later, we will see that to represent an
arbitrary implication (without respecting the openness condition), even the
full KN-frames (without using any topology) are sufficient.

######  Definition 5.2.

A _KN-space_ is a KN-frame $\mathcal{X}=(X,\leq,R,C(X),N)$, where $(X,\leq)$
is a Priestley space and $C(X)$ is the set of all clopens of $X$. Spelling
out, it means that:

  * $\bullet$

$R$ is compatible with the order, i.e., if $x\leq y$ and $(y,z)\in R$, then
$(x,z)\in R$,

  * $\bullet$

for any $x\in X$ and any clopen upsets $U$ and $V$, if $U\subseteq V$ and
$U\in N(x)$ then $V\in N(x)$,

  * $\bullet$

$\lozenge_{R}(U)=\\{x\in X\mid\exists y\in U,(x,y)\in R\\}$ is clopen, for any
clopen $U$,

  * $\bullet$

$j(U)=\\{x\in X\mid U\in N(x)\\}$ is clopen, for any clopen upset $U$.

######  Theorem 5.3.

(KN-space Representation) For any strong algebra
$(\mathcal{A},\to_{\mathcal{A}})$, there exist a KN-space $\mathcal{X}$ and an
embedding $i:(\mathcal{A},\to_{\mathcal{A}})\to\mathfrak{A}(\mathcal{X})$.
Moreover, if $\mathcal{A}$ is open, then so is $\mathcal{X}$.

###### Proof.

By Lemma 5.1, w.l.o.g., we assume that $\mathcal{A}$ is a locale and there are
join preserving map $\nabla:\mathcal{A}\to\mathcal{A}$ and order-preserving
map $F:\mathcal{A}\to\mathcal{A}$ such that $\nabla c\wedge F(a)\leq F(b)$ iff
$c\leq a\to_{\mathcal{A}}b$, for any $a,b,c\in\mathcal{A}$. Set $(X,\leq)$ as
the Priestley space $(\mathcal{F}_{p}(\mathcal{A}),\subseteq)$ and define $R$
as $\\{(P,Q)\in X^{2}\mid\nabla[P]\subseteq Q\\}$. Set $i(a)=\\{P\in X\mid
a\in P\\}$ and define $N(P)=\\{U\in CU(X,\leq)\mid\exists
a\in\mathcal{A}\;[i(a)\subseteq U\;\text{and}\;F(a)\in P]\\}$. First, it is
clear that $R$ is compatible with the inclusion, $N(P)$ is upward closed on
the clopen upsets of $(X,\leq)$, for any $P\in X$ and $i$ is a bounded lattice
embedding. Secondly, observe that $i(a)\in N(P)$ iff $F(a)\in P$, for any
$a\in\mathcal{A}$ and $P\in\mathcal{F}_{p}(\mathcal{A})$. One direction is
obvious from the definition of $N$. For the other, if $i(a)\in N(P)$, then
there exists $b\in\mathcal{A}$ such that $F(b)\in P$ and $i(b)\subseteq i(a)$.
Hence, $b\leq a$ which implies $F(b)\leq F(a)$. As $P$ is a filter, $F(a)\in
P$. Now, to prove that $j$ maps the clopen upsets to the clopens, if $U$ is a
clopen upset, there is $a\in\mathcal{A}$ such that $U=i(a)$. Therefore,
$j(U)=j(i(a))=\\{P\in X\mid i(a)\in N(P)\\}=i(F(a))$ which is clopen.
For the closure of the clopens under $\Diamond_{R}$, we need an auxiliary
implication. First, notice that $\mathcal{A}$ is a locale and $\nabla$ is join
preserving. Hence, it has a right adjoint $\Delta$. Define
$a\to_{\nabla}b=\Delta(a\Rightarrow b)$, where $\Rightarrow$ is the Heyting
implication of $\mathcal{A}$. By Example 1.3, it is clear that $\to_{\nabla}$
is an implication. Also notice that $c\leq a\to_{\nabla}b$ iff $\nabla c\wedge
a\leq b$, for any $a,b,c\in\mathcal{A}$. The important property of
$\to_{\nabla}$ for us is that $\Diamond_{R}(i(a)\cap
i(b)^{c})=i(a\to_{\nabla}b)^{c}$, for any $a,b\in\mathcal{A}$. To prove it, we
have to show that $a\to_{\nabla}b\in P$ iff for any $Q\in X$ such that
$\nabla[P]\subseteq Q$, if $a\in Q$ then $b\in Q$. The forward direction is
easy, as $a\to_{\nabla}b\in P$ implies $\nabla(a\to_{\nabla}b)\in Q$. Hence,
if $a\in Q$, as $a\wedge\nabla(a\to_{\nabla}b)\leq b$, we reach $b\in P$. For
the converse, if $a\to_{\nabla}b\notin P$, then define $G$ and $I$ as the
filter generated by $\nabla[P]\cup\\{a\\}$ and the ideal generated by $b$,
respectively. It is clear that $I\cap G=\varnothing$. As otherwise, if $x\in
I\cap G$, there are $p_{1},\ldots,p_{n}\in P$ such that
$\bigwedge_{i=1}^{n}\nabla p_{i}\wedge a\leq x\leq b$. As $\nabla$ is order-
preserving, $\nabla(\bigwedge_{i=1}^{n}p_{i})\wedge a\leq b$ which implies
$\bigwedge_{i=1}^{n}p_{i}\leq a\to_{\nabla}b$. As $p_{i}$’s are in $P$ and $P$
is a filter, $a\to_{\nabla}b\in P$ which is a contradiction. Hence, $G\cap
I=\varnothing$. Now, by the prime filter theorem, there is a prime filter $Q$
such that $G\subseteq Q$ and $Q\cap I=\varnothing$. By the former, we see that
$\nabla[P]\subseteq Q$ and $a\in Q$. By the latter, we see $b\notin Q$.
Now, having the identity $\Diamond_{R}(i(a)\cap
i(b)^{c})=i(a\to_{\nabla}b)^{c}$, we prove the closure of the clopens of $X$
under $\Diamond_{R}$. Assume that $U$ is clopen. Then, there are finite sets
$\\{a_{1},\ldots,a_{n}\\},\\{b_{1},\ldots,b_{n}\\}\subseteq\mathcal{A}$ such
that $U=\bigcup_{r=1}^{n}i(a_{r})\cap i(b_{r})^{c}$. As $\Diamond_{R}$
preserves the unions, we have
$\Diamond_{R}(U)=\bigcup_{r=1}^{n}\Diamond_{R}(i(a_{r})\cap i(b_{r})^{c})$. As
$\Diamond_{R}(i(a_{r})\cap i(b_{r})^{c})=i(a_{r}\to_{\nabla}b_{r})^{c}$, we
conclude that $\Diamond_{R}(i(a_{r})\cap i(b_{r})^{c})$ and hence
$\Diamond_{R}(U)$ is clopen.
The only remaining part is showing
$i(a\to_{\mathcal{A}}b)=i(a)\to_{\mathcal{X}}i(b)$, for any
$a,b\in\mathcal{A}$. To prove $i(a\to_{\mathcal{A}}b)\subseteq
i(a)\to_{\mathcal{X}}i(b)$, assume $P\in i(a\to_{\mathcal{A}}b)$ which implies
$a\to_{\mathcal{A}}b\in P$. Assume $(P,Q)\in R$ and $i(a)\in N(Q)$, for some
$Q\in X$. By definition of $R$, we have $\nabla[P]\subseteq Q$. As we observed
above, $i(a)\in N(Q)$ implies $F(a)\in Q$. Hence, we also have $F(a)\in Q$. To
prove $F(b)\in Q$, as $a\to_{\mathcal{A}}b\in P$, we have
$\nabla(a\to_{\mathcal{A}}b)\in\nabla[P]\subseteq Q$. As $Q$ is a filter and
$\nabla(a\to_{\mathcal{A}}b)\wedge F(a)\leq F(b)$, we reach $F(b)\in Q$. For
the converse, assume $a\to_{\mathcal{A}}b\notin P$. We intend to provide a
prime filter $Q\in X$ such that $\nabla[P]\subseteq Q$ and $F(a)\in Q$ but
$F(b)\notin Q$. Set $I$ and $G$ as the ideal generated by $F(b)$ and the
filter generated by $\nabla[P]\cup\\{F(a)\\}$, respectively. We claim $I\cap
G=\varnothing$. Otherwise, if $x\in I\cap G$, there are $p_{1},\ldots,p_{n}\in
P$ such that $\bigwedge_{i=1}^{n}\nabla p_{i}\wedge F(a)\leq x\leq F(b)$. As,
$\nabla$ is order-preserving, $\nabla(\bigwedge_{i=1}^{n}p_{i})\wedge F(a)\leq
F(b)$ which implies $\bigwedge_{i=1}^{n}p_{i}\leq a\to_{\mathcal{A}}b$. As
$p_{i}$’s are in $P$ and it is a filter, $a\to_{\mathcal{A}}b\in P$ which is a
contradiction. Hence, $G\cap I=\varnothing$. Now, by the prime filter theorem,
there is a prime filter $Q\in X$ such that $G\subseteq Q$ and $Q\cap
I=\varnothing$. By the former, we reach $\nabla[P]\subseteq Q$ and $F(a)\in
Q$. By the latter, we prove $F(b)\notin Q$.
Finally, note that as all clopen upsets are in the image of $i$, if
$\to_{\mathcal{A}}$ is open, for any clopen upsets $U$ and $V$, there are
$a,b\in\mathcal{A}$ such that $U=i(a)$ and $V=i(b)$. Therefore, as $a\leq
b\to_{\mathcal{A}}a\wedge b$ and $i$ preserves the meet and the implication,
we reach $U\subseteq V\to_{\mathcal{X}}U\cap V$. Hence, for any $P,Q\in X$ and
any clopen upsets $U$ and $V$, if $P\in U$, $(P,Q)\in R$ and $V\in N(Q)$, then
$U\cap V\in N(Q)$. ∎

######  Corollary 5.4.

For any strong algebra $(\mathcal{A},\to_{\mathcal{A}})$, there exist a full
KN-frame $\mathcal{K}$ and an embedding
$i:(\mathcal{A},\to_{\mathcal{A}})\to\mathfrak{A}(\mathcal{K})$.

###### Proof.

It is an easy consequence of Theorem 5.3 and Remark 3.7. ∎

Acknowledgments: The support by the FWF project P 33548 and the Czech Academy
of Sciences (RVO 67985840) are gratefully acknowledged.

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