††

# On maximal nowhere dense sublocales

Mbekezeli Nxumalo  Department of Mathematical Sciences, University of South
Africa, P.O. Box 392, 0003 Pretoria, SOUTH AFRICA.  Department of Mathematics,
Rhodes University, P.O. Box 94, Grahamstown 6140, South Africa.
<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to study some variants of nowhere dense sublocales
called maximal nowhere dense and homogeneous maximal nowhere dense sublocales.
These concepts were initially introduced by Veksler in classical topology. We
give some general properties of these sublocales and further examine their
relationship with both inaccessible sublocales and remote sublocales. It turns
out that a locale has all of its non-void nowhere dense sublocales maximal
nowhere dense precisely when all of its its non-void nowhere dense sublocales
are inaccessible. We show that the Booleanization of a locale is inaccessible
with respect to every dense and open sublocale. In connection to remote
sublocales, we prove that, if the supplement of an open dense sublocale $S$ is
homogeneous maximal nowhere dense, then every $S^{\\#}$-remote sublocale is
∗-remote from $S$. Every open localic map that sends dense elements to dense
elements preserves and reflects maximal nowhere dense sublocales. If such a
localic map is further injective, then it sends homogeneous maximal nowhere
dense sublocales back and forth.

###### Key words and phrases:

nowhere dense sublocale, maximal nowhere dense sublocale, homogeneous maximal
nowhere dense sublocale, inaccessible sublocale, almost inaccessible
sublocale, remote sublocale

###### 2010 Mathematics Subject Classification:

06D22

This paper is part of a Ph.D thesis written under the supervision of Prof.
Themba Dube. The author acknowledges and highly appreciates the support he
received from his supervisor, and also acknowledges funding from the National
Research Foundation of South Africa under Grant 134159

## Introduction

Veksler [14], in 1975, introduced the concepts of a maximal nowhere dense
subset and a homogeneous maximal nowhere dense subset of a Tychonoff space. He
defined a closed subset of a Tychonoff space to be maximal nowhere dense if it
is not a nowhere dense subset of any other closed nowhere dense subset of the
space, and homogeneous maximal nowhere dense provided that each of its non-
void regular-closed subsets is maximal nowhere dense in the space. These
subsets have since appeared in several articles such as [7] and [13]. In [7],
Koldunov considered connections of these subsets with $\theta$-sets, $P$-sets
and $P^{\prime}$-sets, and in [13], Veksler applied maximal nowhere dense
subsets to problems of the existence of remote points and of weak $P$-points.

Nowhere dense sublocales were initially introduced by Plewe in [12] as those
sublocales missing the Booleanization of a locale. In locale theory, these
sublocales appear in a number of papers, for instance in [4], where the author
used them to characterize submaximal locales. In [8], we used nowhere dense
sublocales to define remote sublocales which are those sublocales missing all
nowhere dense sublocales. We further verified that the Booleanization of a
locale is a largest remote sublocale. Knowing how nowhere dense sublocales
relate with the Booleanization of a locale, it is of interest to know how some
types of nowhere dense sublocales also relate with the Booleanization and to
also know how these types of nowhere dense sublocales fit in the point-free
context independently to their already documented use in classical topology.
In this paper, we introduce and study two variants of nowhere density called
maximal nowhere dense and homogeneous maximal nowhere dense sublocales. These
notions are transferred from classical topology. Here though, we generalize
the scope to arbitrary sublocales instead of closed sublocales. We provide
general results of these sublocales and show among other things that maximal
nowhere dense sublocales are precisely those sublocales that are not nowhere
in any nowhere dense sublocale. It turns out that in the category of TD-
spaces, the localic definitions of maximal nowhere dense as well as
homogeneous maximal nowhere sublocales are conservative in locales in the
sense that a subset of a TD-space is maximal nowhere dense (resp. homogeneous
maximal nowhere dense) if and only the sublocale it induces is maximal nowhere
dense (resp. homogeneous maximal nowhere dense) in the locale of opens. In a
non-Boolean strongly submaximal locale, the supplement of the Booleanization
is maximal nowhere dense. We study connnections between these sublocales and
inaccessible as well as remote sublocales. It turns out that a locale has all
of its non-void nowhere dense sublocales maximal nowhere dense precisely when
all of its its non-void nowhere dense sublocales are inaccessible.
Furthermore, the Booleanization of a locale is inaccessible with respect to a
dense and open sublocale. In relation to remote sublocales, we show that if
the supplement of an open dense sublocale $S$ is homogeneous maximal nowhere
dense, then every $S^{\\#}$-remote sublocale is ∗-remote from $S$.

This paper is organized as follows. The first section introduces the basic
background. In the second section, we define maximal nowhere dense sublocales.
We characterize these sublocales where we show that a sublocale is maximal
nowhere dense if and only if its closure is maximal nowhere dense. We use this
property of maximal nowhere density to prove among other things that the
localic definition of maximal nowhere density is conservative in locales. As
far as maximality of objects is concerned, one would expect a sublocale to be
maximal nowhere dense if it is not contained in any other nowhere dense
sublocale different from itself. This is not the concept that we use in this
paper, however we show that for a locale to have a nowhere dense sublocale
that is not contained in any other nowhere dense sublocale, it is necessary
and sufficient for the join of all of its nowhere dense sublocales to be
nowhere dense.

In the third section, we introduce the concept of a homogeneous maximal
nowhere dense sublocale and show that the introduced localic definition is
conservative in locales. We prove that homogeneous maximal nowhere density is
regular-closed hereditary.

In section four, we explore a relationship between maximal nowhere density and
inaccessibility. Veksler [14] introduced inaccessible and almost inaccessible
points of a Tychonoff space. We transfer these concepts to completely regular
locales and further define inaccessible and almost inaccessible sublocales of
arbitrary locales. We use these sublocales to characterize locales in which
all of their non-void nowhere dense sublocales are maximal nowhere dense.

The fifth section discusses a connection between maximal nowhere density and
remoteness.

The last section studies preservation and reflection of maximal nowhere
density. We prove that every open localic map that sends dense elements to
dense elements preserves and reflects maximal nowhere dense sublocales and if
such a localic map is further injective, then it sends homogeneous maximal
nowhere dense sublocales back and forth.

##  1\. Preliminaries

The book [10] is our main reference for the theory of locales and sublocales.

###  1.1. Locales

We recall that a locale $L$ is a complete lattice satisfying the following
infinite distributive law:

$a\wedge\bigvee B=\bigvee\\{a\wedge b:b\in B\\}$

for every $a\in L$, $B\subseteq L$. The top element and the bottom element of
a locale $L$ are denoted by $1_{L}$ and $0_{L}$, respectively, with subscripts
dropped if there is no possibility of confusion. An element $p\in L$ is called
a point if $p<1$ and $a\wedge b\leq p$ implies $a\leq p$ or $b\leq p$, for all
$a,b\in L$. The pseudocomplement of an element $a\in L$ is the element
$a^{\ast}=\bigvee\\{x\in L:x\wedge a=0\\}$. An element $a\in L$ is dense if
$a^{\ast}=0$, rather below $b\in L$, denoted by $a\prec b$, if $a^{*}\vee
b=1$, complemented if $a\vee a^{\ast}=1$ and completely below $b\in L$,
denoted by $a\prec\\!\\!\prec b$, if there is a sequence $(x_{q})$ of elements
of $L$ indexed by $\mathbb{Q}\cap[0,1]$ such that $a=0$, $b=1$ and $x_{q}\prec
x_{r}$ whenever $q<r$. A locale $L$ is Boolean if all of its elements are
complemented and completely regular in case $a={\bigvee}\\{x\in
L:x\prec\\!\\!\prec a\\}$ for all $a\in L$.

A localic map is an infimum-preserving function $f:L\rightarrow M$ between
locales such that its left adjoint $f^{*}$, called the associated frame
homomorphism, preserves binary meets.

###  1.2. Sublocales

A sublocale of a locale $L$ is a subset $S\subseteq L$ such that (1)
$\bigwedge A\in S$ for all $A\subseteq S$, and (2) for all $x\in L$ and $s\in
S$, $x\rightarrow s\in S$, where $\rightarrow$ is a Heyting operation on $L$
satisfying:

$a\leq b\rightarrow c\quad\text{if and only if}\quad a\wedge b\leq c$

for every $a,b,c\in L$. We denote by $\mathcal{S}(L)$ the collection of all
sublocales of a locale $L$. A sublocale $S$ of a locale $L$ is void if
$S=\mathsf{O}=\\{1\\}$, non-void provided that $S\neq\mathsf{O}$, and
complemented if it has a complement in $\mathcal{S}(L)$. It misses a sublocale
$T\subseteq L$ provided that $S\cap T=\mathsf{O}$. Every $S\in\mathcal{S}(L)$
has a supplement, denoted by $L\smallsetminus S$ or $S^{\\#}$, which is the
smallest sublocale $T$ of $L$ such that $S\vee T=L$. A sublocale $S\subseteq
L$ is linear if

$S\cap\bigvee\\{C_{i}:i\in I\\}=\bigvee\\{S\cap C_{i}:i\in I\\}$

for each family $\\{C_{i}:i\in I\\}\subseteq\mathcal{S}(L)$. Complemented
sublocales are linear. The sublocales

${\mathfrak{c}}(a)=\\{x\in L:a\leq
x\\}\quad\text{and}\quad\mathfrak{o}(a)=\\{a\rightarrow x:x\in L\\},$

of a locale $L$ are respectively the closed and open sublocales induced by
$a\in L$. They are complements of each other. A sublocale is clopen if it is
both closed and open. For any $S\in\mathcal{S}(L)$, its closure, denoted by
$\overline{S}$, is the smallest closed sublocale containing $S$ and its
interior, denoted by $\operatorname{int}(S)$, is the largest open sublocale
contained in $S$. By a dense sublocale we mean a sublocale $S$ of $L$ such
that $\overline{S}=L$. In fact, a sublocale is dense if and only if it
contains the bottom element of the locale. The sublocale
$\mathfrak{B}(L)=\\{x\rightarrow 0:x\in L\\}$ of a locale $L$ is the smallest
dense sublocale of $L$ and is referred to as the Booleanization of $L$. In
classical topology, a subset of a topological space is nowhere dense if the
interior of its closure is empty. By a nowhere dense sublocale we mean a
sublocale that misses the smallest dense sublocale and by a regular-closed
sublocale we refer to a sublocale which is the closure of some open sublocale.
We denote by $\operatorname{Nd}(L)$ the join of all nowhere dense sublocales
of a locale $L$. We shall use the prefix $S$\- for localic properties defined
on a sublocale $S$ of $L$.

By a nucleus we mean a function $\nu:L\rightarrow L$ such that (1)
$a\leq\nu(a)$, (2) $a\leq b$ $\Longrightarrow$ $\nu(a)\leq\nu(b)$, (3)
$\nu\nu(a)=\nu(a)$, and (4) $\nu(a\wedge b)=\nu(a)\wedge\nu(b)$ for every
$a,b\in L$. For each sublocale $S\subseteq L$ there is an onto frame
homomorphism $\nu_{S}:L\rightarrow S$ defined by

$\nu_{S}(a)={\bigwedge}\\{s\in S:a\leq s\\}.$

Open sublocales and closed sublocales of a sublocale $S$ of $L$ are given in
terms of nucleus as

$\mathfrak{o}_{S}(\nu_{S}(a))=S\cap\mathfrak{o}(a)\quad\text{and}\quad{\mathfrak{c}_{S}}(\nu_{S}(a))=S\cap\mathfrak{c}(a),$

respectively, for $a\in L$. For any $S\in\mathcal{S}(L)$ and $x\in L$,
$S\subseteq\mathfrak{o}(x)$ if and only if $\nu_{S}(x)=1$.

A localic map $f:L\rightarrow M$ induces the following functions: (1)
$f[-]:\mathcal{S}(L)\rightarrow\mathcal{S}(M)$ given by the set-theoretic
image of each sublocale of $L$ under $f$, and (2)
$f_{-1}[-]:\mathcal{S}(M)\rightarrow\mathcal{S}(L)$ given by

$f_{-1}[T]={\bigvee}\\{A\in\mathcal{S}(L):A\subseteq f^{-1}(T)\\}.$

For a localic map $f:L\rightarrow M$, $x\in M$ and $A\in\mathcal{S}(L)$,

$f_{-1}[\mathfrak{c}_{M}(x)]=\mathfrak{c}_{L}(f^{*}(x));\quad
f_{-1}[\mathfrak{o}_{M}(x)]=\mathfrak{o}_{L}(f^{*}(x))\quad\text{and}\quad
f[\overline{A}]\subseteq\overline{f[A]}.$

By an open localic map we refer to a localic map $f:L\rightarrow M$ such that
$f[A]$ is open for each open $A\in\mathcal{S}(L)$.

We denote by $\mathfrak{O}X$ the locale of open subsets of a topological space
$X$, and denote by $\widetilde{A}$ or $S_{A}$ a sublocale of $\mathfrak{O}X$
induced by a subset $A$ of a topological space $X$. For a topological space
$X$ and $A,B\subseteq X$, we have that
$\widetilde{A}\vee\widetilde{B}=\widetilde{A\cup B}$ and $\widetilde{A\cap
B}\subseteq\widetilde{A}\cap\widetilde{B}$.

##  2\. Maximal nowhere dense sublocales

Veksler [14] says that a closed nowhere dense subset of a Tychonoff space is
maximal nowhere dense if it is not nowhere dense in any other closed nowhere
dense subset of the space. We broaden our study to arbitrary nowhere dense
subsets of any topological space. We give the following definition.

######  Definition 2.1.

A nowhere dense subset $N$ of a topological space $X$ is maximal nowhere dense
in case there is no nowhere dense subset $K$ of $X$ such that $N$ is nowhere
dense in $K$.

We aim to introduce maximal nowhere dense sublocales such that a subset of a
topological space $X$ is maximal nowhere dense if and only if the sublocale it
induces is maximal nowhere dense in $\mathfrak{O}X$.

We transfer maximal nowhere density to locales by replacing subsets with
sublocales from Definition 2.1.

######  Definition 2.2.

Let $L$ be a locale. A nowhere dense sublocale $N$ of $L$ is maximal nowhere
dense (m.n.d) if there is no nowhere dense sublocale $S$ of $L$ such that $N$
is nowhere dense in $S$.

We consider some examples. We remind the reader that a locale $L$ is nowhere
dense as a sublocale of itself if and only if $L=\\{1\\}$.

######  Example 2.3.

(1) In a non-Boolean strongly submaximal locale $L$ (according to [4], a
locale is strongly submaximal if each of its dense sublocales is open),

$\operatorname{Nd}(L)=\bigvee\\{N\in\mathcal{S}(L):N\text{ is nowhere dense in
}L\\}$

is maximal nowhere dense. To see this, we start by showing that in a strongly
submaximal locale $L$, $\operatorname{Nd}(L)$ is nowhere dense. Indeed,
observe that in a strongly submaximal locale $L$, the dense sublocale
$\mathfrak{B}L$ is open (in particular, complemented), so

$\displaystyle\mathfrak{B}L\cap\operatorname{Nd}(L)$
$\displaystyle\quad=\quad\mathfrak{B}L\cap\bigvee\\{S:S\ \text{is nowhere
dense}\\}$ $\displaystyle\quad=\quad\bigvee\left\\{\mathfrak{B}L\cap S:S\
\text{is nowhere dense}\right\\}$
$\displaystyle\quad=\quad\bigvee\\{\mathsf{O}\\}=\mathsf{O}$

making $\operatorname{Nd}(L)$ nowhere dense.

Now, let $A\in\mathcal{S}(L)$ be nowhere dense such that
$\operatorname{Nd}(L)$ is nowhere dense in $A$. By the nature of
$\operatorname{Nd}(L)$, $A=\operatorname{Nd}(L)$, making
$\operatorname{Nd}(L)$ nowhere dense as a sublocale of itself. Hence
$\operatorname{Nd}(L)=\mathsf{O}$. This means that $\mathsf{O}$ is the only
nowhere dense sublocale of $L$, which contradicts that $L$ is non-Boolean.

An easy computation may show that
$\operatorname{Nd}(L)=L\smallsetminus\mathfrak{B}L$ whenever
$\operatorname{Nd}(L)$ is nowhere dense in a locale $L$. So, in a non-Boolean
strongly submaximal locale $L$, $L\smallsetminus\mathfrak{B}L$ is maximal
nowhere dense.

(2) $\mathsf{O}$ is not maximal nowhere dense. This follows since $\mathsf{O}$
is nowhere dense as a sublocale of itself. As a result, we get that a Boolean
locale does not have a maximal nowhere dense sublocale. This also tells us
that $N$ in Definition 2.2 cannot be open. Otherwise,
$L\smallsetminus\overline{N}$ is dense (since a sublocale $N$ is nowhere dense
if and only if $L\smallsetminus\overline{N}$ is nowhere dense, see [8]),
making $L\smallsetminus N$ dense because $L\smallsetminus\overline{N}\subseteq
L\smallsetminus N$. Since $N$ is non-void (since it is maximal nowhere dense,
making it different from $\mathsf{O}$) and open, we must have that
$(L\smallsetminus N)\cap N\neq\mathsf{O}$ which is not possible because $N$ is
complemented.

In Proposition 2.5 below, we give a characterization of maximal nowhere dense
sublocales some part of which will be used in calculations that involve
maximal nowhere dense sublocales.

Denote by $x^{*A}$ the pseudocomplement of an $x\in A$, calculated in
$A\in\mathcal{S}(L)$. Recall that for a dense $A\in\mathcal{S}(L)$, we have
the equalities

$x^{*A}=x\rightarrow_{A}0_{A}=x\rightarrow 0_{L}=x^{*}$

so that $\mathfrak{B}A=\mathfrak{B}L$.

For use below, we prove the following lemma.

######  Lemma 2.4.

A sublocale $N$ of a locale $L$ is nowhere dense in $K\in\mathcal{S}(L)$ iff
$N$ is nowhere dense in $\overline{K}$.

###### Proof.

Recall that $\mathfrak{B}S=\mathfrak{B}L$ for every dense
$S\in\mathcal{S}(L)$. Because every sublocale is dense in its closure, we have
$N\cap\mathfrak{B}K=N\cap\mathfrak{B}\overline{K}$, which implies $N$ is
nowhere dense in $K$ if and only if $N$ is nowhere dense in $\overline{K}$. ∎

We shall use $\operatorname{\mathfrak{ND}}(L)$ to denote the collection of all
nowhere dense sublocales of a locale $L$.

Recall from [8] that a sublocale is nowhere dense if and only if its closure
is nowhere dense.

######  Proposition 2.5.

Let $N$ be a nowhere dense sublocale of a locale $L$. The following statements
are equivalent.

  1. (1)

$N$ is maximal nowhere dense.

  2. (2)

There is no closed nowhere dense sublocale of $L$ having $N$ as its nowhere
dense sublocale.

  3. (3)

$\overline{N}$ is maximal nowhere dense.

  4. (4)

$\overline{N}$ is not a nowhere dense sublocale of any closed nowhere dense
sublocale of $L$.

  5. (5)

There is no dense $y\in L$ such that $y\leq\bigwedge N$ and $\left(\bigwedge
N\right)^{*\mathfrak{c}(y)}=y$.

###### Proof.

$(1)\Longrightarrow(2)$: Follows since there is no nowhere dense
(particularly, closed nowhere dense) sublocale $A$ of $L$ in which
$N\in\operatorname{\mathfrak{ND}}(A)$.

$(2)\Longrightarrow(3)$: Nowhere density of $\overline{N}$ follows since a
sublocale is nowhere dense if and only if its closure is nowhere dense, see
[8].

Now, suppose that there is a nowhere dense sublocale $K$ such that
$\overline{N}$ is nowhere dense in $K$. Since every sublocale of a nowhere
dense sublocale is nowhere dense, $N$ is nowhere dense in $K$. By Lemma 2.4,
$N$ is nowhere dense in the closed nowhere dense sublocale $\overline{K}$,
which contradicts the hypothesis in condition (2). Thus $\overline{N}$ is
maximal nowhere dense.

$(3)\Longrightarrow(4)$: Trivial.

$(4)\Longrightarrow(5)$: Let $y\in L$ be dense such that $y\leq\bigwedge N$
and $\left(\bigwedge N\right)^{*\mathfrak{c}(y)}=y$. This means that
$\bigwedge N$ is $\mathfrak{c}(y)$-dense which implies that
$\mathfrak{c}_{\mathfrak{c}(y)}\left(\bigwedge N\right)$ is
$\mathfrak{c}(y)$-nowhere dense. Because
$\mathfrak{c}_{\mathfrak{c}(y)}\left(\bigwedge
N\right)=\overline{N}\cap\mathfrak{c}(y)=\overline{N}$, we have that
$\overline{N}$ is nowhere dense in the closed sublocale $\mathfrak{c}(y)$,
which contradicts the hypothesis.

$(5)\Longrightarrow(1)$: Assume that $N$ is nowhere dense in a nowhere dense
sublocale $K$ of $L$. By Lemma 2.4, $N$ is nowhere dense in the closed nowhere
dense sublocale $\overline{K}$. Therefore $\bigwedge K$ is a dense element of
$L$ such that $\bigwedge K\leq\bigwedge N$ and $\left(\bigwedge
N\right)^{*\overline{K}}=\bigwedge K$, which is a contradiction. ∎

In terms of closed nowhere dense sublocales, we have the following result
which holds since for every $a,b\in L$,

$\displaystyle a^{*\mathfrak{c}(b)}$
$\displaystyle\quad=\quad\bigvee_{\mathfrak{c}(b)}\\{x\in\mathfrak{c}(b):a\wedge
x=0_{\mathfrak{c}(b)}=b\\}$
$\displaystyle\quad=\quad\nu_{\mathfrak{c}(b)}\left(\bigvee\\{x\in\mathfrak{c}(b):a\wedge
x=b\\}\right)$ $\displaystyle\quad=\quad b\vee\left(\bigvee\\{x\in L:a\wedge
x=b\\}\right)$ $\displaystyle\quad=\quad b\vee(a\rightarrow b)$
$\displaystyle\quad=\quad a\rightarrow b.$

######  Corollary 2.6.

Let $L$ be a locale and $\mathfrak{c}(x)\in\mathcal{S}(L)$. Then
$\mathfrak{c}(x)$ is maximal nowhere dense if and only if $x$ is dense and
there is no dense $y\in L$ such that $y\leq x$ and $x\rightarrow y=y$.

Proposition 2.5 suggests that, when doing calculations about maximal nowhere
dense sublocales, there is no loss of generality with restricting to closed
nowhere dense sublocales.

We give the following result regarding binary intersections of induced
sublocales which will be used below. Recall that for any open subset $U$ of a
topological space $X$, $\mathfrak{o}(U)=\widetilde{U}$ and
$\mathfrak{O}X\smallsetminus\widetilde{U}=\widetilde{X\smallsetminus U}$.

######  Lemma 2.7.

Let $X$ be a topological space. For any $A,B\subseteq X$ with $\widetilde{B}$
complemented in $\mathfrak{O}X$, $\widetilde{A\cap
B}=\widetilde{A}\cap\widetilde{B}$.

###### Proof.

It is clear that $\widetilde{A\cap B}\subseteq\widetilde{A}\cap\widetilde{B}$.

On the other hand, $A\cap B\subseteq A\cap B$ implies that $A\subseteq(A\cap
B)\cup(X\smallsetminus B)$. Because $\widetilde{S\cup
T}=\widetilde{S}\vee\widetilde{T}$ for all $S,T\subseteq X$, we have that
$\widetilde{A}\subseteq\widetilde{A\cap B}\vee\widetilde{X\smallsetminus
B}=\widetilde{A\cap B}\vee(\mathfrak{O}X\smallsetminus\widetilde{B})$.
Therefore

$\widetilde{A}\cap\widetilde{B}=\widetilde{A}\cap(\mathfrak{O}X\smallsetminus(\mathfrak{O}X\smallsetminus\widetilde{B}))\subseteq\widetilde{A\cap
B}$

since $\widetilde{B}$ is complemented. ∎

We show below that a subset is nowhere dense in a subspace of a TD-space
precisely when the sublocale it induces is nowhere dense in the sublocale
induced by the subspace. We shall make use of [2, Proposition 4.1.] which
states that a sublocale is nowhere dense if and only if its closure has a void
interior. We remind the reader that open sublocales of a sublocale $S$ of a
locale $L$ are the $\mathfrak{o}_{S}(a)=S\cap\mathfrak{o}(a)$ for $a\in S$.
So, in $\mathfrak{O}X$, open sublocales of $S\in\mathcal{S}(\mathfrak{O}X)$
are the $\mathfrak{o}_{S}(U)=S\cap\mathfrak{o}(U)=S\cap\widetilde{U}$ for
$U\in S$.

We give the following lemma which we shall use below.

######  Lemma 2.8.

Let $X$ be a TD-space. For each subset $N$ such that $\widetilde{N}$ is
complemented in $\mathfrak{O}X$, $N\cap A=\emptyset$ if and only if
$\widetilde{N}\cap\widetilde{A}=\mathsf{O}$.

###### Proof.

$(\Longrightarrow)$: Follows from Lemma 2.7.

$(\Longleftarrow)$: Follows from the fact that $B\subseteq C$ if and only if
$\widetilde{B}\subseteq\widetilde{C}$, for all $B,C\subseteq X$ where $X$ is a
TD-space. See [10]. ∎

######  Lemma 2.9.

Let $X$ be a TD-space and $F\subseteq X$. Then $A\subseteq F$ is $F$-nowhere
dense iff $\widetilde{A}$ is $\widetilde{F}$-nowhere dense.

###### Proof.

$(\Longrightarrow):$ Let $U\in\widetilde{F}$ be such that
$\mathfrak{o}_{\widetilde{F}}(U)\subseteq\overline{\widetilde{A}}^{\widetilde{F}}$.
Then $U\in\mathfrak{O}X$ and
$\widetilde{U}\cap\widetilde{F}\subseteq\overline{\widetilde{A}}^{\widetilde{F}}=\overline{\widetilde{A}}\cap\widetilde{F}$
so that $\widetilde{U\cap
F}\subseteq\widetilde{U}\cap\widetilde{F}\subseteq\widetilde{\overline{A}}$.
Therefore $U\cap F\subseteq\overline{A}$ making $U\cap
F\subseteq\overline{A}^{F}$. Since $A$ is $F$-nowhere dense, $U\cap
F=\emptyset$. By Lemma 2.8, $\widetilde{U}\cap\widetilde{F}=\mathsf{O}$ making
$\operatorname{int}_{\widetilde{F}}\left(\overline{\widetilde{A}}^{\widetilde{F}}\right)=\mathsf{O}$.
Thus $\widetilde{A}$ is $\widetilde{F}$-nowhere dense.

$(\Longleftarrow):$ Let $U\in\mathfrak{O}X$ be such that $U\cap
F\subseteq\overline{A}^{F}=\overline{A}\cap F$. Then $U\cap
F\subseteq\overline{A}$. Since $U$ is open, it follows from Lemma 2.7 that
$\widetilde{U}\cap\widetilde{F}\subseteq\widetilde{\overline{A}}$ which gives
$\widetilde{U}\cap\widetilde{F}\subseteq\widetilde{\overline{A}}\cap\widetilde{F}$.
Because $\widetilde{A}$ is $\widetilde{F}$-nowhere dense,
$\widetilde{U}\cap\widetilde{F}=\mathsf{O}$ implying that $U\cap F=\emptyset$.
Therefore $A$ is $F$-nowhere dense. ∎

We are now in a position to show that the notion of maximal nowhere density
introduced in Definition 2.2 is conversative in locales. Recall that for a
subset $N$ of a topological space $X$:

  1. (1)

$N$ is nowhere dense in $X$ if and only if $\widetilde{N}$ is nowhere dense in
$\mathfrak{O}X$, and

  2. (2)

If $X$ is a TD-space, then
$\overline{\widetilde{N}}=\widetilde{\overline{N}}$.

######  Proposition 2.10.

Let $X$ be a TD-space. A subset $F$ of $X$ is maximal nowhere dense in $X$ iff
$\widetilde{F}$ is maximal nowhere dense in $\mathfrak{O}X$.

###### Proof.

$(\Longrightarrow)$: Let $F\subseteq X$ be such that $\widetilde{F}$ is not
maximal nowhere dense in $\mathfrak{O}X$. It follows from Proposition 2.5 that
the sublocale $\overline{\widetilde{F}}=\widetilde{\overline{F}}$ is not
maximal nowhere dense in $\mathfrak{O}X$. Therefore, there is a closed
sublocale $\mathcal{K}$ of $\mathfrak{O}X$ such that
$\widetilde{\overline{F}}$ in nowhere dense in $\mathcal{K}$. Because
$\mathcal{S}$ is a closed sublocale of $\mathfrak{O}X$, choose a closed set
$K\subseteq X$ such that $\mathcal{K}=\mathfrak{c}(X\smallsetminus
K)=\widetilde{K}$. It follows from Lemma 2.9 that $\overline{F}$ is
$K$-nowhere dense, making $F$ not maximal nowhere dense in $X$.

$(\Longleftarrow)$: If $F$ is not maximal nowhere dense in $X$, then $F$ is
nowhere dense in some nowhere dense subset $K$ of $X$. It follows from Lemma
2.9 that $\widetilde{F}$ is nowhere dense in the nowhere dense sublocale
$\widetilde{K}$ of $\mathfrak{O}X$. Thus $\widetilde{F}$ is not maximal
nowhere dense in $\mathfrak{O}X$. ∎

We close this section by considering some results about maximal nowhere dense
sublocales.

Recall from [6] that for any sublocale $S$ of $L$,

$\operatorname{int}(S)=\mathfrak{o}\left(\bigwedge(L\smallsetminus
S)\right)=L\smallsetminus\mathfrak{c}\left(\bigwedge(L\smallsetminus
S)\right)=L\smallsetminus\overline{L\smallsetminus S}.$

This can be applied to any complemented sublocale $A$ of a locale $L$ and
arbitrary $F\in\mathcal{S}(L)$ as follows:

$\displaystyle\operatorname{int}_{A}\left(\overline{F}\cap A\right)$
$\displaystyle\quad=\quad
A\smallsetminus\overline{(A\smallsetminus\left(\overline{F}\cap
A\right))}^{A}$ $\displaystyle\quad=\quad
A\smallsetminus\left(\overline{A\smallsetminus\left(\overline{F}\cap
A\right)}\cap A\right)$ $\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\left(\overline{A\smallsetminus\left(\overline{F}\cap
A\right)}\cap A\right)\right)$ $\displaystyle\quad=\quad
A\cap\left(\left(L\smallsetminus\left(\overline{A\smallsetminus\left(\overline{F}\cap
A\right)}\right)\right)\vee(L\smallsetminus A)\right)$
$\displaystyle\quad=\quad\left(A\cap\left(L\smallsetminus\left(\overline{A\smallsetminus\left(\overline{F}\cap
A\right)}\right)\right)\right)\vee(A\cap(L\smallsetminus A))$
$\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\left(\overline{A\smallsetminus\left(\overline{F}\cap
A\right)}\right)\right)$ $\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\overline{\left(A\cap\left(L\smallsetminus\left(\overline{F}\cap
A\right)\right)\right)}\right)$ $\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\overline{\left(A\cap\left(L\smallsetminus\overline{F}\right)\right)\vee\left(A\cap\left(L\smallsetminus
A\right)\right)}\right)$ $\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\overline{\left(A\cap\left(L\smallsetminus\overline{F}\right)\right)}\right).$

######  Observation 2.11.

From the preceding paragraph we get that a sublocale $F$ of a complemented
sublocale $A$ of $L$ is $A$-nowhere dense if and only if
$A\subseteq\overline{A\cap(L\smallsetminus\overline{F})}$. This follows since

$\displaystyle F\text{ is nowhere dense in }A$
$\displaystyle\quad\Longleftrightarrow\quad\operatorname{int}_{A}\left(\overline{F}^{A}\right)=\mathsf{O}$
$\displaystyle\quad\Longleftrightarrow\quad
A\cap\left(L\smallsetminus\overline{\left(A\cap\left(L\smallsetminus\overline{F}\right)\right)}\right)=\mathsf{O}$
$\displaystyle\quad\Longleftrightarrow\quad
A\subseteq\overline{A\cap\left(L\smallsetminus\overline{F}\right)}.$

######  Proposition 2.12.

Let $L$ be a locale and $F$ a non-void nowhere dense sublocale of $L$. Then

  1. (1)

If $A\in\operatorname{\mathfrak{ND}}(L)$, $F$ is maximal nowhere dense in $L$
and $F\subseteq A$, then $A$ is maximal nowhere dense in $L$.

  2. (2)

If $F\cap(L\smallsetminus\overline{N})\neq\mathsf{O}$ for all
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus F)$, then $F$ is maximal
nowhere dense in $L$.

###### Proof.

(1) If $A\in\operatorname{\mathfrak{ND}}(L)$, $F$ is maximal nowhere dense in
$L$, $F\subseteq A$ and $N\in\operatorname{\mathfrak{ND}}(L)$ such that
$A\in\operatorname{\mathfrak{ND}}(N)$, then
$F\in\operatorname{\mathfrak{ND}}(N)$, which is a contradiction. Thus $A$ is
maximal nowhere dense in $L$.

(2) We prove this statement by contradiction: Suppose that $F$ is not maximal
nowhere dense, i.e., there is a nowhere dense sublocale $\mathfrak{c}(x)$ such
that $F\in\operatorname{\mathfrak{ND}}(\mathfrak{c}(x))$. We get that
$\mathfrak{c}(x)\cap\left(L\smallsetminus\overline{\mathfrak{c}(x)\cap(L\smallsetminus
F)}\right)=\mathsf{O}$.

Claim: $\mathfrak{c}(x)\cap(L\smallsetminus
F)\in\operatorname{\mathfrak{ND}}(L\smallsetminus F)$. To verify this, assume
that $\mathfrak{c}(x)\cap(L\smallsetminus F)$ is not $(L\smallsetminus
F)$-nowhere dense. Then there is a non-void $(L\smallsetminus F)$-open
sublocale $S$ such that $S\subseteq\overline{(L\smallsetminus
F)\cap\mathfrak{c}(x)}$. Such $S$ is of the form
$S=\mathfrak{o}(a)\cap(L\smallsetminus F)$ for some $a\in L$. Therefore

$\mathfrak{o}(a)\cap(L\smallsetminus\overline{F})\subseteq\mathfrak{o}(a)\cap(L\smallsetminus
F)\subseteq\overline{(L\smallsetminus
F)\cap\mathfrak{c}(x)}\subseteq\overline{(L\smallsetminus
F)}\cap\overline{\mathfrak{c}(x)}\subseteq\mathfrak{c}(x).$

But $\mathfrak{c}(x)\in\operatorname{\mathfrak{ND}}(L)$ and
$\mathfrak{o}(a)\cap(L\smallsetminus\overline{F})$ is open in $L$, so
$\mathfrak{c}(a)\cap(L\smallsetminus\overline{F})=\mathsf{O}$. Therefore
$\mathfrak{o}(a)\subseteq\overline{F}$. Since $\overline{F}$ is nowhere dense
in $L$, $\mathfrak{o}(a)=\mathsf{O}$ making
$\mathsf{O}=S=\mathfrak{o}(a)\cap(L\smallsetminus F)$, which is impossible.
Thus $\mathfrak{c}(x)\cap(L\smallsetminus F)$ is $(L\smallsetminus F)$-nowhere
dense.

Now, by hypothesis,
$F\cap(L\smallsetminus\overline{\mathfrak{c}(x)\cap(L\smallsetminus
F)})\neq\mathsf{O}$. Since $F\subseteq\mathfrak{c}(x)$,
$\mathfrak{c}(x)\cap(L\smallsetminus\overline{(L\smallsetminus
F)\cap\mathfrak{c}(x)})\neq\mathsf{O}$ which contradicts that
$\mathfrak{c}(x)\cap\left(L\smallsetminus\overline{\mathfrak{c}(x)\cap(L\smallsetminus
F)}\right)=\mathsf{O}$. Thus $F$ is maximal nowhere dense in $L$. ∎

######  Observation 2.13.

Using the fact that non-empty finite joins of nowhere dense sublocales are
nowhere dense, Proposition 2.12(1) tells us that any finite join of maximal
nowhere dense sublocales is maximal nowhere dense.

As far as maximality of objects with some property P is concerned, one expects
it to refer to an object $A$ with property P in which no other object with
property P, other than $A$ itself, contains $A$. So, we spare some time to
show that a nowhere dense sublocale with this maximality exists precisely when
a locale has a largest nowhere dense sublocale.

Let us call a nowhere dense sublocale $N$ of a locale $L$ strongly maximal
nowhere dense if, for any nowhere dense sublocale $A$, $N\subseteq A$ implies
$A=N$.

######  Proposition 2.14.

Let $L$ be a locale. The following statements are equivalent.

  1. (1)

$L$ has a strongly maximal nowhere dense sublocale.

  2. (2)

$\operatorname{Nd}(L)$ is nowhere dense.

If $L$ is non-Boolean, this is further equivalent to:

  1. (3)

$\operatorname{Nd}(L)$ is maximal nowhere dense.

###### Proof.

$(1)\Longrightarrow(2)$: Let $A\in\mathcal{S}(L)$ be strongly maximal nowhere
dense. We show that $\operatorname{Nd}(L)\subseteq A$ which will make
$\operatorname{Nd}(L)$ nowhere dense. Choose a nowhere dense
$N\in\mathcal{S}(L)$. Then $N\vee A$ is nowhere dense in $L$. But $A\subseteq
N\vee A$ and $A$ is strongly maximal nowhere dense, so $A=N\vee A$. Therefore
$N\subseteq N\vee A=A$. Since $N$ was arbitrary,
$\operatorname{Nd}(L)=\bigvee\\{S\in\mathcal{S}(L):S\in\operatorname{\mathfrak{ND}}(L)\\}\subseteq
A$. Thus $\operatorname{Nd}(L)\subseteq A$ implying that
$\operatorname{Nd}(L)$ is nowhere dense.

$(2)\Longrightarrow(1)$: If $\operatorname{Nd}(L)$ is nowhere dense, then
there is no other nowhere dense sublocale containing $\operatorname{Nd}(L)$
other than itself. Thus $\operatorname{Nd}(L)$ is a strongly maximal nowhere
dense sublocale of $L$.

Assume that $L$ is non-Boolean. The equivalence $(2)\Longleftrightarrow(3)$
follows since $\operatorname{Nd}(L)$ contains every nowhere dense sublocale of
$L$ and $\operatorname{Nd}(L)\neq\mathsf{O}$. ∎

##  3\. Homogeneous maximal nowhere dense sublocales

Related to maximal nowhere dense subsets are homogeneous maximal nowhere dense
subsets which were defined for spaces by Veksler in [14] as closed nowhere
dense subsets $F$ of a Tychonoff space $X$ in which each non-empty
$F$-regular-closed subset is maximal nowhere dense in $X$. In this paper, we
do not only focus on Tychonoff spaces, but all topological spaces.

We extend Veksler’s definition of a homogeneous maximal nowhere dense subset
of any topological space to locales as follows.

######  Definition 3.1.

A closed nowhere dense sublocale $N$ of a locale $L$ is homogeneous maximal
nowhere dense (h.m.n.d) if each non-void $N$-regular-closed sublocale is
maximal nowhere dense in $L$.

Without the closedness requirement in Definition 3.1, we shall call $F$ weakly
homogeneous maximal nowhere dense.

######  Observation 3.2.

We note that regular-closed sublocales of a closed nowhere dense sublocale
$\mathfrak{c}(x)$ of $L$ are of the form $\mathfrak{c}(a\rightarrow x)$ for
some $a\in L$. Indeed, $A$ is $\mathfrak{c}(x)$-regular-closed if and only if
$A=\overline{\mathfrak{o}(a)\cap\mathfrak{c}(x)}\cap\mathfrak{c}(x)$ for some
$a\in L$. We have

$\displaystyle\overline{\mathfrak{o}(a)\cap\mathfrak{c}(x)}\cap\mathfrak{c}(x)$
$\displaystyle\quad=\quad\mathfrak{c}\left(\bigwedge\left(\mathfrak{o}(a)\cap\mathfrak{c}(x)\right)\right)\cap\mathfrak{c}(x)$
$\displaystyle\quad=\quad\mathfrak{c}\left(\bigwedge\left(\mathfrak{c}_{\mathfrak{o}(a)}(\nu_{\mathfrak{o}(a)}(x))\right)\right)\cap\mathfrak{c}(x)$
$\displaystyle\quad=\quad\mathfrak{c}\left(\nu_{\mathfrak{o}(a)}(x)\right)\cap\mathfrak{c}(x)$
$\displaystyle\quad=\quad\mathfrak{c}\left(a\rightarrow
x\right)\cap\mathfrak{c}(x)$
$\displaystyle\quad=\quad\mathfrak{c}\left(a\rightarrow
x\right)\quad\text{since }x\leq a\rightarrow x.$

In light of Observation 3.2 and the characterizations of maximal nowhere
nowhere dense sublocales given in Proposition 2.5, we get the following
characterizations of homogeneous maximal nowhere dense sublocales. The proof
is straight forward and shall be omitted.

######  Proposition 3.3.

Let $L$ be a locale and $x\in L$. The following statements are equivalent.

  1. (1)

$\mathfrak{c}(x)$ is homogeneous maximal nowhere dense.

  2. (2)

For each $a\in L$, $\mathfrak{c}(a\rightarrow x)$ is maximal nowhere dense in
$L$.

  3. (3)

For each $a\in L$, there is no dense $y\in L$ such that $y\leq a\rightarrow x$
and $(a\rightarrow x)^{*\mathfrak{c}(y)}=y$.

In the next result, we show that the definition of a homogeneous maximal
nowhere dense sublocale given in Definition 3.1 is conservative in locales.
Prior to that, we give the following lemmas where the proof for Lemma 3.4(1)
is a straightforward application of Lemma 2.7 and Lemma 2.9, and shall be
omitted. We mentioned in the Preliminaries section that we shall sometimes use
the notation $S_{A}$ instead of $\widetilde{A}$ for the sublocale of
$\mathfrak{O}X$ induced by the subset $A$ of a topological space $X$.

######  Lemma 3.4.

Let $X$ be a TD-space and $A,K,F,U\subseteq X$ be such that $\widetilde{F}$ is
complemented in $\mathfrak{O}X$. Then:

  1. (1)

If $U\cap F\subseteq K$, then $U\cap F$ is $K$-nowhere dense iff
$\widetilde{U}\cap\widetilde{F}$ is $\widetilde{K}$-nowhere dense.

  2. (2)

$A=\overline{F\cap U}\cap F$ if and only if
$\widetilde{A}=\overline{\widetilde{F}\cap\widetilde{U}}\cap\widetilde{F}$.

###### Proof.

(2) We have that

$\displaystyle A=\overline{F\cap U}\cap F$
$\displaystyle\quad\Longleftrightarrow\quad\widetilde{A}=S_{\overline{F\cap
U}\cap F}$
$\displaystyle\quad\Longleftrightarrow\quad\widetilde{A}=\widetilde{\overline{F\cap
U}}\cap\widetilde{F}\text{ since }\widetilde{F}\text{ is complemented in
}\mathfrak{O}X$
$\displaystyle\quad\Longleftrightarrow\quad\widetilde{A}=\overline{\widetilde{F\cap
U}}\cap\widetilde{F}$
$\displaystyle\quad\Longleftrightarrow\quad\widetilde{A}=\overline{\widetilde{F}\cap\widetilde{U}}\cap\widetilde{F}\text{
since }\widetilde{F}\text{ is complemented in }\mathfrak{O}X$

which proves the result. ∎

Using Lemma 2.7(2), we get the following result, whose proof shall be omitted.

######  Corollary 3.5.

Let $X$ be a TD-space and $F\subseteq X$. Then a subset $A\subseteq F$ is
$F$-regular-closed iff $\widetilde{A}$ is $\widetilde{F}$-regular-closed.

######  Proposition 3.6.

Let $X$ be TD-space. A closed set $F\subseteq X$ is homogeneous maximal
nowhere dense iff $\widetilde{F}\subseteq\mathfrak{O}X$ is homogeneous maximal
nowhere dense.

###### Proof.

Follows from Proposition 2.10, Lemma 3.4(1) and Corollary 3.5. ∎

The following result tells us that homogeneous maximal nowhere density is
regular-closed hereditary.

######  Proposition 3.7.

Let $L$ be a locale and $F$ be a closed nowhere dense sublocale of $L$. If $F$
is homogeneous maximal nowhere dense and $A$ is a non-void $F$-regular-closed
sublocale, then $A$ is homogeneous maximal nowhere dense.

###### Proof.

Let $N$ be a non-void regular-closed sublocale of $A$ and suppose that there
is $B\in\operatorname{\mathfrak{ND}}(L)$ such that
$N\in\operatorname{\mathfrak{ND}}(B)$. Because $A$ is $F$-regular-closed and
$N$ is $A$-regular-closed, $A=\overline{\mathfrak{o}(x)\cap F}\cap F$ and
$N=\overline{\mathfrak{o}(y)\cap A}\cap A$ for some $x,y\in L$. Since both $F$
and $A$ are closed, $A=\overline{\mathfrak{o}(x)\cap F}$ and
$N=\overline{\mathfrak{o}(y)\cap A}$ so that
$N=\overline{\mathfrak{o}(y)\cap\overline{\mathfrak{o}(x)\cap F}}$. Therefore

$\overline{\mathfrak{o}(y)\cap\mathfrak{o}(x)\cap
F}=\overline{\mathfrak{o}(y\wedge x)\cap F}\subseteq N.$

The sublocale $\overline{\mathfrak{o}(y)\cap\mathfrak{o}(x)\cap
F}\neq\mathsf{O}$, otherwise $\mathfrak{o}(y)\cap\mathfrak{o}(x)\cap
F=\mathsf{O}$ making $\mathfrak{o}(y)\cap\overline{\mathfrak{o}(x)\cap
F}=\mathsf{O}$ which is not possible. Since $N$ is nowhere dense in $B$, we
get that $\overline{\mathfrak{o}(y\wedge x)\cap F}$ is nowhere dense in $B$
making $\overline{\mathfrak{o}(y\wedge x)\cap F}\cap
F\in\operatorname{\mathfrak{ND}}(B)$. This is not possible because
$\overline{\mathfrak{o}(y\wedge x)\cap F}\cap F$ is non-void and regular-
closed in $F$ which must be maximal nowhere dense in $L$. Thus $A$ is
homogeneous maximal nowhere dense. ∎

We close this section by considering a relationship between maximal nowhere
dense sublocales and (strongly) homogeneous maximal nowhere dense sublocales.

######  Proposition 3.8.

Every (strongly) homogeneous maximal nowhere dense sublocale is maximal
nowhere dense.

###### Proof.

Follows since every locale is a regular-closed sublocale of itself. ∎

##  4\. Maximal nowhere density and Inaccessibility

The aim of this section is to explore a relationship between the variations of
nowhere dense sublocales discussed in the previous two sections (which are
maximal nowhere dense sublocales and homogeneous maximal nowhere dense
sublocales) and sublocales called inaccessible sublocales and almost
inaccessible sublocales.

En route to introducing inaccessible and almost inaccessible sublocales, we
consider Veksler’s notions of an inaccessible point and an almost inaccessible
point of a Tychonoff space which he introduced in [14]. He calls a point $x\in
E\subseteq X$, where $X$ is a Tychonoff space, $E$-inaccessible (resp. almost
$E$-inaccessible) if $x\notin\overline{N}$ (resp.
$x\notin\operatorname{int}_{E}(\overline{N}\cap E)$) for all $(X\smallsetminus
E)$-closed nowhere dense $N$. We shall transfer these notions to locales.

Our journey to introducing localic notions of inaccessible and almost
inaccessible points will start with inaccessible points and end with almost
inaccessible points.

In a Tychonoff space $X$, we have that $x\notin\overline{N}$ if and only if
$\overline{\\{x\\}}\cap\overline{N}=\emptyset$ if and only if
$(X\smallsetminus\overline{\\{x\\}})\cup(X\smallsetminus\overline{N})=X$ for
every $x\in X$, $N\subseteq X$. So the definition of an $E$-inaccessible point
$x\in E\subseteq X$ is equivalent to:

> _$(X\smallsetminus\overline{\\{x\\}})\cup(X\smallsetminus\overline{N})=X$
> for all $(X\smallsetminus E)$-closed nowhere dense $N$._

Recall that $X\smallsetminus\overline{A}=0_{\widetilde{A}}$ for any subset $A$
of a space $X$. This and the preceding paragraph motivate the following
localic definition of an inaccessible point.

######  Definition 4.1.

A point $p$ of a sublocale $S$ of a completely regular locale $L$ is
$S$-inaccessible if for each $(L\smallsetminus S)$-closed nowhere dense
sublocale $N$, $0_{N}\vee p=1$, where the join is calculated in $L$.

Recall from [1, Lemma 3.2.1] that for a subset $A$ of a TD-space $X$,

$\bigvee\\{\\{X\smallsetminus\overline{\\{x\\}},X\\}:x\notin
A\\}=\widetilde{X\smallsetminus A}$

is the supplement of $\widetilde{A}$, i.e., $\widetilde{X\smallsetminus
A}=\widetilde{X}\smallsetminus\widetilde{A}$.

According to [5], a regular locale is T1 in the sense that every point is a
maximal element. Hence a point $p$ of a regular locale $L$ has a property that
$a\vee p=1$ if and only if $a\nleq p$ for every $a\in L$.

In what follows, we show that a point $x$ of a Tychonoff space $X$ is
inaccessible if and only if $\widetilde{x}$ is inaccessible. We shall need the
following lemma. We shall make use of the fact that
$\widetilde{A}=\\{\operatorname{int}((X\smallsetminus A)\cup
G):G\in\mathfrak{O}X\\}$ for every subset $A$ of a topological space $X$.

######  Lemma 4.2.

Let $X$ be a topological space, $F$ a closed subset of $X$ and $A\subseteq X$.
Then
$\bigwedge(\widetilde{F}\cap\widetilde{A})=\operatorname{int}((X\smallsetminus
A)\cup(X\smallsetminus F))$

###### Proof.

We have that $X\smallsetminus F\subseteq(X\smallsetminus
A)\cup(X\smallsetminus F)$, making

$X\smallsetminus F=\operatorname{int}(X\smallsetminus
F)\subseteq\operatorname{int}((X\smallsetminus A)\cup(X\smallsetminus F))$

so that $\operatorname{int}((X\smallsetminus A)\cup(X\smallsetminus
F))\in\widetilde{F}$. Also, since $X\smallsetminus F$ is open,
$\operatorname{int}((X\smallsetminus A)\cup(X\smallsetminus
F))\in\widetilde{A}.$ Therefore $\operatorname{int}((X\smallsetminus
A)\cup(X\smallsetminus F))\in\widetilde{F}\cap\widetilde{A}$, making
$\bigwedge(\widetilde{F}\cap\widetilde{A})\leq\operatorname{int}((X\smallsetminus
A)\cup(X\smallsetminus F))$.

On the other hand, let $V\in\widetilde{F}\cap\widetilde{A}$. Then
$X\smallsetminus F\subseteq V$ and $V=\operatorname{int}((X\smallsetminus
A)\cup G)$ for some $G\in\mathfrak{O}X$. We get that

$\displaystyle\operatorname{int}((X\smallsetminus A)\cup(X\smallsetminus F))$
$\displaystyle\quad\subseteq\quad\operatorname{int}((X\smallsetminus A)\cup
V)$ $\displaystyle\quad=\quad\operatorname{int}\left((X\smallsetminus
A)\cup\operatorname{int}((X\smallsetminus A)\cup G)\right)$
$\displaystyle\quad\subseteq\quad\operatorname{int}((X\smallsetminus A)\cup
G)$ $\displaystyle\quad=\quad V.$

Therefore $\operatorname{int}((X\smallsetminus A)\cup(X\smallsetminus
F))\leq\bigwedge(\widetilde{F}\cap\widetilde{A})$ and hence
$\bigwedge(\widetilde{F}\cap\widetilde{A})=\operatorname{int}((X\smallsetminus
A)\cup(X\smallsetminus F))$. ∎

######  Proposition 4.3.

Let $X$ be a Tychonoff space and $x\in E\subseteq X$. Then $x$ is
$E$-inaccessible iff $\widetilde{x}$ is $\widetilde{E}$-inaccessible.

###### Proof.

$(\Longrightarrow):$ Let $K$ be an
$(\widetilde{X}\smallsetminus\widetilde{E})$-closed nowhere dense sublocale.
Choose a closed subset $F$ of $X$ such that
$K=\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})$. Since
$\widetilde{X}\smallsetminus\widetilde{E}=\widetilde{X\smallsetminus E}$,
$K=\widetilde{F}\cap\widetilde{X\smallsetminus E}\supseteq
S_{\left(F\cap(X\smallsetminus E)\right)}$. It follows from Lemma 2.9 that
$F\cap(X\smallsetminus E)$ is an $(X\smallsetminus E)$-closed nowhere dense
subset. Since $x$ is $E$-inaccessible,

$\left(X\smallsetminus\overline{\\{x\\}}\right)\cup\left(X\smallsetminus\overline{F\cap(X\smallsetminus
E)}\right)=X.$

Because $\widetilde{x}=X\smallsetminus\overline{\\{x\\}}$ is a point and every
completely regular locale is regular and hence T1, it follows that
$X\smallsetminus\overline{F\cap(X\smallsetminus E)}\nleq\widetilde{x}$. Since
$X\smallsetminus\overline{F\cap(X\smallsetminus
E)}=\operatorname{int}(E\cup(X\smallsetminus F))$ and
$\operatorname{int}(E\cup(X\smallsetminus
F))=\bigwedge(\widetilde{F}\cap\widetilde{X\smallsetminus E})$ by Lemma 4.2,
$\bigwedge(\widetilde{F}\cap\widetilde{X\smallsetminus E})\nleq\widetilde{x}$
so that

$\bigwedge(\widetilde{F}\cap\widetilde{X\smallsetminus
E})\vee\widetilde{x}=0_{K}\vee\widetilde{x}=1_{\mathfrak{O}X}$

because $\mathfrak{O}X$ is T1. Thus $\widetilde{x}$ is a
$\widetilde{E}$-inaccessible point.

$(\Longleftarrow):$ Let $C$ be an $(X\smallsetminus E)$-closed nowhere dense
subset. Set $C=F\cap(X\smallsetminus E)$ for some closed $F\subseteq X$. If
follows from Proposition 3.4 that $\widetilde{F}\cap\widetilde{X\smallsetminus
E}$ is $(\widetilde{X\smallsetminus E})$-closed nowhere dense. But
$\widetilde{X}\smallsetminus\widetilde{E}=\widetilde{X\smallsetminus E}$, so
the $(\widetilde{X}\smallsetminus\widetilde{E})$-closed sublocale
$\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})$ is
$(\widetilde{X}\smallsetminus\widetilde{E})$-nowhere dense. By hypothesis,
$\widetilde{x}\vee
0_{\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})}=1_{\mathfrak{O}X}$.
But

$0_{\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})}=\bigwedge\left(\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})\right)=\operatorname{int}(E\cup(X\smallsetminus
F))=X\smallsetminus\overline{F\cap(X\smallsetminus E)}$

where the second equality follows from Lemma 4.2. Therefore
$\left(X\smallsetminus\overline{\\{x\\}}\right)\cup\left(X\smallsetminus\overline{F\cap(X\smallsetminus
E)}\right)=X$ which implies that

$\emptyset=\overline{\\{x\\}}\cap\overline{F\cap(X\smallsetminus
E)}=\\{x\\}\cap\overline{F\cap(X\smallsetminus E)}.$

Thus $x\notin\overline{F\cap(X\smallsetminus E)}=\overline{C}$, making $x$
$E$-inaccessible. ∎

To transfer almost inaccessibility to locales, we recall that for $x\in
E\subseteq X$,

$\displaystyle x\notin\operatorname{int}_{E}(\overline{N}\cap E)$
$\displaystyle\quad\Longleftrightarrow\quad x\notin
E\cap\left(X\smallsetminus\overline{E\cap(X\smallsetminus\overline{N})}\right)$
$\displaystyle\quad\Longleftrightarrow\quad
x\in\overline{E\cap(X\smallsetminus\overline{N})}.$

The above equivalence motivates the following localic definition of an almost
inaccessible point.

######  Definition 4.4.

A point $p$ of a sublocale $S$ of a completely regular locale $L$ is almost
$S$\- inaccessible if for each $(L\smallsetminus S)$-closed nowhere dense
sublocale $N$,
$p\in\operatorname{cl}_{S}\left(S\cap\left(L\smallsetminus\overline{N}\right)\right)$.

In what follows we prove that a point $x$ of $X$ is almost inaccessible
precisely when $\widetilde{x}$ is almost inaccessible.

######  Proposition 4.5.

Let $X$ be a Tychonoff space. A point $x$ of a subset $E$ of $X$ is almost
$E$-inaccessible iff $\widetilde{x}$ is almost $\widetilde{E}$-inaccessible.

###### Proof.

$(\Longrightarrow):$ It is clear that $\widetilde{x}$ is a point belonging to
the closed sublocale $\widetilde{\overline{E}}=\overline{\widetilde{E}}$.
Choose an $(\widetilde{X}\smallsetminus\widetilde{E})$-closed nowhere dense
$K$. Then $K=\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})$ for
some closed $F\subseteq X$. Since
$\widetilde{X}\smallsetminus\widetilde{E}=\widetilde{X\smallsetminus E}$,
$K=\widetilde{F}\cap\widetilde{X\smallsetminus E}$. It follows from
Proposition 3.4 that $F\cap(X\smallsetminus E)$ is $(X\smallsetminus
E)$-closed nowhere dense. Therefore
$x\notin\operatorname{int}_{E}(\overline{F\cap(X\smallsetminus E)}\cap E)$
which means
$x\in\overline{E\cap(X\smallsetminus\overline{F\cap(X\smallsetminus E)})}$. We
show that
$\widetilde{x}\in\overline{\widetilde{E}\cap(\widetilde{X}\smallsetminus\overline{K})}$.
Let $U\in\mathfrak{O}X$ be such that
$\widetilde{E}\cap(\widetilde{X}\smallsetminus\overline{K})\subseteq\mathfrak{c}(U)$.
Then
$\widetilde{E}\cap\left(\widetilde{X}\smallsetminus\overline{\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})}\right)\subseteq\mathfrak{c}(U)$,
i.e.,
$\widetilde{E}\cap\left(\widetilde{X}\smallsetminus\overline{\widetilde{F}\cap\widetilde{X\smallsetminus
E}}\right)\subseteq\mathfrak{c}(U)$. We get that

$\displaystyle\widetilde{E}$
$\displaystyle\quad\subseteq\quad\mathfrak{c}(U)\vee\overline{\widetilde{F}\cap\widetilde{X\smallsetminus
E}}$
$\displaystyle\quad=\quad\mathfrak{c}(U)\vee\mathfrak{c}\left(\bigwedge\left(\widetilde{F}\cap(\widetilde{X\smallsetminus
E})\right)\right)$
$\displaystyle\quad=\quad\mathfrak{c}(U)\vee\mathfrak{c}(X\smallsetminus\overline{F\cap(X\smallsetminus
E)})\quad\text{since
}\bigwedge\left(\widetilde{F}\cap(\widetilde{X\smallsetminus
E})\right)=X\smallsetminus\overline{F\cap(X\smallsetminus E)}$
$\displaystyle\quad=\quad\mathfrak{c}\left(U\cap(X\smallsetminus\overline{F\cap(X\smallsetminus
E)})\right)$ $\displaystyle\quad=\quad S_{\left((X\smallsetminus
U)\cup\overline{F\cap(X\smallsetminus E)}\right)}\quad\text{by
\cite[cite]{[\@@bibref{}{N}{}{}, Lemma 2.10.]}}.$

Therefore $E\subseteq(X\smallsetminus U)\cup\overline{F\cap(X\smallsetminus
E)}$ so that $E\cap(X\smallsetminus\overline{F\cap(X\smallsetminus
E)})\subseteq X\smallsetminus U$. Because $U$ is open, $X\smallsetminus U$ is
closed, making
$x\in\overline{E\cap(X\smallsetminus\overline{F\cap(X\smallsetminus
E)})}\subseteq X\smallsetminus U$. Therefore
$\widetilde{x}\in\widetilde{X\smallsetminus U}=\mathfrak{c}(U)$ implying that
$\widetilde{x}\in\overline{\widetilde{E}\cap(\widetilde{X}\smallsetminus\overline{K})}$.
Thus

$\widetilde{x}\in\overline{\widetilde{E}\cap(\widetilde{X}\smallsetminus\overline{K})}\cap\widetilde{E}=cl_{\widetilde{E}}\left(\widetilde{E}\cap(\widetilde{X}\smallsetminus\overline{K})\right)$

which means that $\widetilde{x}$ is almost $\widetilde{E}$-inaccessible.

$(\Longleftarrow):$ Let $N$ be an $(X\smallsetminus E)$-closed nowhere dense
subset and set $N=F\cap(X\smallsetminus E)$ for some closed $F\subseteq X$. It
follows from Proposition 3.4 that $\widetilde{F}\cap\widetilde{X\smallsetminus
E}=\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})$ is
$(\widetilde{X\smallsetminus
E}=\widetilde{X}\smallsetminus\widetilde{E})$-closed nowhere dense. Therefore

$\widetilde{x}\in\operatorname{cl}_{\widetilde{E}}\left(\widetilde{E}\cap\left(\widetilde{X}\smallsetminus\overline{\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})}\right)\right).$

We show that $x\in\overline{E\cap(X\smallsetminus\overline{N})}$. Let $K$ be a
closed set such that $E\cap(X\smallsetminus\overline{N})\subseteq K$. Then
$E\subseteq\overline{N}\cup K$ so that

$\widetilde{E}\subseteq\widetilde{\overline{N}}\vee\widetilde{K}=\overline{\widetilde{N}}\vee\widetilde{K}=\overline{S_{(F\cap(X\smallsetminus
E))}}\vee\widetilde{K}\subseteq\overline{\widetilde{F}\cap\left(\widetilde{X}\smallsetminus\widetilde{E}\right)}\vee\widetilde{K}.$

Therefore

$\widetilde{E}\cap\left(\widetilde{X}\smallsetminus\overline{\widetilde{F}\cap\left(\widetilde{X}\smallsetminus\widetilde{E}\right)}\right)\subseteq\widetilde{K}.$

Because $\widetilde{K}$ is a closed sublocale,

$\operatorname{cl}_{\widetilde{E}}\left(\widetilde{E}\cap\left(\widetilde{X}\smallsetminus\overline{\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})}\right)\right)=\widetilde{E}\cap\overline{\left(\widetilde{E}\cap\left(\widetilde{X}\smallsetminus\overline{\widetilde{F}\cap(\widetilde{X}\smallsetminus\widetilde{E})}\right)\right)}\subseteq\widetilde{K}$

so that $\widetilde{x}\in\widetilde{K}$. Therefore $x\in K$. Thus
$x\in\overline{E\cap(X\smallsetminus\overline{N})}$ which implies
$x\in\operatorname{cl}_{E}\left(E\cap(X\smallsetminus\overline{N})\right)$. As
a result, $x\notin\operatorname{int}_{E}\left(E\cap\overline{N}\right)$. Hence
$x$ is almost $E$-inaccessible. ∎

In terms of sublocales, we define inaccessibility and almost inaccessibility
on arbitrary locales. We give the following definition.

######  Definition 4.6.

Let $S$ be a sublocale of $L$. A sublocale $T\in\mathcal{S}(S)$ is
$S$-inaccessible (resp. almost $S$-inaccessible) if for all $(L\smallsetminus
S)$-nowhere dense sublocale $N$, $T\cap\overline{N}=\mathsf{O}$ (resp.
$T\subseteq\operatorname{cl}_{S}\left[S\cap\left(L\smallsetminus\overline{N}\right)\right]$).

We introduce the following notations for any locale $L$ and
$S\in\mathcal{S}(L)$:

$\mathcal{S}_{\operatorname{Inac}}(S)=\\{A\in\mathcal{S}(L):A\ \text{is}\
S\text{-inaccessible}\\},$

and

$\mathcal{S}_{\operatorname{Ainac}}(S)=\\{A\in\mathcal{S}(L):A\ \text{is}\
\text{almost}\ S\text{-inaccessible}\\}.$

We shall drop prefix $S$\- if the sublocale is clear from the context.

In what follows, we characterize inaccessible sublocales. The proof is similar
to that of [8, Proposition 3.10.] and shall be omitted.

######  Proposition 4.7.

The following are equivalent for a sublocale of $S$ of $L$.

  1. (1)

$T\in\mathcal{S}(S)$ is $S$-inaccessible.

  2. (2)

$T\cap\mathfrak{c}(x)=\mathsf{O}$ for each $(L\smallsetminus S)$-dense $x$.

  3. (3)

$T\subseteq\mathfrak{o}(x)$ for every $(L\smallsetminus S)$-dense $x$.

  4. (4)

$\nu_{T}(x)=1$ for each $(L\smallsetminus S)$-dense $x$.

For sublocales $F$ and $A$ of a locale $L$, we have that

$F\subseteq\operatorname{cl}_{F}(F\cap
A)\;\Longleftrightarrow\;F=\operatorname{cl}_{F}(F\cap A)\;\Longleftrightarrow
F\cap A\text{ is }F\text{-dense}.$

As a result of this, we have the following observation regarding almost
inaccessible sublocales.

######  Observation 4.8.

Let $F$ be a sublocale of a locale $L$. The following statements are
equivalent.

  1. (1)

$F\in\mathcal{S}_{\operatorname{Ainac}}(F)$.

  2. (2)

$F=\operatorname{cl}_{F}(F\cap(L\smallsetminus\overline{N}))$ for every
$(L\smallsetminus F)$-nowhere dense $N$.

  3. (3)

$F\cap(L\smallsetminus\overline{N})$ is $F$-dense for every $(L\smallsetminus
F)$-nowhere dense $N$.

We give the following lemma which we shall use below.

######  Lemma 4.9.

Let $L$ be a locale, $N\in\mathcal{S}(L)$, $S$ a complemented sublocale of $L$
and $T\in\mathcal{S}(S)$. Then
$T\subseteq\operatorname{cl}_{S}(S\cap\left(L\smallsetminus\overline{N}\right))$
iff $T\cap\operatorname{int}_{S}\left(S\cap\overline{N}\right)=\mathsf{O}.$

###### Proof.

We have that

$\displaystyle
T\subseteq\operatorname{cl}_{S}(S\cap\left(L\smallsetminus\overline{N}\right))$
$\displaystyle\quad\Longleftrightarrow\quad
T\subseteq\overline{S\cap\left(L\smallsetminus\overline{N}\right)}$
$\displaystyle\quad\Longleftrightarrow\quad
T\cap\left(L\smallsetminus\overline{S\cap(L\smallsetminus\overline{N})}\right)=\mathsf{O}$
$\displaystyle\quad\Longleftrightarrow\quad T\cap
S\cap\left(L\smallsetminus\overline{S\cap(L\smallsetminus\overline{N})}\right)=\mathsf{O}$
$\displaystyle\quad\Longleftrightarrow\quad
T\cap\operatorname{int}_{S}\left(S\cap\overline{N}\right)=\mathsf{O}$

which proves the result. ∎

The preceding lemma leads us to the following characterization of almost
inaccessible sublocales of complemented sublocales. We only prove the
equivalences $(2)\Longleftrightarrow(3)$.

######  Proposition 4.10.

The following are equivalent for a complemented sublocale $S$ of a locale $L$
and $T\in\mathcal{S}(S)$.

  1. (1)

$T$ is almost $S$-inaccessible.

  2. (2)

$T\cap\operatorname{int}_{S}(S\cap\overline{N})=\mathsf{O}$ for each
$(L\smallsetminus S)$-nowhere dense sublocale $N$.

In particular, if $S$ is closed, this is further equivalent to:

  1. 3.

$a\rightarrow\left(\bigwedge S\right)\leq\bigwedge T$ for every
$(L\smallsetminus S)$-dense $a$.

###### Proof.

$(2)\Longleftrightarrow(3)$: Let $a$ be $(L\smallsetminus S)$-dense. Then
$\mathfrak{c}_{(L\smallsetminus S)}(a)$ is $(L\smallsetminus S)$-nowhere
dense. By $(2)$,

$\displaystyle
T\cap\operatorname{int}_{S}(S\cap\overline{\mathfrak{c}_{(L\smallsetminus
S)}(a)})=\mathsf{O}$ $\displaystyle\quad\Longleftrightarrow\quad
T\subseteq\operatorname{cl}_{S}\left(S\cap\left(L\smallsetminus\overline{\mathfrak{c}_{(L\smallsetminus
S)}(a)}\right)\right)\text{ since }S\text{ is complemented}$
$\displaystyle\quad\Longleftrightarrow\quad
T\subseteq\overline{S\cap\left(L\smallsetminus\overline{\mathfrak{c}_{(L\smallsetminus
S)}(a)}\right)}$ $\displaystyle\quad\Longleftrightarrow\quad
T\subseteq\overline{S\cap\left(L\smallsetminus\mathfrak{c}(a)\right)}\quad\text{since
}\overline{\mathfrak{c}_{(L\smallsetminus S)}(a)}=\mathfrak{c}(a)$
$\displaystyle\quad\Longleftrightarrow\quad
T\subseteq\mathfrak{c}\left(\bigwedge\left(S\cap\mathfrak{o}(a)\right)\right)$
$\displaystyle\quad\Longleftrightarrow\quad
T\subseteq\mathfrak{c}\left(a\rightarrow\left(\bigwedge
S\right)\right)\quad\text{since }S\text{ is closed}$
$\displaystyle\quad\Longleftrightarrow\quad\overline{T}\subseteq\mathfrak{c}\left(a\rightarrow\left(\bigwedge
S\right)\right)$ $\displaystyle\quad\Longleftrightarrow\quad
a\rightarrow\left(\bigwedge S\right)\leq\bigwedge T.$

Starting the above argument with an $(L\smallsetminus S)$-nowhere dense $N$
and using the fact that $N$ is $(L\smallsetminus S)$-nowhere dense if and only
if $\bigwedge N$ is $(L\smallsetminus S)$-dense, gives the desired
equivalence. ∎

Next, we collect into one proposition some results about inaccessible
sublocales and almost inaccessible sublocales.

######  Proposition 4.11.

Let $L$ be a locale and $S\in\mathcal{S}(L)$.

  1. (1)

Every $S$-inaccessible sublocale is almost $S$-inaccessible.

  2. (2)

If $A\in\mathcal{S}(L)$ is $S$-inaccessible (resp. almost $S$-inaccessible)
and $\mathcal{S}(L)\ni B\subseteq A$, then $B$ is $S$-inaccessible (resp.
almost $S$-inaccessible).

  3. (3)

If $S$ is open in $L$, then $S$ is $S$-inaccessible.

  4. (4)

$L$ is $L$-inaccessible and hence by (2) every sublocale of $L$ is
$L$-inaccessible.

  5. (5)

If $S$ is complemented, then every sublocale $T$ of $L\smallsetminus S$ which
is open in $L$ is $(L\smallsetminus S)$-inaccessible.

  6. (6)

A join of $S$-inaccessible (resp. almost $S$-inaccessible) sublocales is
$S$-inaccessible (resp. almost $S$-inaccessible).

###### Proof.

(1) Let $T\in\mathcal{S}(S)$ be such that $T$ is $S$-inaccessible. Then
$T\cap\overline{N}=\mathsf{O}$ for all
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus S)$, which implies that
$T\subseteq L\smallsetminus\overline{N}$. Therefore $T\subseteq
S\cap(L\smallsetminus\overline{N})\subseteq\operatorname{cl}_{S}(S\cap(L\smallsetminus\overline{N}))$
which proves the result.

(2) Straightforward.

(3) Assume that $S$ is open in $L$ and choose
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus S)$. It is clear that
$\overline{N}\subseteq L\smallsetminus S$ since $L\smallsetminus S$ is closed.
Because $S$ is complemented, we have that $S\cap(L\smallsetminus
S)=\mathsf{O}$ so that $S\cap\overline{N}=\mathsf{O}$. Thus
$S\in\mathcal{S}_{\operatorname{Inac}}(S)$.

(4) Because $L$ is open as a sublocale of itself, it follows from (3) that $L$
is $L$-inaccessible. Therefore, by (2), every sublocale of $L$ is
$L$-inaccessible.

(5) Let $T$ be a sublocale of $L\smallsetminus S$ which is open in $L$. We
must show that $T\cap\overline{N}=\mathsf{O}$ for every
$(L\smallsetminus(L\smallsetminus S))$-nowhere dense $N$, i.e., for every
$S$-nowhere dense $N$. Since $T\subseteq L\smallsetminus S$, $T\cap
S=\mathsf{O}$ because $S$ is complemented. So, for any $S$-nowhere dense $N$,
$T\cap N=\mathsf{O}$. But $T$ is open in $L$ so
$T\cap\overline{N}=\mathsf{O}$.

(6) We only verify the case of $S$-inaccessible. Let
$U_{i}\in\mathcal{S}_{\operatorname{Inac}}(S)$ (for $i\in I$) and choose
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus S)$. Since $\overline{N}$ is
complemented, $\overline{N}\cap\bigvee U_{i}=\bigvee\left(\overline{N}\cap
U_{i}\right)=\bigvee\\{\mathsf{O}\\}=\mathsf{O}$. Thus $\bigvee
U_{i}\in\mathcal{S}_{\operatorname{Inac}}(S)$. ∎

######  Remark 4.12.

We note from Proposition 4.11(4) that since every sublocale $S$ of a locale
$L$ is a locale in its own right, it is therefore $S$-inaccessible as a
sublocale of itself. However, in this paper, the notion
$S\in\mathcal{S}_{\operatorname{Inac}}(S)$ for $S\in\mathcal{S}(L)$, read as
$S$ is inaccessible as a sublocale of $L$ with respect to itself, shall mean
$S\cap\overline{N}=\mathsf{O}$ for every $(L\smallsetminus S)$-nowhere dense
sublocale $N$. This also applies to
$S\in\mathcal{S}_{\operatorname{Ainac}}(S)$.

We note the following example.

######  Example 4.13.

In a completely regular locale $L$, a point $p$ of $L$ is
$\mathfrak{c}(p)$-inaccessible (resp. almost $\mathfrak{c}(p)$-inaccessible)
if and only if $\mathfrak{c}(p)$ is $\mathfrak{c}(p)$-inaccessible (resp.
almost $\mathfrak{c}(p)$-inaccessible).

We give the following theorem in which some of its statements prepare us for a
relationship between maximal nowhere density and remoteness in Proposition
5.6.

######  Theorem 4.14.

Let $L$ be a locale and $F$ be a non-void and closed nowhere dense sublocale
of $L$. Then each of the following statements holds.

  1. (1)

If $F\in\mathcal{S}_{\operatorname{Ainac}}(F)$, then $F$ is maximal nowhere
dense in $L$.

  2. (2)

If $L$ is compact, then $F$ is maximal nowhere dense in $L$ implies that there
is $x\in F$ such that $x\notin\operatorname{int}_{F}(\overline{N}\cap F)$ for
every $(L\smallsetminus F)$-nowhere dense $N$.

  3. (3)

$F$ is homogeneous maximal nowhere dense implies that every sublocale of $F$
is almost $F$-inaccessible.

###### Proof.

(1) Let $F=\mathfrak{c}(b)$ for some $b\in L$ and choose
$\mathfrak{c}(c)\in\operatorname{\mathfrak{ND}}(L)$ such that
$F\subseteq\mathfrak{c}(c)$. We show that $F$ is not nowhere dense in
$\mathfrak{c}(c)$. Observe that
$\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b)\in\operatorname{\mathfrak{ND}}(\mathfrak{o}(b))$.
Indeed, if

$\mathfrak{o}(x)\cap\mathfrak{o}(b)\subseteq\overline{\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b)}^{\mathfrak{o}(b)}=\overline{\mathfrak{c}(c)\cap\mathfrak{o}(b)}\cap\mathfrak{o}(b)=\mathfrak{c}(c)\cap\mathfrak{o}(b),$

then $\mathfrak{o}(x)\cap\mathfrak{o}(b)=\mathfrak{o}(x\wedge
b)\subseteq\mathfrak{c}(c)$. But
$\mathfrak{c}(c)\in\operatorname{\mathfrak{ND}}(L)$, so $\mathfrak{o}(x\wedge
b)=\mathsf{O}$. Thus
$\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b)\in\operatorname{\mathfrak{ND}}(\mathfrak{o}(b))$.

Since $\mathfrak{c}(b)\in\mathcal{S}_{\operatorname{Ainac}}(\mathfrak{c}(b))$,
we have that

$\mathfrak{c}(b)=\operatorname{cl}_{\mathfrak{c}(b)}\left(\mathfrak{c}(b)\cap\left(L\smallsetminus\overline{(\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b))}\right)\right)=\operatorname{cl}_{\mathfrak{c}(b)}\left(\mathfrak{c}(b)\smallsetminus\overline{(\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b))}\right).$

Because $\mathfrak{c}(b)$ is non-void,
$\mathfrak{c}(b)\smallsetminus\overline{(\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b))}\neq\mathsf{O}$.
Since $\mathfrak{c}(b)\subseteq\mathfrak{c}(c)$,
$\mathfrak{c}(c)\smallsetminus\overline{(\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b))}\neq\mathsf{O}$.

We must have that
$\mathfrak{c}(b)\notin\operatorname{\mathfrak{ND}}(\mathfrak{c}(c))$,
otherwise

$\displaystyle\mathsf{O}$
$\displaystyle\quad=\quad\operatorname{int}_{\mathfrak{c}(c)}\left(\overline{\mathfrak{c}(b)}^{\mathfrak{c}(c)}\right)$
$\displaystyle\quad=\quad\mathfrak{c}(c)\cap\left(L\smallsetminus\overline{(\mathfrak{c}(c)\cap
L\smallsetminus\overline{\mathfrak{c}(b)})}\right)$
$\displaystyle\quad=\quad\mathfrak{c}(c)\cap\left(L\smallsetminus\overline{(\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b))}\right)$
$\displaystyle\quad=\quad\mathfrak{c}(c)\smallsetminus\overline{(\mathfrak{c}(c)\smallsetminus\mathfrak{c}(b))}$

which is not possible. So there is no closed
$K\in\operatorname{\mathfrak{ND}}(L)$ such that
$F\in\operatorname{\mathfrak{ND}}(K)$. Thus $F$ is maximal nowhere dense in
$L$.

(2) Assume that $L$ is compact, $F=\mathfrak{c}(b)$ is maximal nowhere dense
in $L$ and suppose that for each $x\in F$, there is an
$\mathfrak{o}(b)$-nowhere dense $N_{x}$ such that
$x\in\operatorname{int}_{\mathfrak{c}(b)}(\mathfrak{c}(b)\cap\overline{N_{x}})$.
Set
$\operatorname{int}_{\mathfrak{c}(b)}(\mathfrak{c}(b)\cap\overline{N_{x}})=\mathfrak{o}(a_{x})\cap\mathfrak{c}(b)$.
Then

$\mathfrak{c}(b)\subseteq\bigvee_{x\in\mathfrak{c}(b)}(\mathfrak{o}(a_{x})\cap\mathfrak{c}(b))\subseteq\mathfrak{o}\left(\bigvee_{x\in\mathfrak{c}(b)}a_{x}\right).$

Therefore $b\vee\left(\bigvee_{x\in\mathfrak{c}(b)}a_{x}\right)=1$. By
compactness of $L$, there is a finite set $B\subseteq\mathfrak{c}(b)$ such
that $b\vee\left(\bigvee_{x\in B}a_{x}\right)=1.$ We get that

$\displaystyle\mathfrak{c}(b)$
$\displaystyle\quad\subseteq\quad\mathfrak{c}(b)\cap\mathfrak{o}\left(\bigvee_{x\in
B}a_{x}\right)$ $\displaystyle\quad=\quad\mathfrak{c}(b)\cap\bigvee_{x\in
B}\mathfrak{o}(a_{x})$ $\displaystyle\quad=\quad\bigvee_{x\in
B}(\mathfrak{c}(b)\cap\mathfrak{o}(a_{x}))$
$\displaystyle\quad=\quad\bigvee_{x\in
B}\left(\operatorname{int}_{\mathfrak{c}(b)}(\mathfrak{c}(b)\cap\overline{N_{x}})\right)$
$\displaystyle\quad\subseteq\quad\bigvee_{x\in
B}\left(\mathfrak{c}(b)\cap\overline{N_{x}}\right)$
$\displaystyle\quad=\quad\mathfrak{c}(b)\cap\bigvee_{x\in B}\overline{N_{x}}$
$\displaystyle\quad\subseteq\quad\bigvee_{x\in
B}\overline{N_{x}}=\overline{\bigvee_{x\in B}N_{x}}.$

Observe that $N_{x}\in\operatorname{\mathfrak{ND}}(L)$. This is so because
$N_{x}$ is nowhere dense in a dense sublocale $\mathfrak{o}(b)$ of $L$ making
it nowhere dense in $L$. Therefore
$\overline{N_{x}}\in\operatorname{\mathfrak{ND}}(L)$. Since finite joins of
closed nowhere dense sublocales are nowhere dense, $\bigvee_{x\in
B}\overline{N_{x}}=\overline{\bigvee_{x\in B}N_{x}}$ is nowhere dense in $L$.
We show that $\mathfrak{c}(b)$ is nowhere dense in $\overline{\bigvee_{x\in
B}N_{x}}$ which will contradict that $\mathfrak{c}(b)$ is maximal nowhere
dense in $L$.

Set $A=\overline{\bigvee_{x\in B}N_{x}}$. Observe that

$\displaystyle\operatorname{int}_{A}\left(\overline{\mathfrak{c}(b)}^{A}\right)$
$\displaystyle\quad=\quad A\cap\left(L\smallsetminus\overline{(A\cap
L\smallsetminus\overline{\mathfrak{c}(b)})}\right)$ $\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\overline{(A\cap\mathfrak{o}(b)}\right)$
$\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\overline{\left(\overline{\bigvee_{x\in
B}N_{x}}\cap\mathfrak{o}(b)\right)}\right)$ $\displaystyle\quad\subseteq\quad
A\cap\left(L\smallsetminus\overline{\left(\left(\bigvee_{x\in
B}N_{x}\right)\cap\mathfrak{o}(b)\right)}\right)$ $\displaystyle\quad=\quad
A\cap\left(L\smallsetminus\overline{\bigvee_{x\in
B}N_{x}}\right)\quad\text{since}\quad\bigvee_{x\in
B}N_{x}\subseteq\mathfrak{o}(b)$ $\displaystyle\quad=\quad\mathsf{O}.$

Thus $\mathfrak{c}(b)$ is nowhere dense in $A$ which is a contradiction.

(3) Suppose that there is $B\in\mathcal{S}(F)$ such that
$B\notin\mathcal{S}_{\operatorname{Ainac}}(F)$. Then, by Proposition 4.10,
$B\cap\operatorname{int}_{F}(\overline{N}\cap F)\neq\mathsf{O}$ for some
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus F)$. We get that
$\operatorname{int}_{F}(\overline{N}\cap F)\neq\mathsf{O}$. Set
$A=\operatorname{int}_{F}(\overline{N}\cap F)$. Then
$\overline{A}^{F}=\overline{A}$ is a non-void $F$-regular-closed sublocale.
Since $F$ is homogeneous maximal nowhere dense, $\overline{A}$ is maximal
nowhere dense in $L$. We show that $\overline{A}$ is nowhere dense in
$\overline{N}$ which will contradict that it is a maximal nowhere dense
sublocale. It is clear that $\overline{A}\subseteq\overline{N}$. Furthermore,
observe that

$\displaystyle\operatorname{int}_{\overline{N}}\left(\overline{A}^{\overline{N}}\right)$
$\displaystyle\quad=\quad\operatorname{int}_{\overline{N}}(\overline{A})$
$\displaystyle\quad=\quad\overline{N}\cap\left(L\smallsetminus\overline{\overline{N}\cap
L\smallsetminus\overline{A}}\right)$
$\displaystyle\quad\subseteq\quad\overline{N}\cap\left(L\smallsetminus\overline{\overline{N}\cap
L\smallsetminus F}\right)\quad\text{since}\quad\overline{A}\subseteq F$
$\displaystyle\quad\subseteq\quad\overline{N}\cap\left(L\smallsetminus\overline{N\cap
L\smallsetminus F}\right)$
$\displaystyle\quad=\quad\overline{N}\cap\left(L\smallsetminus\overline{N}\right)\quad\text{since}\quad
N\subseteq L\smallsetminus F$ $\displaystyle\quad=\quad\mathsf{O}.$

Thus $\overline{A}$ is nowhere dense in $\overline{N}$ which is not possible.
Hence every sublocale of $F$ belongs to
$\mathcal{S}_{\operatorname{Inac}}(F)$. ∎

If we consider $F$-clopen sublocales, we get the following result.

######  Proposition 4.15.

Let $F$ be a non-void nowhere dense sublocale of $L$. If $F$ is homogeneous
maximal nowhere dense, then each $F$-clopen sublocale is almost inaccessible
as a sublocale of $L$ with respect to itself.

###### Proof.

Let $A$ be an $F$-clopen sublocale and assume that
$A\cap\operatorname{int}_{A}(\overline{N}\cap A)\neq\mathsf{O}$ for some
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus A)$. Then
$\operatorname{int}_{A}(\overline{N}\cap A)$ is $F$-open so that
$\overline{\operatorname{int}_{A}(\overline{N}\cap A)}\cap
F=\overline{\operatorname{int}_{A}(\overline{N})}$ is a non-void regular-
closed sublocale of $F$. By hypothesis,
$\overline{\operatorname{int}_{A}(\overline{N}\cap A)}$ is maximal nowhere
dense. Following the argument used in last part of the proof of Theorem
4.14(3) and using the fact that $A$ is $F$-closed, we get that
$\overline{\operatorname{int}_{A}(\overline{N}\cap A)}$ is nowhere dense in
$\overline{N}$ which is not possible. ∎

We note the following example.

######  Example 4.16.

(1) If $X$ is a Hausdorff space with no isolated point, then every one-point
sublocale of $\mathfrak{O}X$ which is almost inaccessible as a sublocale of
$\mathfrak{O}X$ with respect to itself is maximal nowhere dense. To see this,
it suffices to show that such sublocales are non-void closed nowhere dense in
$\mathfrak{O}X$. Since every Hausdorff space is sober, each point $p$ of
$\mathfrak{O}X$ is of the form $p=X\smallsetminus\overline{\\{x\\}}$ for some
$x\in X$. Applying Hausdorffness again gives $p=X\smallsetminus\\{x\\}$ which
is open and dense in $X$, making $\\{x\\}$ closed nowhere dense in $X$. It
follows from [8, Lemma 2.11.] that $\widetilde{\\{x\\}}$ is closed and nowhere
dense in $\mathfrak{O}X$. But
$\widetilde{\\{x\\}}=\\{X\smallsetminus\\{x\\},1_{\mathfrak{O}X}\\}=\\{p,1_{\mathfrak{O}X}\\}$,
so the one-point sublocale $\\{p,1_{\mathfrak{O}X}\\}$ is non-void closed
nowhere dense. Now, if such a one-point sublocale $\\{p,1_{\mathfrak{O}X}\\}$
is an almost inaccessible sublocale of $L$ with respect to itself, it follows
from Proposition 4.14(1) that $\\{p,1_{\mathfrak{O}X}\\}$ is maximal nowhere
dense.

(2) The sublocales described in (1) are homogeneous maximal nowhere dense.
This is so because for any point $p\in\mathfrak{O}X$,
$\\{p,1_{\mathfrak{O}X}\\}$ is the only non-void sublocale contained in
$\\{p,1_{\mathfrak{O}X}\\}$. Therefore all non-void
$\\{p,1_{\mathfrak{O}X}\\}$-regular-closed sublocales are maximal nowhere
dense in $\mathfrak{O}X$.

We include the following result where we make use of Theorem 4.14(3) to show
that a homogeneous maximal nowhere dense sublocale is regular-closed in every
complemented nowhere dense sublocale containing it.

######  Proposition 4.17.

Let $L$ be a locale and $F$ a non-void closed nowhere dense sublocale of $L$.
If $F$ is homogeneous maximal nowhere dense and $F\subseteq A$, where $A$ is a
complemented nowhere dense sublocale of $L$, then $F$ is an $A$-regular-closed
sublocale.

###### Proof.

Assume that $F\neq\overline{\operatorname{int}_{A}(F)}$. Because it is always
true that $\overline{\operatorname{int}_{A}(F)}\subseteq F$, this assumption
says that $F\nsubseteq\overline{\operatorname{int}_{A}(F)}$. Therefore the
$F$-open sublocale
$F\smallsetminus\overline{\operatorname{int}_{A}(F)}=F\cap\left(L\smallsetminus\overline{\operatorname{int}_{A}(F)}\right)$
is non-void. Also,
$F\smallsetminus\overline{\operatorname{int}_{A}(F)}\subseteq\overline{A\smallsetminus
F}$. Indeed,

$\displaystyle F\smallsetminus\overline{\operatorname{int}_{A}(F)}$
$\displaystyle\quad\subseteq\quad
F\cap(L\smallsetminus\operatorname{int}_{A}(F))$ $\displaystyle\quad=\quad
F\cap\left(L\smallsetminus\left(A\cap(L\smallsetminus\overline{A\cap(L\smallsetminus
F)})\right)\right)$ $\displaystyle\quad=\quad F\cap\left((L\smallsetminus
A)\vee(L\smallsetminus(L\smallsetminus\overline{A\cap(L\smallsetminus
F)}))\right)$ $\displaystyle\quad=\quad F\cap\left((L\smallsetminus
A)\vee\overline{A\cap(L\smallsetminus F)}\right)$
$\displaystyle\quad=\quad(F\cap(L\smallsetminus
A))\vee(F\cap\overline{A\cap(L\smallsetminus F)})$
$\displaystyle\quad=\quad\mathsf{O}\vee(F\cap\overline{A\cap(L\smallsetminus
F)})\ \text{since}\ A\ \text{is complemented and }F\subseteq A$
$\displaystyle\quad=\quad F\cap\overline{A\cap(L\smallsetminus F)}$
$\displaystyle\quad\subseteq\quad\overline{A\cap(L\smallsetminus
F)}=\overline{A\smallsetminus F}.$

This makes
$\left(F\smallsetminus\overline{\operatorname{int}_{A}(F)}\right)\cap\operatorname{int}_{F}(F\cap\overline{A\smallsetminus
F})\neq\mathsf{O}$. Observe that $A\smallsetminus
F\in\operatorname{\mathfrak{ND}}(L\smallsetminus F)$. To see this, let $U$ be
an open sublocale of $L\smallsetminus F$ contained in
$\overline{A\smallsetminus F}^{(L\smallsetminus F)}=\overline{A\smallsetminus
F}\cap(L\smallsetminus F)$. Then $U\subseteq\overline{A}$. But
$A\in\operatorname{\mathfrak{ND}}(L)$ and an open sublocale of
$L\smallsetminus F$ is open in $L$, so we have that $U=\mathsf{O}$. Thus
$A\smallsetminus F\in\operatorname{\mathfrak{ND}}(L\smallsetminus F)$. We have
found a sublocale $F\smallsetminus\overline{\operatorname{int}_{A}(F)}$ of $F$
and $A\smallsetminus F\in\operatorname{\mathfrak{ND}}(L\smallsetminus F)$ such
that
$\left(F\smallsetminus\overline{\operatorname{int}_{A}(F)}\right)\cap\operatorname{int}_{F}\left(F\cap\overline{A\smallsetminus
F}\right)\neq\mathsf{O}$, i.e, a sublocale of $F$ which is not almost
$F$-inaccessible. By Theorem 4.14(3), $F$ is not homogeneous maximal nowhere
dense, which is a contradiction. ∎

In what follows, we characterize locales in which every non-void nowhere dense
sublocale is maximal nowhere dense. Recall from [4] that the boundary of a
sublocale $S$ of a locale $L$ is given by
$\operatorname{bd}(S)=\overline{S}\smallsetminus\operatorname{int}(S)$.

######  Theorem 4.18.

Let $L$ be a locale. The following statements are equivalent.

  1. (1)

Every non-void nowhere dense sublocale of $L$ is maximal nowhere dense.

  2. (2)

Every non-void closed nowhere dense sublocale is maximal nowhere dense.

  3. (3)

Every non-void open sublocale induced by a non-complemented element of $L$ has
a maximal nowhere dense boundary.

  4. (4)

Every non-void closed nowhere dense sublocale of $L$ is homogeneous maximal
nowhere dense.

  5. (5)

Every non-void closed nowhere dense sublocale is almost inaccessible as a
sublocale of $L$ with respect to itself.

  6. (6)

Every non-void closed nowhere dense sublocale is inaccessible as a sublocale
of $L$ with respect to itself.

###### Proof.

$(1)\Longleftrightarrow(2)$: Follows from Proposition 2.5.

$(2)\Longrightarrow(3)$: Let $\mathfrak{o}(x)\in\mathcal{S}(L)$ be non-void
with $x$ non-complemented. Then $x\vee x^{*}\neq 1$ making
$\operatorname{bd}(\mathfrak{o}(x))=\mathfrak{c}(x\vee x^{*})$ a non-void
closed nowhere dense sublocale. It follows from (2) that
$\operatorname{bd}(\mathfrak{o}(x))$ is maximal nowhere dense.

$(3)\Longrightarrow(4)$: Let $\mathfrak{c}(x)$ be a non-void nowhere dense
sublocale of $L$ and choose $y\in L$ such that
$\overline{\mathfrak{o}(y)\cap\mathfrak{c}(x)}\cap\mathfrak{c}(x)\neq\mathsf{O}$.
Then

$\overline{\mathfrak{o}(y)\cap\mathfrak{c}(x)}\cap\mathfrak{c}(x)=\overline{\mathfrak{o}(y)\cap\mathfrak{c}(x)}=\mathfrak{c}(y\rightarrow
x)\neq\mathsf{O}$

implying that $y\rightarrow x\neq 1$. But $\mathfrak{c}(y\rightarrow
x)\subseteq\mathfrak{c}(x)\in\operatorname{\mathfrak{ND}}(L)$, so
$\mathfrak{c}(y\rightarrow x)\in\operatorname{\mathfrak{ND}}(L)$ making
$\mathfrak{o}(y\rightarrow x)$ non-void, open and

$(y\rightarrow x)\vee(y\rightarrow x)^{*}=(y\rightarrow x)\vee 0=y\rightarrow
x\neq 1.$

It follows from (3) that

$\operatorname{bd}(\mathfrak{o}(y\rightarrow x))=\mathfrak{c}((y\rightarrow
x)\vee(y\rightarrow x)^{*})=\mathfrak{c}(y\rightarrow
x)=\overline{\mathfrak{o}(y)\cap\mathfrak{c}(x)}$

is maximal nowhere dense. Thus $\mathfrak{c}(x)$ is homogeneous maximal
nowhere dense.

$(4)\Longrightarrow(5)$: Follows from Theorem 4.14(3).

$(5)\Longrightarrow(6)$: Let $\mathfrak{c}(x)$ be a non-void nowhere dense
sublocale and assume that $\overline{N}\cap\mathfrak{c}(x)\neq\mathsf{O}$ for
some
$N\in\operatorname{\mathfrak{ND}}((L\smallsetminus\mathfrak{c}(x))=\mathfrak{o}(x))$.
Since $\mathfrak{o}(x)$ is dense in $L$, $N$ is nowhere dense in $L$ so that
the non-void sublocale $\overline{N}\cap\mathfrak{c}(x)$ is closed nowhere
dense in $L$. It follows from (5) that $\overline{N}\cap\mathfrak{c}(x)$ is
almost $\left(\overline{N}\cap\mathfrak{c}(x)\right)$-inaccessible. Observe
that
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x)))$.
To see this, let $a\in L$ be such that
$\mathfrak{o}(a)\cap\left(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x))\right)\subseteq\overline{N}\cap\left(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x))\right)$.
Then
$\mathfrak{o}(a)\cap\left(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x))\right)\subseteq\overline{N}$.
Because $N\in\operatorname{\mathfrak{ND}}(L)$ and
$\mathfrak{o}(a)\cap\left(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x))\right)$
is open,
$\mathfrak{o}(a)\cap\left(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x))\right)=\mathsf{O}$.
Therefore $\mathfrak{o}(a)\subseteq\overline{N}\cap\mathfrak{c}(x)$ making
$\mathfrak{o}(a)=\mathsf{O}$ since
$\overline{N}\cap\mathfrak{c}(x)=\overline{\overline{N}\cap\mathfrak{c}(x)}$
is nowhere dense in $L$. Thus
$N\in\operatorname{\mathfrak{ND}}(L\smallsetminus(\overline{N}\cap\mathfrak{c}(x)))$.

Therefore
$\overline{N}\cap\mathfrak{c}(x)\subseteq\overline{\left(\overline{N}\cap\mathfrak{c}(x)\right)\cap\left(L\smallsetminus\overline{N}\right)}=\overline{\mathsf{O}}=\mathsf{O}$
which is not possible. Thus $\mathfrak{c}(x)\cap\overline{N}=\mathsf{O}$
making $\mathfrak{c}(x)\in\mathcal{S}_{\operatorname{Inac}}(\mathfrak{c}(x))$.

$(6)\Longrightarrow(2)$: Let $F$ be a non-void closed nowhere dense sublocale
of $L$. It follows from $(6)$ that $F\in\mathcal{S}_{\operatorname{Inac}}(F)$.
Because
$\mathcal{S}_{\operatorname{Inac}}(F)\subseteq\mathcal{S}_{\operatorname{Ainac}}(F)$
by Proposition 4.11(1), it follows from Theorem 4.14(1) that $F$ is maximal
nowhere dense. ∎

In the following example we show that there is a locale having the properties
described in Theorem 4.18.

######  Example 4.19.

Consider the three-element chain $\operatorname{\textbf{3}}=\\{1,0,a\\}$.
Clearly, $\operatorname{\textbf{3}}$ is non-Boolean and the only non-void
closed nowhere dense sublocale of $\operatorname{\textbf{3}}$ is
$\mathfrak{c}(a)$ which is maximal nowhere dense because it is not nowhere
dense as a sublocale of itself.

##  5\. Maximal nowhere density and Remoteness

In this section, we consider a connection between remoteness and maximal
nowhere density. Inaccessibility will be useful here.

We recall the following notations from [8, 9]. Let $L$ be a locale, $S$ a
dense sublocale of $L$ and $T\in\mathcal{S}(L)$. Then

  1. (1)

$T$ is a remote sublocale of $L$ in case $T\cap N=\mathsf{O}$ for every
nowhere dense $N\in\mathcal{S}(L)$.

  2. (2)

$\mathcal{S}_{\text{rem}}(L)$ denotes the collection of all remote sublocales
of $L$.

  3. (3)

$T$ is remote from $S$, denoted $T\in\mathcal{S}_{\text{rem}}(L\ltimes S)$, if
$T\cap\overline{N}=\mathsf{O}$ for each $S$-nowhere dense $N$.

  4. (4)

$T$ is ${}^{*}\\!$remote from $S$, denoted
$T\in^{*}\\!\mathcal{S}_{\text{rem}}(L\ltimes S)$, provided that
$T\in\mathcal{S}_{\text{rem}}(L\ltimes S)$ and $T\subseteq L\smallsetminus S$.

######  Observation 5.1.

A simple case where inaccessibility differs from remoteness is that of
Proposition 4.11(4). Recall from [8] that a necessary and sufficient condition
for a locale $L$ to be remote as a sublocale of itself is that it must be
Boolean. Yet, by Proposition 4.11(4), every locale $L$ (not necessarily
Boolean) is $L$-inaccessible as a sublocale of itself.

Observe that for each dense and complemented $S\in\mathcal{S}(L)$,
$T\in\mathcal{S}(L\smallsetminus S)$ is $(L\smallsetminus S)$-inaccessible if
and only if $T\cap\overline{N}=\mathsf{O}$ for all
$(L\smallsetminus(L\smallsetminus S))$-nowhere dense sublocale $N$, i.e., for
all $S$-nowhere dense sublocale $N$. This shows that sublocales of
$L\smallsetminus S$ which are ∗remote from a dense and complemented sublocale
$S$ of $L$ are precisely the $(L\smallsetminus S)$-inaccessible sublocales. We
formalise this in the following proposition.

######  Proposition 5.2.

A sublocale $T\in\mathcal{S}(L\smallsetminus S)$ where $S$ is dense and
complemented in a locale $L$, is $(L\smallsetminus S)$-inaccessible iff it is
${}^{*}\\!$remote from $S$.

In the following result, we codify the variants of remoteness and
inaccessibility for dense and complemented sublocales.

######  Proposition 5.3.

Let $L$ be a locale, $S$ a dense and complemented sublocale of $L$ and
$T\in\mathcal{S}(L\smallsetminus S)$. Consider the following statements:

  1. (1)

$T\in\mathcal{S}_{\text{rem}}(L)$.

  2. (2)

$T\in\mathcal{S}_{\text{rem}}(L\ltimes S)$.

  3. (3)

$T\in{{}^{*}\\!\mathcal{S}_{\text{rem}}(L\ltimes S)}$.

  4. (4)

$T\in\mathcal{S}_{\operatorname{Inac}}(L\smallsetminus S)$.

  5. (5)

$T\in\mathcal{S}_{\operatorname{Ainac}}(L\smallsetminus S)$.

Then
$(1)\Longrightarrow(2)\Longleftrightarrow(3)\Longleftrightarrow(4)\Longrightarrow(5)$.

###### Proof.

$(1)\Longrightarrow(2)$: Follows from [8, Proposition 3.5.(4).]

$(2)\Longleftrightarrow(3)$: This is a combination of [8, Proposition 3.2.]
and the fact that $T\subseteq L\smallsetminus S$.

$(3)\Longleftrightarrow(4)$: Follows from Proposition 5.2.

$(4)\Longrightarrow(5)$: Follows from Proposition 4.11(1). ∎

######  Example 5.4.

[13] In Top, we have that a point $p\in\beta X\smallsetminus X$ is remote if
and only if $p$ is $(\beta X\smallsetminus X)$-inaccessible.

We also include the following result which shows inaccessibility of the least
dense sublocale.

######  Proposition 5.5.

Let $L$ be a locale and $S\in\mathcal{S}(L)$. If $S$ is open, then
$S\cap\operatorname{Rs}(L\ltimes S)$ is $S$-inaccessible, where
$\operatorname{Rs}(L\ltimes S)=\bigvee\\{A\in\mathcal{S}(L):A\text{ is remote
from }S\\}$. Moreover, if $S$ is dense and open, then $\mathfrak{B}L$ is
$S$-inaccessible.

###### Proof.

The first part follows from Proposition 4.11(3) and (2).

Since, according to Observation [8, Observation 3.1.],
$\mathfrak{B}L=T\cap\operatorname{Rs}(L\ltimes T)$ for any dense
$T\in\mathcal{S}(L)$, we have that
$\mathfrak{B}L=S\cap\operatorname{Rs}(L\ltimes S)$ is $S$-inaccessible for
dense and open $S$. ∎

Using Proposition 5.3 and the fact if $S\in\mathcal{S}(L)$ is open and dense,
then $S^{\\#}=L\smallsetminus S$ is nowhere dense, we get the following result
about remoteness and maximal nowhere density.

######  Proposition 5.6.

Let $S\neq L$ be an open dense sublocale of $L$.

  1. (1)

If $S^{\\#}\in{{}^{*}\\!\mathcal{S}_{\text{rem}}(L\ltimes S)}$, then $S^{\\#}$
is maximal nowhere dense in $L$.

  2. (2)

If $S^{\\#}$ is homogeneous maximal nowhere dense, then every $S^{\\#}$-remote
sublocale is ∗remote from $S$.

###### Proof.

(1) If $S^{\\#}\in{{}^{*}\\!\mathcal{S}_{\text{rem}}(L\ltimes S)}$, then, by
Proposition 5.3, $S^{\\#}\in\mathcal{S}_{\operatorname{Ainac}}(S^{\\#})$. It
follows from Theorem 4.14(1) that $S^{\\#}$ is maximal nowhere dense in $L$.

(2) Let $A\in\mathcal{S}_{\text{rem}}(S^{\\#})$ and choose an $S$-nowhere
dense $N$. Since, by Theorem 4.14(3), sublocales of homogeneous maximal
nowhere dense sublocales are almost inaccessible as sublocales of $L$ with
respect to themselves, $S^{\\#}$ is almost $S^{\\#}$-inaccessible, i.e.,
$\operatorname{int}_{S^{\\#}}(S^{\\#}\cap\overline{N})=\mathsf{O}$. Since
$S^{\\#}\cap\overline{N}=\overline{S^{\\#}\cap\overline{N}}^{S^{\\#}}$, we get
that $S^{\\#}\cap\overline{N}$ is $S^{\\#}$-nowhere dense. Because
$A\in\mathcal{S}_{\text{rem}}(S^{\\#})$,

$\mathsf{O}=A\cap S^{\\#}\cap\overline{N}=A\cap\overline{N}.$

Thus $A\in\mathcal{S}_{\text{rem}}(L\ltimes S)$ making
$A\in{{}^{*}\\!\mathcal{S}_{\text{rem}}(L\ltimes S)}$ since $A\subseteq
S^{\\#}$. ∎

##  6\. Preservation of maximal nowhere density

We end this paper with a discussion of localic maps that send maximal nowhere
density back and forth. We shall also include results about inaccessibility.

######  Proposition 6.1.

Let $f:L\rightarrow M$ be a localic map such that both $f$ and $f^{*}$ send
dense elements to dense elements. Then $f_{-1}$ preserves maximal nowhere
dense sublocales.

###### Proof.

Let $N\in\mathcal{S}(M)$ be maximal nowhere dense in $M$. Then $\overline{N}$
is maximal nowhere dense in $L$. Since $f^{*}$ is weakly open, it follows from
[8, theorem 2.29] that $f_{-1}$ preserves closed nowhere dense sublocales so
that $f_{-1}[\overline{N}]$ (which is equal to $\mathfrak{c}(f^{*}(\bigwedge
N))$) is nowhere dense in $L$. It is left to show that it is maximal nowhere
dense. Suppose not, that is, there exists a nowhere dense sublocale
$\mathfrak{c}(b)$ of $L$ such that $f_{-1}[\overline{N}]$ is nowhere dense in
$\mathfrak{c}(b)$. Since $f$ sends dense elements to dense elements, $f(b)$ is
dense in $M$ making $\mathfrak{c}(f(b))$ nowhere dense in $M$. The sublocale
$\overline{N}$ is nowhere dense in the nowhere dense sublocale
$\mathfrak{c}(f(b))$. Indeed, if
$\mathfrak{o}(x)\cap\mathfrak{c}(f(b))\subseteq\overline{N}^{\mathfrak{c}(f(b))}=\overline{N}\cap\mathfrak{c}(f(b))$
for some $x\in M$, then
$\mathfrak{o}(x)\cap\mathfrak{c}(f(b))\subseteq\overline{N}$ so that

$\mathfrak{o}(f^{*}(x))\cap\mathfrak{c}(b)\subseteq\mathfrak{o}(f^{*}(x))\cap\mathfrak{c}(f^{*}(f(b)))=f_{-1}[\mathfrak{o}(x)\cap\mathfrak{c}(f(b))]\subseteq
f_{-1}[\overline{N}].$

Therefore $\mathfrak{o}(f^{*}(x))\cap\mathfrak{c}(b)\subseteq
f_{-1}[\overline{N}]\cap\mathfrak{c}(b)$. Since
$f_{-1}[\overline{N}]\in\operatorname{\mathfrak{ND}}(\mathfrak{c}(b))$ and
$\mathfrak{o}(f^{*}(x))\cap\mathfrak{c}(b)$ is open in $\mathfrak{c}(b)$,
$\mathfrak{o}(f^{*}(x))\cap\mathfrak{c}(b)=\mathsf{O}$ which implies that
$\mathfrak{c}(b)\subseteq\mathfrak{c}(f^{*}(x))=f_{-1}[\mathfrak{c}(x)]$.
Therefore $f[\mathfrak{c}(b)]\subseteq
f[f_{-1}[\mathfrak{c}(x)]]\subseteq\mathfrak{c}(x)$ implying that
$\mathfrak{c}(f(b))=\overline{f[\mathfrak{c}(b)]}\subseteq\overline{\mathfrak{c}(x)}=\mathfrak{c}(x)$.
This makes $\mathfrak{c}(f(b))\cap\mathfrak{o}(x)=\mathsf{O}$. Therefore
$\overline{N}$ is nowhere dense in $\mathfrak{c}(f(b))$ which contradicts that
$\overline{N}$ is maximal nowhere dense in $M$. Therefore $f_{-1}[N]$ is
maximal nowhere dense in $L$. ∎

In the next result, we discuss localic maps that preserve maximal nowhere
dense sublocales.

Recall that a frame homomorphism $h:M\rightarrow L$ is open if it has a left
adjoint $h_{!}$ satisfying the Frobenius identity:

$h_{!}(h(a)\wedge b)=h_{!}(a)\wedge b$

for every $a\in M$ and $b\in L$. This is equivalent to saying that $h_{*}$ is
an open localic map. Because, by [3], every open frame homomorphism is weakly
open, it follows that for each open localic map $f$, $f^{*}$ is weakly open.

######  Observation 6.2.

Not every open localic map sends dense elements to dense elements. Consider
the localic map $f:L\rightarrow\operatorname{\textbf{2}}$ where $L$ is non-
Boolean and $\operatorname{\textbf{2}}$ is the two-element locale. Since
$\operatorname{\textbf{2}}$ is Boolean, every sublocale of
$\operatorname{\textbf{2}}$ is open making the localic image of each open
sublocale of $L$ to be open in $\operatorname{\textbf{2}}$. Hence $f$ is open.
However, $f$ does not send all dense elements to dense elements since the only
(dense) element of $L$ that is mapped to $1_{\operatorname{\textbf{2}}}$ (the
only dense element of $\operatorname{\textbf{2}}$) is $1$. But $1$ is not the
only dense element of $L$ otherwise $L$ is Boolean.

We recall from [11] that if a localic map $f:L\rightarrow M$ is open, then
$f_{-1}[\overline{A}]=\overline{f_{-1}[A]}$ for each $A\in\mathcal{S}(M)$.

######  Proposition 6.3.

Let $f:L\rightarrow M$ be an open localic map that sends dense elements to
dense elements. Then $f$ preserves maximal nowhere dense sublocales.

###### Proof.

Let $N$ be a maximal nowhere dense sublocale of $L$. We show that
$\overline{f[N]}$ is maximal nowhere dense in $M$. By [8, Lemma 2.33], $f[N]$
is nowhere dense in $M$ so that $\overline{f[N]}=\mathfrak{c}\left(f(\bigwedge
N)\right)$ is nowhere dense in $M$. Suppose that
$\mathfrak{c}\left(f(\bigwedge
N)\right)\in\operatorname{\mathfrak{ND}}(\mathfrak{c}(y))$ for some
$\mathfrak{c}(y)\in\operatorname{\mathfrak{ND}}(M)$. By Observation 2.11,
$\mathfrak{c}(y)\subseteq\overline{\mathfrak{c}(y)\cap\mathfrak{o}\left(f(\bigwedge
N)\right)}$. Therefore
$f_{-1}[\mathfrak{c}(y)]=\mathfrak{c}(f^{*}(y))\subseteq
f_{-1}\left[\overline{\mathfrak{c}(y)\cap\mathfrak{o}\left(f(\bigwedge
N)\right)}\right]$. Openness of $f$ gives
$f_{-1}\left[\overline{\mathfrak{c}(y)\cap\mathfrak{o}\left(f(\bigwedge
N)\right)}\right]=\overline{f_{-1}[\mathfrak{c}(y)\cap\mathfrak{o}\left(f(\bigwedge
N)\right)]}$, so that

$\displaystyle\mathfrak{c}(f^{*}(y))$
$\displaystyle\quad\subseteq\quad\overline{f_{-1}\left[\mathfrak{c}(y)\cap\mathfrak{o}\left(f(\bigwedge
N)\right)\right]}$
$\displaystyle\quad=\quad\overline{\mathfrak{c}(f^{*}(y))\cap\mathfrak{o}\left(f^{*}\left(f\left(\bigwedge
N\right)\right)\right)}$
$\displaystyle\quad\subseteq\quad\overline{\mathfrak{c}(f^{*}(y))\cap\mathfrak{o}\left(\bigwedge
N\right)}$
$\displaystyle\quad=\quad\overline{\mathfrak{c}(f^{*}(y))\cap\left(L\smallsetminus\overline{N}\right)}.$

Therefore
$\mathfrak{c}(f^{*}(y))\cap\left(L\smallsetminus\overline{\mathfrak{c}(f^{*}(y))\cap\left(L\smallsetminus\overline{N}\right)}\right)=\mathsf{O}$.
This makes $N\in\operatorname{\mathfrak{ND}}(\mathfrak{c}(f^{*}(y)))$, where
$\mathfrak{c}(f^{*}(y))\in\operatorname{\mathfrak{ND}}(L)$, contradicting that
$N$ is maximal nowhere dense in $L$. Therefore $\overline{f[N]}$ is maximal
nowhere dense so that by Proposition 2.5, $f[N]$ is maximal nowhere dense
sublocales. ∎

Since in Proposition 6.1 we only needed a condition that both $f$ and $f^{*}$
send dense elements to dense elements and because the left adjoint of an open
localic map is weakly open, we have the following result.

######  Corollary 6.4.

Every open localic map that sends dense elements to dense elements preserves
and reflects maximal nowhere dense sublocales.

In the next result, we discuss preservation and reflection of strongly
homogeneous maximal nowhere dense sublocales by localic maps.

######  Proposition 6.5.

Let $f:L\rightarrow M$ be an open localic map that sends dense elements to
dense elements.

  1. (1)

Then $f$ preserves weakly homogeneous maximal nowhere dense sublocales.

  2. (2)

If $f$ is injective, then it reflects (strongly) homogeneous maximal nowhere
dense sublocales.

###### Proof.

(1) Let $F$ be a weakly homogeneous maximal nowhere dense sublocale of $L$ and
choose a non-void sublocale $\overline{\mathfrak{o}(y)\cap f[F]}\cap f[F]$
where $y\in M$. Such a sublocale is $f[F]$-regular-closed. The $F$-regular-
closed sublocale $\overline{\mathfrak{o}(f^{*}(y))\cap F}\cap F$ is non-void
otherwise, $\mathfrak{o}(f^{*}(y))\cap F=\mathsf{O}$ so that $f[F]\subseteq
f[\mathfrak{c}(f^{*}(y))]=f[f_{-1}[\mathfrak{c}(y)]]\subseteq\mathfrak{c}(y)$.
Therefore $f[F]\cap\mathfrak{o}(y)=\mathsf{O}$ which is not possible. Since
$F$ is strongly h.m.n.d, $\overline{\mathfrak{o}(f^{*}(y))\cap F}\cap F$ is
m.n.d. Because open localic maps that send dense elements to dense elements
preserve maximal nowhere dense sublocales (by Proposition 6.3),
$f\left[\overline{\mathfrak{o}(f^{*}(y))\cap F}\cap F\right]$ is m.n.d in $M$.
Since

$\displaystyle f\left[\overline{\mathfrak{o}(f^{*}(y))\cap F}\cap F\right]$
$\displaystyle\quad\subseteq\quad f\left[\overline{\mathfrak{o}(f^{*}(y))\cap
F}\right]\cap f[F]$
$\displaystyle\quad\subseteq\quad\overline{f[\mathfrak{o}(f^{*}(y))\cap
F]}\cap f[F]$
$\displaystyle\quad\subseteq\quad\overline{f[\mathfrak{o}(f^{*}(y))]\cap
f[F]}\cap f[F]$ $\displaystyle\quad\subseteq\quad\overline{\mathfrak{o}(y)\cap
f[F]}\cap f[F]$

and because $\overline{\mathfrak{o}(y)\cap f[F]}\cap f[F]$ is nowhere dense in
$M$, it follows from Proposition 2.12(2) that $\overline{\mathfrak{o}(y)\cap
f[F]}\cap f[F]$ is m.n.d. Thus $f[F]$ is weakly homogeneous maximal nowhere
dense in $M$.

(2) We only prove reflection of weakly homogeneous maximal nowhere dense
sublocales. That of homogeneous maximal nowhere dense sublocales follows the
same sketch. Let $K$ be a weakly homogeneous maximal nowhere dense sublocale
of $M$ and consider a non-void sublocale $\overline{\mathfrak{o}(x)\cap
f_{-1}[K]}\cap f_{-1}[K]$ where $x\in L$. We must show that this
$f_{-1}[K]$-regular-closed sublocale is m.n.d. We have that
$\overline{\mathfrak{o}(f(x))\cap K}\cap K$ is a non-void $K$-regular-closed
sublocale. To see that is it non-void, observe that having
$\overline{\mathfrak{o}(f(x))\cap K}\cap K=\mathsf{O}$ implies that
$\mathfrak{o}(f(x))\cap K=\mathsf{O}$ so that

$\mathsf{O}=f_{-1}[\mathfrak{o}(f(x))]\cap
f_{-1}[K]=\mathfrak{o}(f^{*}(f(x)))\cap f_{-1}[K]=\mathfrak{o}(x)\cap
f_{-1}[K]$

where the latter equality follows from injectivity of $f$. This cannot be
true, so $\overline{\mathfrak{o}(f(x))\cap K}\cap K$ is non-void. Since $K$ is
weakly homogeneous maximal nowhere dense, $\overline{\mathfrak{o}(f(x))\cap
K}\cap K$ is m.n.d in $M$. Because open localic maps are weakly open and $f$
sends dense elements to dense elements, it follows from Proposition 6.1 that
$f_{-1}[\overline{\mathfrak{o}(f(x))\cap K}\cap K]$ is m.n.d. Observe that

$f_{-1}[\overline{\mathfrak{o}(f(x))\cap K}\cap
K]=f_{-1}[\overline{\mathfrak{o}(f(x))\cap K}]\cap
f_{-1}[K]=\overline{f_{-1}[\mathfrak{o}(f(x))]\cap f_{-1}[K]}\cap f_{-1}[K]$

where the latter equality follows from openness of $f$. By injectivity of $f$,

$f_{-1}[\overline{\mathfrak{o}(f(x))\cap K}\cap
K]=\overline{\mathfrak{o}(x)\cap f_{-1}[K]}\cap f_{-1}[K]$

making $\overline{\mathfrak{o}(x)\cap f_{-1}[K]}\cap f_{-1}[K]$ m.n.d. in $L$.
Thus $f_{-1}[K]$ is weakly homogeneous maximal nowhere dense in $L$. ∎

######  Observation 6.6.

For the preservation of homogeneous maximal nowhere dense sublocales, the
localic map $f$ in Proposition 6.5 must also preserve closed sublocales. That
is, it must also be closed which is a rather too stringent condition.

Open localic maps also allow us to study, under certain conditions,
preservation and reflection of inaccessible and almost inaccessible sublocales
as presented below.

######  Proposition 6.7.

Let $f:L\rightarrow M$ be an open and injective localic map. Then for all open
$S\in\mathcal{S}(L)$,

  1. (1)

$f[\mathcal{S}_{\operatorname{Inac}}(S)]\subseteq\mathcal{S}_{\operatorname{Inac}}(f[S])$,
and

  2. (2)

$f[\mathcal{S}_{\operatorname{Ainac}}(S)]\subseteq\mathcal{S}_{\operatorname{Ainac}}(f[S])$.

###### Proof.

Follows from Proposition 4.11(3), (2) and (1). ∎

######  Proposition 6.8.

Let $f:L\rightarrow M$ be localic map such that both $f$ and $f^{*}$ send
dense elements to dense elements and let $T\in\mathcal{S}(M)$ be closed
nowhere dense. Then

  1. (1)

$f_{-1}[\mathcal{S}_{\operatorname{Inac}}(T)]\subseteq\mathcal{S}_{\operatorname{Inac}}(f_{-1}[T])$,
and

  2. (2)

If $f$ is open, then
$f_{-1}[\mathcal{S}_{\operatorname{Ainac}}(T)]\subseteq\mathcal{S}_{\operatorname{Inac}}(f_{-1}[T])$.

###### Proof.

(1) Let $T\in\mathcal{S}(M)$ be closed and nowhere dense and choose
$A\in\mathcal{S}_{\operatorname{Inac}}(T)$. Then $f_{-1}[A]\subseteq
f_{-1}[T]$. Let $N\in\operatorname{\mathfrak{ND}}(L\smallsetminus f_{-1}[T])$.
Since, by [6, (5.1)], $P\smallsetminus g_{-1}[C]\subseteq
g_{-1}[R\smallsetminus C]$ for every localic map $g:P\rightarrow R$ with
$C\in\mathcal{S}(R)$, we get that $N\subseteq L\smallsetminus
f_{-1}[T]\subseteq f_{-1}[M\smallsetminus T]$. Therefore $f[N]\subseteq
f[f_{-1}[M\smallsetminus T]]\subseteq M\smallsetminus T$. We show that
$f[N]\in\operatorname{\mathfrak{ND}}(M\smallsetminus T)$. Because $f^{*}$ is
weakly open, it follows from [8, Theorem 2.29.] that
$f_{-1}[-]:\mathcal{S}(M)\rightarrow\mathcal{S}(L)$ preserves closed nowhere
dense sublocales so that $f_{-1}[T]$ is closed nowhere dense in $L$. Therefore
$L\smallsetminus f_{-1}[T]$ is open and dense. Now, $N$ being nowhere dense in
$L\smallsetminus f_{-1}[T]$ implies $N$ is nowhere dense in $L$. Since $f$
sends dense elements to dense elements, it follows from [8, Lemma 2.33.] that
$f[-]$ preserves nowhere dense sublocales so that $f[N]$ is nowhere dense in
$M$. Observe that $f[N]$ is nowhere dense in $M\smallsetminus T$. To see this,
let $y\in M$ be such that $\mathfrak{o}(y)\cap(M\smallsetminus
T)\subseteq\overline{f[N]}\cap(M\smallsetminus T)$. Then
$\mathfrak{o}(y)\cap(M\smallsetminus T)\subseteq\overline{f[N]}$. Because
$M\smallsetminus T$ is open in $M$ and
$f[N]\in\operatorname{\mathfrak{ND}}(M)$, we have that
$\mathfrak{o}(y)\cap(M\smallsetminus T)=\mathsf{O}$ making
$\mathfrak{o}(y)\subseteq T$. But $T$ is nowhere dense in $M$, so
$\mathfrak{o}(y)=\mathsf{O}$ implying that

(6.1) $f[N]\in\operatorname{\mathfrak{ND}}(M\smallsetminus T).$

$S$-inaccessibility of $A$ implies $A\cap\overline{f[N]}=\mathsf{O}$.
Therefore

$\mathsf{O}=f_{-1}\left[A\cap\overline{f[N]}\right]=f_{-1}[A]\cap
f_{-1}\left[\overline{f[N]}\right]=f_{-1}[A]\cap\mathfrak{c}\left(h\left(f\left(\bigwedge
N\right)\right)\right)\supseteq f_{-1}[A]\cap\overline{N}.$

Thus $f_{-1}[A]\in\mathcal{S}_{\operatorname{Inac}}(f_{-1}[T])$.

(2) Assume that $f$ is open. Set $T=\mathfrak{c}(b)$ for some $b\in M$ and
choose $A\in\mathcal{S}_{\operatorname{Ainac}}(T)$. (6.1) still holds, so

$A\subseteq\mathfrak{c}(b)\cap\overline{\mathfrak{c}(b)\cap\left(M\smallsetminus\overline{f[N]}\right)}\\\
=\mathfrak{c}(b)\cap\overline{\mathfrak{c}(b)\cap\mathfrak{o}\left(\bigwedge
f[N]\right)}.$

Therefore

$\displaystyle f_{-1}[A]$
$\displaystyle\quad\subseteq\quad\mathfrak{c}(f^{*}(b))\cap
f_{-1}\left[\overline{\mathfrak{c}(b)\cap\mathfrak{o}\left(\bigwedge
f[N]\right)}\right]$
$\displaystyle\quad=\quad\mathfrak{c}(f^{*}(b))\cap\overline{f_{-1}\left[\mathfrak{c}(b)\cap\mathfrak{o}\left(\bigwedge
f[N]\right)\right]}\quad\text{since }f\text{ is open}$
$\displaystyle\quad=\quad\mathfrak{c}(f^{*}(b))\cap\overline{\mathfrak{c}(f^{*}(b))\cap\mathfrak{o}\left(h\left(f\left(\bigwedge
N\right)\right)\right)}$
$\displaystyle\quad\subseteq\quad\mathfrak{c}(f^{*}(b))\cap\overline{\mathfrak{c}(f^{*}(b))\cap\mathfrak{o}\left(\bigwedge
N\right)}$ $\displaystyle\quad=\quad
f_{-1}[T]\cap\overline{f_{-1}[T]\cap\left(L\smallsetminus\overline{N}\right)}$
$\displaystyle\quad=\quad\operatorname{cl}_{f_{-1}[T]}\left(f_{-1}[T]\cap\left(L\smallsetminus\overline{N}\right)\right).$

Thus $f_{-1}[A]\in\mathcal{S}_{\operatorname{Ainac}}(f_{-1}[T])$. ∎

## References

  * [1] Baboolal D., Picado J., Pillay P., Pultr A., Hewitt’s irresolvability and induced sublocales in spatial frames, Quaestiones Mathematicae, 43 (2020), 1601–1612.
  * [2] Baboolal, D., Pillay P., Irreducible locales, Filomat, 32(10) (2018), 3443–3453.
  * [3] Banaschewski, B., Pultr, A., Variants of openness, Applied Categorical Structures, 2 (1994), 331–350.
  * [4] Dube T., Submaximality in locales, Topology Proceedings, 29 (2005), 431–444.
  * [5] Dube T., Mugochi M.M., Localic remote points revisited, Filomat, 29(1) (2015), 111–120.
  * [6] Ferreira M.J., Picado J., Pinto S.M., Remainders in pointfree topology, Topology and its Applications, 245 (2018), 21–45.
  * [7] Koldunov A.V.E., Maximal nowhere dense sets, Siberian Mathematical Journal, 21(2) (1980), 558–564.
  * [8] Nxumalo M., Remote Sublocales, Quaestiones Mathematicae, (2023), 1–18.
  * [9] Nxumalo M.S., Remoteness in the category of locales, Ph.D. Thesis, University of South Africa, (2023), pp 154.
  * [10] Picado J., Pultr A., Frames and Locales: topology without points, Springer Science and Business Media, (2011).
  * [11] Picado J., Pultr A., Tozzi A., Locales. Categorical Foundations: Special Topics in Order, Topology, Algebra and Sheaf Theory. Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge 97 (2004), 49–101.
  * [12] Plewe T., Higher order dissolutions and Boolean coreflections of locales, Journal of Pure and Applied Algebra, 154 (1-3) (2000), 273–293.
  * [13] Veksler A.I., Maximal nowhere dense sets and their applications to problems of existence of remote points and of weak p‐points, Mathematische Nachrichten, 150(1) (1991), 263–275.
  * [14] Veksler A.I., Maximal nowhere dense sets in topological spaces, IIzv. Vys. Ucheb. Zav. Mat., (5) (1975), 9–16.