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${}$
\vskip10mm

\centerline{\Kapitola Nearness, subfitness and sequential regularity }
          \vskip7mm

\centerline{\it In honour of Bernhard Banaschewski on the occasion of his 70th
 birthday}


\bigskip\bigskip


\heading H. Herrlich
\
\ and \  A.  Pultr\footnote  "*"
{The support of the Alexander von
Humboldt-Stiftung and the Grant Agency
of the
Czech Republic under Grant 201/93/950 is gratefully acknowledged.}
\
 \endheading
 \vskip5mm               \eightpoint
\roster

\item""{\bf Abstract :} In the point-free context, the structure of
nearness has been so far studied in the regular case only. Here we
answer the question as to how far beyond that one can go. It turns out
that a frame (locale) admits a nearness iff it is subfit. Unlike in the
case of spaces, where admitting nearness is a hereditary property,
subfitness is not; therefore, also the hereditary subfitness (here
called {\it sequential regularity\/} for reasons obvious from the
properties presented) is studied. It is weaker than regularity and
seems to be of some interest also in the spatial case.

\vskip2mm
\noindent {\bf 1991 Mathematics Subject Classification :}  54A05,
06D10, 54E17, 54D10

\vskip2mm
\noindent {\bf Keywords :} nearness space, nearnes frame (locale),
subfit frame, regular, sequentially regular
      \endroster \tenpoint

\vskip10mm


Nearness structures (see, e.g., \cite{5}, \cite{6},
$T$-uniformities in \cite{11}) have been studied in the
point-free context so far only in the regular case (see, e.g.,
\cite{3}), where it has particularly
natural properties. In this paper we are concerned with the question
 as to
how far beyond the regularily one can
go, that is, what is the weakest separation property under which
 an
admissible nearness makes a reasonable
sense. The answer is the {\it subfitness} (also studied under the
 name of
{\it conjunctivity} by Simmons
in \cite{15},  see also \cite{12}) as we show in Section 2 below.

Unlike the separation axiom of symmetry $(x\in\overline{\{ y\}}
\Longrightarrow y \in\overline{\{ x\}})$,
necessary for admitting a nearness in spaces, subfitness is not hereditary.
Thus, the question naturally
arises as to which locales (point-free spaces) have the property
 that all
their sublocales admit a nearness
structure.
This question is treated  in Sections 3 and 4, and a necessary and
sufficient condition is found (Theorem 4.5), in a
form of a ``separation axiom'' which we call {\it sequential regularity}.
Finally, in Section 5 we compare
the sequential regularity with more common
separation axioms, and add a few further remarks.

\vskip 10mm
           \newpage
\centerline{\kapitola 1. Preliminaries}

\bigskip

{\bf 1.1.} A {\it frame} is a complete lattice $L$ satisfying the distributive
 law
$$a\wedge\bigvee S = \bigvee \{a\wedge s\mid s\in S\}$$
for all $a\in L$ and $S\subseteq L$,
and a {\it frame homomorphism} $\varphi\colon L\longrightarrow M$
is a map preserving finite meets including the top (usually denoted
 by 1),
and arbitrary joins including
the bottom (usually denoted by 0). The resulting category will be
 denoted by
$$\bold {Frm}.$$

\medskip
{\bf Example:} The frame ${\frak O} (X)$ of open sets of a topological
 space
$X$, the homomorphism ${\frak O} (f)
\colon {\frak O} (Y) \longrightarrow {\frak O} (X)$ obtained from a
continuous function
$f\colon X\longrightarrow Y$ by
putting ${\frak O} (f) (u) = f^{- 1} (u)$. Thus a contravariant
  functor
$${\frak O} \colon\  \bold {Top} \longrightarrow \bold {Frm}$$
is obtained; the dual of {\bf Frm} is called the {\it category of
 locales},
usually denoted by  {\bf Loc}, and
${\frak O}$ becomes a covariant functor $\bold {Top} \longrightarrow
\bold {Loc}$.

For details on frames see \cite{8}, \cite{16}.

\bigskip

{\bf 1.2.} For $x, y \in L$ write
$$x\prec y$$
if there is a  $z\in L$ such that $x\wedge z = 0$ and $z\vee y
 = 1$. $L$ is
said to be {\it regular} if
$$a = \bigvee \{ x\mid x\prec a\}$$
for each $a\in L$. Note that ${\frak O} (X)$ is regular iff $X$ is
 a regular
space.

\bigskip
{\bf 1.3.} A {\it sublocale} ({\it part} in \cite{7}) of a frame $L$
 is a
surjective frame  homomorphism
\newline $\varphi\colon
L\longrightarrow M$. We write
$$\varphi\sqsubseteq\psi$$
if there is a homomorphism $\gamma$ such that $\gamma\circ\psi =
 \varphi$. Sublocales $\varphi, \psi$
are considered to be
equivalent if $\varphi\sqsubseteq \psi$ and $\psi\sqsubseteq\varphi$.
 The
(equivalence classes of) sublocales
can be represented by the associated congruences
$$x E_\varphi y\ \  \equiv\ \  \varphi (x) = \varphi (y)$$
and one has $\varphi \sqsubseteq\psi$ iff $E_\varphi \supseteq E_\psi$.
 In
consequence, the preorder
$\sqsubseteq$ is complete; the suprema and infima will be denoted
 by
$$\bigsqcup\varphi_i,\ \varphi_1 \sqcup \varphi_2,\
\varphi_1\sqcap \varphi_2\ \
\text{etc.}$$
In particular, the congruence associated with $\bigsqcup\varphi_i$
 is
$\bigcap E_{\varphi_i}$.

The elements $a\in L$ induce the {\it open}\  resp.\  {\it closed}
 sublocales
$$\hat{a} = (x\longmapsto a\wedge x)\colon L\longrightarrow\dr
 a\ \
\text{resp.}\ \ \check{a} =
(x\longmapsto
a\vee x)\colon L\longrightarrow \ur a$$
(where $\dr a = \{ x\mid x\le a\},\ \ur a = \{ x\mid
 x\ge a\}$),
complementing each other.

The closure of a sublocale $\varphi\colon L\longrightarrow M$ is
$$\ol{\varphi} = \check{c} \colon L\longrightarrow\ur c$$
where $c = \bigvee \{ x\mid\varphi (x) = 0\}$, and it is easy to
 see that this
is the smallest closed $\check{a} $ such
that $\varphi\sqsubseteq\check{a}$.

The meet $\varphi\sqcap\hat{a}$  is obtained from the pushout
$$\CD L@>\varphi >>M \\
@V\hat aVV     @VV{(\varphi (a))\ \hat{}}V\\
\dr a@>{}>>\dr\varphi (a).  \endCD$$
In particular,
$$
  \align
  &\varphi\sqcap\hat{a}\not= 0 \quad \text{iff} \quad\varphi (a) \not=
  0, \tag{1.3.1}\\
  &\check{b} \sqcap\hat{a} \not= 0\quad \text{iff}\quad a\not\le
b.
  \endalign
$$
Another formula we will use is that
$$
\check{b} \sqsubseteq\hat{a}\quad \text{ iff}\quad a\vee b = 1.
\tag{1.3.2}
$$

\bigskip
{\bf  1.4.} A {\it cover} of a frame $L$ is a subset $A\subseteq L$ such
 that
$\bigvee A = 1$. We say that $A$
refines $B$ and write $A\le B$ if
$$\forall a\in A\ \ \exists\  b\in B,\ a\le b.$$
For $x\in L$ and a cover $A$ we write
$$A x = \bigvee \{ a\in A\mid a\wedge x\not= 0\}$$
(see, e.g., \cite{13}, \cite{3}). Obviously,
$$A\le B\ \ \& \ \ x\le y \Longrightarrow Ax \le By$$
and
$$
 Ax \le y \quad \Longrightarrow \quad x\prec y. \tag{1.4.1}
$$
More generally, for a sublocale $\varphi\colon L\longrightarrow M$,
$$A\varphi = \bigvee \{ a\in A \mid\hat{a} \sqcap \varphi \not=
 0\} =
\bigvee \{ a\in A\mid\varphi (a)
\not= 0\}.$$
In particular (recall (1.3.1))
$$ A \hat{x} = A x,\ \ A\check{x} = \bigvee\{ a\in A\mid
 a\not\le x\}.  \tag{1.4.2}$$
It is easy to check that
$$ A\varphi = A\ol{\varphi}.
\tag{1.4.3}$$
\bigskip
 {\bf 1.5.  Congruences, nuclei and prenuclei}. If $R\subseteq L \times
 L$
is a relation on $L$,
an $s\in L$ is called {\it saturated} if
$$\forall a, b, c \in L\ \ \ a R b\Longrightarrow (a\wedge c\le s\
\text{iff}\ b\wedge c\le s).$$
The system of saturated elements is closed under meets. If we define
$\kappa  (a)$ as the least
saturated $s\ge a$ we obtain the so called {\it nucleus} (associated
 with
$R$) $\kappa \colon L
\longrightarrow L$, a monotone function satisfying the formulas $a\le
 \kappa  (a), \kappa \kappa  (a) =
\kappa  (a)$ and $\kappa  (a\wedge b) = \kappa   (a) \wedge
\kappa  (b)$.
This $\kappa $  can be viewed  as a sublocale
$$\kappa \colon L\longrightarrow L/R = \{a\mid \kappa  (a)
 = a\}$$
and $\{ (a, b) \mid\kappa  (a) = \kappa  (b)\}$ is the least
congruence containing $R$.

If $a R b \Longrightarrow a\le b$, the nucleus $\kappa $  can be constructed
from a {\it pre-nucleus}
$$\kappa _1 (u) = \bigvee \{ x\wedge b\mid x\wedge a \le u\
 \text{for
some}\  (a, b)\in R\}$$
by setting
$$ \kappa _{\alpha + 1} (u) = \kappa _1 \big(\kappa _\alpha
(u)\big),\ \ \kappa _\alpha (u) =
\bigvee\limits_{\beta < \alpha}\kappa _\beta (u)\ \ \text{for
 limit}\
\alpha,$$
and $\kappa  (u) = \kappa _\alpha (u)$ where $\alpha$ is
 such that
$\kappa _{\alpha + 1} (u) =
\kappa _\alpha (u)$. (See \cite{2}, \cite{8}, \cite{9},
 \cite{10}.)

\bigskip
 {\bf 1.6. Convention:} To unify the notation in the spatial and
 in the
general point-free case, we will
denote open sets of spaces by small case letters ($a, u, x$, etc.),
 open
covers by capitals ($A, B$, etc.) and
points by greek letters ($\xi, \eta$, etc.).

\bigskip
{\bf  1.7.} A {\it nearness on a topological space} $X$ (this has appeared,
 in
essence, as $T$-{\it uniformity}
\cite{11},
for a detailed study of this structure see, e.g., \cite{5},
\cite{6}) is a non-void system $\frak N $
of open covers such that
\roster
\item"(a)" if $A\in\frak N $ and $A$ refines $B$ then $B\in\frak N $,
 and
\item"(b)" for any $A, B\in\frak N $ there is a common refinement
$C\in\frak N $.
\endroster
Similarly, a {\it nearness on a frame} $L$ (cf., e.g., \cite{3})
 is a
non-void system $\frak N $ of covers of
$L$ such that
\roster
\item"(a)" $A\in\frak N \  \&\  A\le B \Longrightarrow B\in\frak N $,
\item"(b)" $A, B\in\frak N  \Longrightarrow A\wedge B = \{ a\wedge
 b\mid
a\in A,\  b\in B\} \in\frak N .$
\endroster
A nearness on a space induces a new topology  by declaring  $u$ to
 be an
$\frak N $-neighbourhood of
a point $\xi$ if $A\xi = \bigcup\{ a\in A\mid\xi\in A\} \subseteq
 u$ for
some $A\in\frak N $. We
say that
the given $\frak N $ is {\it admissible} if this new topology coincides
with the original one. The
admissibility (also in the point-free case) will be discussed in
 Section
2; so far we will only prepare
the notation. We write  $a\US _{\frak N } b$ if there is an $A\in\frak N $
such that $A a\le b$ and,
more generally, for sublocales, $\varphi\US _{\frak N }\psi$ if
$(A\varphi)\ \hat{} \sqsubseteq\psi$ for an
$A\in\frak N $. If clear from the context, the subscript in
$\US _{\frak N }$ may be omitted.

Obviously (recall (1.4.1))
$$a\US  b\ \ \Longrightarrow\ \  a\prec b$$
and in case of an $\frak N $ containing all the finite covers,
$$a\US  b\ \ \ \text{iff}\ \ \ a\prec b$$
(as $\{ x, b\} a\le b$ when $x\wedge a = 0$ and $x\vee b = 1$).

\bigskip
{\bf 1.8. Sobriety and $T_D$:} A space $X$ is said to be
 {\it
sober} if
\roster
\item"{\it(Sob)}"
$X$ is $T_0$ and each meet-irreducible open $u\not= X$ (i.e.,
 a $u$ such
that $u =$\newline $ u_1 \cap u_2 \Longrightarrow
u = u_j$ for some $j$) is of the form  $X\smallsetminus\overline{\{\xi\}}$
for some $\xi\in X$.
\endroster
In particular,

\noindent{\it if $X$ is sober then each frame
homomorphism
$\varphi\colon{\frak O}(Y) \longrightarrow
{\frak O}(X)$ is of the form  ${\frak O}(f)$ for  a uniquely defined
continuous map
$f\colon X\longrightarrow Y$}

\noindent (see, e.g.,
\cite{8}).

A space $X$ is said to be a $T_D$-{\it space} (first defined in
 \cite{1}) if
\roster\item"($T_D$)"
for each $\xi\in X$ there is an open $u\subseteq X$ such that $\xi\in
 u$
and $u\smallsetminus\{\xi\}$ is open.
\endroster
In particular,

noindent{\it a space is $T_D$ iff distinct subspaces $Y\subseteq
 X$
generate non-equivalent
sublocales}
$$(u\longmapsto u\cap Y) \colon\ {\frak O} (X) \longrightarrow {\frak
 O}
(Y).
$$
{see \cite{4}, \cite{14})

{\it (Sob)} is between $T_0$ and $T_2$ and incomparable with $T_1$. $T_D$
 is
between $T_0$ and $T_1$
and incomparable with {\it (Sob)}.
\bigskip
{\bf 1.9. Subfit frames:} A frame is said to be  {\it subfit} (see
\cite{7}) if the following condition holds:
          \roster
\item"{\it (Sf1)}" {\it each open sublocale is a join of closed ones}.
\endroster
This is well-known to be equivalent to the
statement
\roster
\item"{\it (Sf)}" {\it whenever $a\not\le b$ there is a $c$ such that}
$$a\vee c = 1 \not= b\vee c.$$\endroster

(The equivalence {\it (Sf1)} $ \equiv$ {\it (Sf)} is very easy to see: in view
 of
(1.3.2), {\it (Sf1)} can be rewritten as
$$( x\wedge a\not=  y\wedge a\ \  \Longrightarrow\ \
  (\exists
  c,\
a\vee c = 1\ \text{and}\
x\vee c \not= y\vee c). \tag{\it Sf2}$$
Now if $a\not\le b$ consider in {\it (Sf2)} $x = a, y = b$ to obtain
{\it (Sf)}. On
the other hand, if $x\wedge a \not= y
\wedge a$ consider in {\it (Sf)} $x\wedge a \not\le x\wedge y \wedge
 a$ to
obtain {\it (Sf2)}.)

It should be noted that {\it (Sf)} was called {\it conjunctivity} in
 \cite{15}
for frames and
{\it weak regularity}
for spaces \cite{11}.
\vskip 10mm

\centerline{\kapitola 2. Admissible nearnesses}

\bigskip

{\bf 2.1.} Recall 1.7. Explicitly, a nearness $\frak N $ is admissible
 on a
space $X$ if

\centerline
{for each open $u\subseteq X,\ u= \{\xi \mid A \{\xi\} \subseteq
 u$ for an
$A\in\frak N \}$,}

\noindent or, in a reformulation more handy for a pointfree treatment:
\roster
\item"(2.1.1)"  for each open $u\subseteq X,\ u = \bigcup\{ \varphi\subseteq
 X\mid
A\varphi\subseteq u$ for
an $A\in\frak N \}$. \endroster

The {\it admissibility in the point-free case} has been so far studied
  as
the condition (see \cite{3}, cf.
\cite{13})
$$\forall u\in L,\ u = \bigvee \{ x\in L\mid A x\le u\ \text{for
 an}\
A\in\frak N \}.$$
This is correct -  that is, in accordance with the admissibility
 in spaces
 - for regular $L$. As the formula
implies regularity (see (1.4.1)), however, it cannot be used in a
 more
general case. Thus, we will use,
extending the formula (2.1.1) above, the following
                                    \medskip
\noindent{\bf Definition:} A nearness $\frak N $ is {\it admissible on a frame
 $L$}
if for each $u\in L$,
$$\hat{u} = \bigsqcup \{\varphi \mid\varphi \ \text{sublocale
 of} \  L,\
A\varphi\le u \ \text{for an} \
A\in\frak N.
 \tag{2.1.2}$$

\bigskip
{\bf 2.2.} As $A\varphi = A\ol{\varphi}$ (recall (1.4.3)), the formula
 (2.1.2)
can be rewritten to
$$\forall \ u\in L,\quad \hat{u} = \bigsqcup \{\check{a} \mid\exists\
\forall\in\frak N ,\
A\check{a}\le u\}, \tag{2.2.1}$$
and since the inclusion $\sqsupseteq$ is trivial, the inclusion
$\sqsubseteq$ is what has to be checked
in a concrete case.

More explicitly, by 1.3, the condition in (2.2.1) can be expressed
 by the
requirement that

\noindent{\it if for all $c\in L$ such that there is an $A\in\frak N $ with
$A\check{c} = \bigvee \{ a\in A,\
a\not\le c\} \le u$ one has $c\vee x = c\vee y$, then $x\wedge
 u = y\wedge u$.}

\medskip
We have a trivial

\proclaim{\bf Observation:}  If $\frak N $ is admissible and $\frak N
 \subseteq
\frak M$ then $\frak M$ is
admissible.\endproclaim

\noindent Thus, $L$ admits a nearness iff the conditions above hold for $\frak N =
\{\text{all covers}\}$.

\bigskip
\proclaim {\hskip2mm{\bf 2.3.} Theorem}  A frame $L$ admits a nearness
 iff it is
 subfit.\endproclaim

\demo{Proof} ($\Longrightarrow$): follows immediately from (2.2.1).

($\Longleftarrow$): We will use the formula {\it (Sf)} from 1.9 for prove
(2.2.1) for the system of all covers. Thus, let $x\vee a = y\vee a$ for any
$a$ such that there is a cover $A$
with $A\check{a} = \bigvee\{ b\in A \mid b\not\le a\} \le u$. We
 want to
prove that then $x\wedge u =
y\wedge u$. Let, say, $x\wedge u\not\le y\wedge u$. By {\it (Sf)} there
 is a
$c$ with $1 = (x\wedge u) \vee c
\not= (y\wedge u) \vee c$ (in particular, $u\vee c = 1$). But
 $\{ x\wedge
u, c\} \check{c} \le x\wedge u \le u$
and hence $x\vee c = y\vee c$ and we get a contradiction
$$(y\wedge u) \vee c = y\vee c = x\vee c = (x\vee c )\wedge
 (u\vee c) =
(x\wedge u) \vee c = 1.$$            \sq
\enddemo
\bigskip

{\bf 2.4.} Recall that a nearness $\frak N $ is said to be {\it regular}
(\cite{6}; {\it strong} in \cite{3}) if
for each $A\in\frak N $ also
$$\wt{A} = \{ a\mid a\US _{\frak N } b\ \text{for some}\ b\in
 A\}
\in\frak N .$$
We have
\proclaim{ Theorem}  $L$ admits a regular nearness iff it is regular.
\endproclaim
\demo{Proof} ($\Longleftarrow$): The system of all covers of a regular
frame is a regular nearness.

($\Longrightarrow$): We have
$$\wt{A}\varphi = \bigvee \{ a\in\wt{A} \mid\varphi (a) \not=
 0\}
\le\bigvee \{a \mid a\US _{\frak N }
A\varphi\}$$
and hence
$$u = \bigvee\{ A\varphi\mid A\in\frak N ,\ A\varphi\le u\} \le\bigvee
 \{
a\mid a
\US _{\frak N } u\}.$$ \sq
\enddemo

\bigskip
{\bf 2.5.} For the spatial case, the equivalence
 from
2.3 is analoguous to that from
\cite{11}, Theorem 1. Another well-known necessary and sufficient
condition for a space $X$ to admit a
nearness (see \cite{5}, \cite{6}) is that  it satisfied
 the
symmetry axiom
$$\xi\in\overline{\{\eta\}}\ \ \ \text{iff}\ \ \ \eta\in\overline{\{\xi\}}$$
(which is the same as saying that the $T_0$-modification of $X$
 is $T_1$,
and hence we will refer to this
property as
$$T_{1-0}.)$$
Now in view of 2.3 one might expect that this $T_{1-0}$ is equivalent
 to
the subfitness of ${\frak O} (X)$.
But this is false, and the discrepancy should be explained.

Recall 1.8: If $X$ were not $T_D$, two subsets may be distinct although
 the
associated congruence coincides.
In particular, this can happen with $u$ and
$\bigcup\{\varphi$ closed $\mid\varphi\subseteq u\}$. On the other
hand, under $T_D,\  T_{1-0}$ is indeed equivalent to subfitness.
 More
exactly, one has                 $$
  \align
T_1 \ &\equiv \  \text{\it (Sf)} \ \&  \ T_D,\quad\text{and}\\
T_{1-0} \ &\equiv \ \text{\it(Sf)} \ \&  \ T_{D-0}
  \endalign
$$
where $T_{D-0}$ stands for the condition that the $T_0$-modification
 of
the space is $T_D$.
           \vskip 10mm

\centerline{\kapitola 3.  Sobriety as a partly hereditary subfitness}

\bigskip

{\bf 3.1.}
{\bf Example:} Consider the following spaces:
\roster
\item"$X$:"the ordinal $\omega$ with the cofinite topology,
\item"$Y$:"$\omega + 1$ with open $\emptyset$ and $Y\smallsetminus K$ where
 $K$
is finite and $\subseteq\omega$,
\item"$S$:"the Sierpi\'nski space represented as the subspace of $Y$ generated
 by
the subset $\{ 0, \omega\}$,
\endroster

\noindent and the mappings:
          \roster
\item"$j$:" the embedding $S\subseteq Y$,
\item"$k$:" the embedding $X\subseteq Y$,
\item "$f,f'$:" the mappings $f\colon X\longrightarrow X$ resp. $f'\colon
Y\longrightarrow Y$ sending $0$ to $0$,
$2 n + 1$ to $2 n + 2,\  2 n + 2$ to $2 n + 1$, and $\omega $ to
 $\omega$,
 \endroster
$$\iota = {\frak O} (j), \kappa  = {\frak O} (k),\ \varphi = {\frak
 O} (f).$$

Obviously
\roster
\item"(3.1.1)" \ \ $\kappa$ is an isomorphism,
\item"(3.1.2)" \ \ $j$ is the equalizer of $f'$ and $id$,
\item"(3.1.3)" \ \ $\iota\circ\kappa ^{-1}$ is the coequalizer of $\varphi$ and
 $id$.     \endroster
((3.1.3) does not follow from (3.1.2) but it is easy to check).

Now one sees that:


{\it Subfitness is not inherited by subspaces} (see $S\subseteq Y$),
 {\it and not even by equalizers with
subfit range} (see (3.1.2)). {\it A $T_1$-space} (although all of
 its
subspaces are $T_1$, and hence subfit)
{\it can have a spatial sublocale that is not subfit and, again,
 even a
coequalizer with a subfit domain}
(see (3.1.3)).


In this section we will investigate the question of subfitness of
 spaces
hereditary with respect to the
spatial sublocales. In the next section, the general hereditarity
 question
will be treated.

\bigskip
{\bf 3.2.} The following is certainly folklore and  can be left to the
 reader:

\proclaim{ Lemma}  A space $Y$ is $T_0$ iff the embeddings $f\colon
X\longrightarrow Y$ are exactly those
continuous maps $f$ for which ${\frak O} (f) \colon {\frak O} (Y)
\longrightarrow {\frak O}  (X)$ are onto.\endproclaim

\bigskip
{\bf 3.3} Recall that an element $a\in L$ is {\it prime} if the following
 holds:
$$a\not= 1 \quad\text{and}\quad (x\wedge y\le a\Longrightarrow
x\le a\ \text{or} \ y\le a).$$

\medskip
\proclaim{ Lemma}  Let $a, b\in L$ be prime and let $a \lvertneqq
 b$. Then
 the Sierpi\'nski space is a sublocale of $L$.\endproclaim

\demo{Proof} Represent the Sierpi\'nski frame (${\frak O} (S)$ of
 a
Sierpi\'nski space $S$) as
$$\Bbb{S} = \{
0 < \frac{1}{2} < 1\}.$$
 Define $\varphi\colon L\longrightarrow \Bbb{S}$
 by
setting
$$
\varphi (x)= \cases 0 \quad\text{if} \ x\le a,\\
\frac{1}{2} \quad \text{if}\  x\not\le a \ \text{and}\ x\le b,\\
1 \quad\text{if} \ x\not\le b.   \endcases
$$
It is easy to check that $\varphi$ is a frame homomorphism.\sq
\enddemo

\bigskip
\proclaim {\hskip2mm{\bf 3.4} Theorem}
 Let $X$ be a $T_1$-space. Then the following
statements are equivalent:
\roster
\item"(i)" $X$ is sober,
\item"(ii)" each $T_0$-spatial sublocale is $T_1$,
\item"(iii)" each spatial sublocale is subfit,
\item"(iv)" $X$ has no Sierpi\'nski sublocale.
\endroster  \endproclaim

\demo{Proof}  (i) $\Longrightarrow$ (ii): Let $X$ be sober.
 Suppose
we have a surjective homomorphism
$\varphi\colon\ {\frak O} (X)
\longrightarrow {\frak O} (Y)$, with $Y$ a $T_0$-space. Then, by
 3.2, $Y$
is a subspace and hence it is $T_1$.

 (ii) $\Longrightarrow$ (iii): Consider a surjective
$\varphi\colon{\frak O} (X) \longrightarrow{\frak O} (Y)$
 and the $T_0$-modification $p\colon Y\longrightarrow\wt{Y}$
 of $Y$ .
Then ${\frak O} (p)$ is an isomorphism. Considering ${\frak O} (p)^{-1}
\circ\varphi$ we see that
$\wt{Y}$ is $T_1$,
hence ${\frak O} (\wt{Y})$ is subfit, and hence ${\frak O} (Y)$
 is subfit.

 (iii) $\Longrightarrow$ (iv) is trivial.

 (iv) $\Longrightarrow$ (i): Let $X$ not be sober. Then there
 is a
prime $u\in{
\frak O} (X)$ unequal to
$X\smallsetminus
\{ \xi\}$ for any $\xi$. Choosing $\xi\in X\smallsetminus u$ we obtain
 prime
$$u \lvertneqq v = X\smallsetminus \{ \xi\}$$
and hence there is a Sierpi\'nski sublocale by 3.3.\sq
\enddemo

\vskip 10mm

\centerline{\kapitola 4. Hereditary subfitness }

\bigskip

{\bf 4.1.}
In this section we will characterize the frames $L$ such that
 each
sublocale of $L$ admits a nearness.
In view of 2.2, this amounts to the {\it hereditary subfitness},
 that is,

{\it (HSf)}\ \  for every sublocale $\varphi\colon L\longrightarrow M,\
 M$ is
subfit.

\proclaim {\hskip2mm{\bf 4.2} Lemma}
 A frame $L$ is hereditarily subfit iff for
 each
sublocale $\varphi\colon L
\longrightarrow M$ the $M$ satisfies
\roster
\item"{\it (WSf)}"
 if $a\not= 0$ in $M$ then there is $x\in L$ such that $x\not=
 1$ and
$x\vee a = 1$.\endroster
\endproclaim

\demo{Proof} Indeed, if $L$ has the property indicated then, by
 force of
the sublocales $\check{b} = (
x\longmapsto x\vee b)\colon L\longrightarrow\ur b$ it is, first,
subfit, and as the property is
hereditary, it is hereditarily subfit.  \sq
\enddemo

{\bf Note:} Obviously, {\it (WSf) can be reformulated as the implication}
$$(x\vee a = 1 \ \ \Longleftrightarrow \ \ x = 1)\ \  \Longrightarrow
  \ \
a = 0.  \tag{4.2.1}$$

\bigskip
 {\bf 4.3. Sequential regularity:} Let $\alpha$ be an ordinal.
We say that $L$ is $\alpha$-{\it sequentially regular} if
                         \roster
\item"{\it (SR$\alpha$)}"
for each $a\in L$ there is a non-decreasing sequence
$(a_\beta)_{\beta\le\alpha}$ such that
$a_0 = 0,\ a_{\beta + 1} \le \bigvee\{ u\mid\exists\  x,\  x\vee
 a = 1\ \&\
 u\wedge x\le a_\beta\}$,
$a_\beta = \bigvee\limits_{\gamma < \beta}a_\gamma$\  for limit
 ordinals
$\beta$, and $a = a_\alpha$.
\endroster
$L$ is {\it sequentially regular} if
          \roster
\item"{\it (SR)}" $L$ is $\alpha$-sequentially regular for some $\alpha$.
\endroster
\medskip
{\bf Note:} One has $a_1 \le\bigvee \{ u\mid\exists\  x,\  x\vee
 a = 1\ \&\
 u\wedge x = 0\}$. Thus, 1-regularity is
the same as regularity.

\bigskip
\proclaim {\hskip2mm{\bf 4.4.} Lemma}  (SR$\alpha$)
 is hereditary and implies (hereditary)
subfitness. \endproclaim

\demo{Proof} Consider a sublocale $\varphi\colon L\longrightarrow
 M$ and
a $\varphi (a) \in L$.
Let $(a_\beta)_{\beta\le\alpha}$ be the sequence from {\it (SR$\alpha$)}.
 Then it
is easy to check that $\big(\varphi
(a_\beta)\big)_{\beta\le\alpha}$ is the required sequence for $\varphi
 (a)$.

As for the second statement, by 4.2 it suffices to show that
{\it(SR$\alpha$)}
implies {\it (WSf)}. Let  $a\not = 0$
and let
$(a_\beta)_{\beta\le\alpha}$ be the associated sequence. Then there
 is a
$\beta$ such that
$$a_\beta \lvertneqq a_{\beta + 1} \le\bigvee \{ u\mid\exists\  x,\
  x\vee
a = 1\  \&\  u\wedge x\le a_\beta\}.$$
In particular there has to be a $u$ and an $x$ such that
$$x\vee a = 1,\ \ u\wedge x \le a_\beta\ \ \text{and}\ \ u\not\le
 a_\beta.$$
Then $x$ cannot be $1$.                                          \sq
\enddemo

\proclaim{\hskip2mm {\bf 4.5.} Theorem}  The following statements
 on a frame $L$ are equivalent:
\roster
\item"(i)" Each sublocale of $L$ admits a nearness,
\item"(ii)" $L$ is hereditarily subfit,
\item"(iii)" $L$ is sequentially regular,
\item"(iv)" each congruence on $L$ is uniquely determined by the class
 containing $1$.
\endroster \endproclaim

\demo{Proof}  (i) $\equiv$ (ii) by 2.2.

(ii) $\Longrightarrow$ (iii): In view of 4.2 and 4.3 it suffices
 to
prove that if each sublocale of $L$
satisfies {\it (WSf)} then $L$ is {\it (SR)}. Consider an $a\in L$ and the
sublocale $L/_\sim$ where $\sim$ is the
congruence generated by
$$R = \{ (x, 1) \mid x\vee a = 1\}.$$
Recall 1.5. $L/_\sim$ consists of all the $R$-saturated elements
 of $L$.

\noindent {\bf Claim:} {\it Each $y\ge a$ is saturated.}

\noindent (Indeed, if $x\vee a = 1$ and $u\wedge x\le y$ we have
$$u = (x\vee a)\wedge u = (x\wedge u) \vee (a\wedge u)\le y\vee
 a = y.)$$
By (4.2.1), if $L/_\sim$ satisfies {\it (WSf)} we have
$$a\sim 0$$
and since by the above Claim $a$ is saturated, this amounts to
$$\kappa  (0) = a$$
where $\kappa $ is the nucleus associated with $R$. The prenucleus of
 $R,\
\kappa _1$, is defined by
$$\kappa _1 (b) = \bigvee \{ u\mid\exists\  x,\ \ x\vee a =
 1,\ u\wedge
x\le b\}.$$
Set $a_0 = 0,\ \ a_\beta = \kappa _\beta (0)\ \text{for} \
 \beta > 1.$

(iii) $\Longrightarrow$ (iv): Let $E_1,\ E_2$ be congruences
 on $L$,
let $\varphi_j \colon\ L
\longrightarrow L/E_j$ be the associated sublocale homomorphisms.
 Let $x
E_1 1$ iff $x E_2 1$ and let
$a E_1 b$. We have $\varphi_2 (a_0) = 0 \le \varphi_2 (b)$. Let
 $\varphi_2
(a_\beta) \le\varphi_2 (b)$.
If $u$ is such that there is an $x$ with $x\vee a = 1$ and $u\wedge
 x \le
a_\beta$, we have $\varphi_1
(x\vee b) = \varphi_1 (x\vee a) = 1$, hence $\varphi_2 (x\vee
 b) = 1$ and
$$\varphi_2 (u)\ \  =\ \  \varphi_2 (u\wedge (x\vee b))\
 \  \le\ \
\varphi_2 (a_\beta) \vee \varphi_2 (b)\ \  \le\ \
\varphi_2 (b).$$
Thus, $\varphi_2 (a_{\beta + 1} )\le \varphi_2 (b)$ and since the
 limit
step is trivial, $\varphi_2 (a) \le
\varphi_2 (b)$. Similarly, $\varphi_2 (b) \le\varphi_2 (a)$ and hence
$\varphi_2 (b) =
 \varphi_2 (a)$.

(iv) $\Longrightarrow$ (ii): It is easy to see that the property
 in
(iv) is hereditary and hence, by
4.2, it suffices to prove that (iv) $\Longrightarrow$ {\it(WSf)}. Recall
(4.2.1). If $x\vee b = 1\ \equiv\
x = 1, \  \check{b}=(x\mapsto x\vee b)$ is the identity and hence $b = 0$.    \sq
\enddemo
          \vskip 10mm

\centerline{\kapitola 5.  More on sequential regularity }

\bigskip

{\bf 5.1.}
 In this section all the concrete examples will be spaces. In
 these we will forget the
convention 1.6 and use the more common notation (small case roman
 letters
for points, capitals for subsets).

The topologies in the examples will be indicated as $\tau$, and the
closures by overline.

 $\Bbb{R}$ (resp. $\Bbb{Q}$, resp. $\Bbb{N}$) is the {\it set} of
 real (resp. rational, resp. natural)
numbers, and the intervals in $\Bbb{R}$ will be denoted by
$$[ a, b],\ \ ] a, b [\  \ \text{etc.}$$
The Urysohn separation axiom (requiring separation of distinct points
 by
{\it closed} neighbourhoods will be
indicated by
$$\text{\it(U)},$$
and regularity by {\it (R)}.

\bigskip
{\bf 5.2.} Let $(a_\beta)_{\beta\le\alpha}$ be the sequence from the definition
of {\it (SR$\alpha$)}. Define
$$\align
&a'_0 = 0,\ \ a'_{\beta + 1} = \bigvee\{ u\mid \exists\  x,\
 x\vee a = 1,\
u \wedge x\le a'_\beta\}\\
 &a'_\beta = \bigvee\limits_{\gamma < \beta}a'_\gamma\ \text{in
 the limit
case}.
\endalign $$
Thus $(a'_\beta)_{\beta < \alpha}$ has again the required property.
 Thus we can reformulate
{\it(SR$\alpha$)} by {\it defining}, for any $a\in L$, by induction
$$
a_0 = 0,\ \ a_{\beta + 1} = \bigvee\{ u\mid\exists\  x,\ x\vee
 a = 1,\
u\wedge x\le a_\beta\}\ \text{etc.},
$$
and stating that $L$ has {\it(SR$\alpha$)}  if for each $a\in L,\ a =
 a_\alpha$.

In a space $X$, the formula (for $A\subseteq X$ open)
$$A_{\beta + 1} = \bigcup \{ U\mid\exists\  V,\ A\cup V = X\
  \&\ V \cap
U\subseteq A_\beta\}$$
is equivalent to a handier
$$A_{\beta + 1} = \bigcup \{ U\mid\overline{U\smallsetminus
A_\beta} \subseteq A\}   \tag{5.2.1}$$
(think of $V = X\smallsetminus (\overline{U\smallsetminus A_\beta})$),
 and
this way it will be viewed in the
constructions to follow.

\bigskip
\proclaim {\hskip2mm{\bf 5.3} Theorem}  If $\alpha < \beta$ then
{\it (SR$\alpha$)}
 is  strictly stronger than {\it (SR$\beta$)}. \endproclaim

\demo{Proof} We will produce spaces $(X_\alpha, \tau_\alpha)$ and
 open
sets $A\in\tau_\alpha$ such that for
$\beta < \alpha,\ A_\beta\not= A$, and $A_\alpha = A$.

Set
$$X_\alpha = \{ (\beta, n) \mid \beta < \alpha,\ n < \omega\}$$
with $B\subseteq X_\alpha$ in $\tau_\alpha$ iff
$$(\beta, n) \in B \Longrightarrow\exists\  m\in\omega,\ \{ (\gamma,
 k)
\mid\gamma < \beta,\ k > m\}
\subseteq B.$$
Thus in particular any neighbourhood of $(\beta, n)$ meets any $\{(\gamma,
m) \mid m\in\omega\}$ with
$\gamma < \beta$ while it does not need to meet $\{ (\beta, m) \mid
m\in\omega\}$ in anything but the
$(\beta, n)$ itself. Hence, if we set $A = \{ (\beta, n) \in X_\alpha
 \mid
n\not= 0\}$ we get
$A_\gamma = \{ (\beta, n) \mid\beta < \gamma,\ n\in\omega\} $ for
 $\gamma <
\alpha$, and $A_\alpha = A$.                                     \sq
\enddemo

\proclaim{\hskip2mm {\bf 5.4.} Theorem} Between the mentioned separation axioms
 the
following implications and
(besides compositions) no others hold:

$$
  \matrix
  && \text{\it (SR2)} \ \& \ T_1&\nadsipku{0}{10mm}{}&\text{\it (SR)}
   \ \& \ T_1&&\\
  &\nadsipku{10}{10mm}{}&&&&\nadsipku{86}{10mm}{}&\\
  \text{\it (R)} \ \&  \ T_1 &&&&&           &          \text{\it (Sob)} \ \& \ T_1\\
    &\nadsipku{86}{10mm}{}&&&&\nadsipku{10}{10mm}{}&\\
      &&\text{\it (U)}&\nadsipku{0}{10mm}{}&T_2&&
      \endmatrix
$$

   \endproclaim
\demo{Proof} The only less standard implication in the scheme
{\it (SR)} $\& \ T_1$ $\Longrightarrow$ {\it (Sob)} $\& \ T_1$
follows from 4.5 and 3.4. Thus, it remains to be proved that
{\it (SR2)} $\&\ T_1$ does not imply $T_2$ and that
{\it (U)} does not imply {\it (SR)}.

{\it (SR2)} $\& \ T_1 \ \Longrightarrow\!\!\!\!\!\!\mid\ \quad
(T_2)$: Consider
 $X =
\Bbb{N} \cup \{ a, b\}$ with
$a\not= b,\  a, b\not\in \Bbb{N}$, and the topology $\tau$,
$$A\in\tau\ \ \ \text{iff}\ \ \ (A\cap \{ a, b\} \not= 0 \Longrightarrow
\exists n\in\Bbb{N},\ \{m\mid m\ge n\}
\subseteq A)$$
Obviously, $(X, \tau)$ is not Hausdorff. Using the trivial observation
 that
the $\{ n\}$  with $n\in\Bbb{N}$ are
clopen we easily deduce that
\roster
\item if $A\subseteq\Bbb{N}$ or $\{ a, b\} \subseteq A,\ \ A =
 A_1$,
\item if $A\cap \{ a, b\} = \{ x\}\ \ (x = a$ or $b)$ then
$A_1 = A\smallsetminus \{ x\}$ and $A_2 = A$.
\endroster
Thus, $( X, \tau)$ satisfies {\it (SR2)}.

$(U) \ \Longrightarrow\!\!\!\!\!\!\mid$ \ \quad
{\it (SR)}: Consider $(\Bbb{R},
 \tau)$ with
  $$
A\in\tau\ \ \text{iff}\ \  \cases
\forall a\in A\cap (\Bbb{R}\smallsetminus\Bbb{Q})\ \ \exists\ \varepsilon
>0 \ \text{s.t.} \ ]a - \varepsilon,
a + \varepsilon [\  \subseteq A, \ \text{and}\\
\forall a\in A\cap\Bbb{Q}\ \ \ \exists\ \varepsilon > 0 \ \text{s.t.}
\ Q\cap ]a -\varepsilon, a + \varepsilon [
\ \subseteq A.
\endcases $$
Obviously $(\Bbb{R}, \tau)$ is Urysohn; but it is not {\it (SR)} as the
 closure
of any $\Bbb{Q}\  \cap\  ] a, b [$ equals
$[ a, b ]$ and hence, in particular, $\Bbb{Q}_\alpha
= \emptyset$
 for all
$\alpha$.   \sq

\enddemo

\vskip10mm
  
\Refs
\ref \no 1\by  C.E. Aull \&\ W.J. Thron\paper
Separation axioms between
 $T_0$
and $T_1$ \jour Indag. Math.\vol 24\yr 1962\pages 26-37 \endref
\ref \no 2\by B. Banaschewski\paper Compactification of frames
\jour Math. Nachr.\vol 149\yr 1990\pages 105-116\endref
\ref \no 3\by B. Banaschewski \&\ A. Pultr
\paper Cauchy points in uniform and nearness frames \paperinfo
to appear in Quaest. Math. \endref
\ref \no 4\by  B.\ Banaschewski \&\ A.\ Pultr\paper Variants of openness
 \jour Appl. Categ. Structures\vol 2\yr 1994\pages 331-350 \endref
\ref \no 5\by H. Herrlich\paper A concept of nearness
\jour Gen. Top. Appl.\vol 5\yr 1974\pages 191-212 \endref
\ref \no 6\by H. Herrlich\paper "Topologie II: Uniforme R\"aume"
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 \ref \no 8\by P.T. Johnstone\paper "Stone Spaces"
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 No. 309 (September 1984)
  \endref
\ref \no 10\by I.Kr{\'\i{}}{\v{z}}\paper A direct description of uniform completion
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Press, Cambridge 1989\endref


\vskip10mm


$$
  \topfoldedtext{Horst Herrlich,
  \newline Fachbereich 3,
  \newline Universit\"at Bremen,
  \newline Bibliothekstr. 1,
  \newline D 28359 BREMEN,
  \newline Germany}
  {}\qquad\qquad{}\qquad\qquad
    \topfoldedtext\foldedwidth{50mm}{Ale{\v{s}} Pultr,
  \newline Department of Applied Mathematics,
  \newline Charles University,
  \newline Malostransk{\'e} n{\'a}m. 25,
  \newline CZ 11800 PRAHA 1,
  \newline Czech Republic}
$$
\enddocument




