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\begin{document}
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\put(143,66){\makebox(0,0)[tl]{\footnotesize Proceedings of the Ninth Prague Topological Symposium}}
\put(143,50){\makebox(0,0)[tl]{\footnotesize Contributed papers from the symposium held in}}
\put(143,34){\makebox(0,0)[tl]{\footnotesize Prague, Czech Republic, August 19--25, 2001}}
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\title{On Tychonoff-type hypertopologies}
\author{Georgi Dimov}
\address{Department of Mathematics\\
University of Sofia\\
Blvd. J. Bourchier 5\\
1126 Sofia, Bulgaria}
\email{gdimov@fmi.uni-sofia.bg}
\author{Franco Obersnel}
\author{Gino Tironi}
\address{Department of Mathematical Sciences and Computer Science\\
University of Trieste\\
Via A. Valerio 12/1\\
34127 Trieste, Italy}
\email{obersnel@mathsun1.univ.trieste.it}
\email{tironi@univ.trieste.it}
\begin{abstract}
In 1975, M. M. Choban \cite{C} introduced a new topology on the set of all
closed subsets of a topological space, similar to the {\em Tychonoff
topology} but weaker than it. In 1998, G. Dimov and D. Vakarelov
\cite{DV} used a generalized version of this new topology, calling it {\em
Tychonoff-type topology}.
The present paper is devoted to a detailed study of Tychonoff-type
topologies on an arbitrary family $\MM$ of subsets of a set $X$.
When $\MM$ contains all singletons, a description of all Tychonoff-type
topologies $\OO$ on $\MM$ is given.
The continuous maps of a special form between spaces of the type
$(\MM,\OO)$ are described in an isomorphism theorem.
The problem of {\em commutability between hyperspaces and subspaces with
respect to a Tychonoff-type topology} is investigated as well.
Some topological properties of the hyperspaces $(\MM,\OO)$ with
Tychonoff-type topologies $\OO$ are briefly discussed.
\end{abstract}
\subjclass[2000]{Primary 54B20, 54B05; Secondary 54B30, 54D10, 54G99}
\keywords{Tychonoff topology, Tychonoff-type topology, T-space,
commutative space, $\OO$-commutative space, $\MM$-cover, $\MM$-closed
family, $P_\infty$-space}
\thanks{The first author was partially supported by a Fellowship for
Mathematics of the NATO-CNR Outreach Fellowships Programme 1999, Bando
219.32/16.07.1999.}
\thanks{The second and the third authors were supported by the National
Group ``Analisi reale'' of the Italian Ministry of the University and
Scientific Research at the University of Trieste.}
\thanks{Georgi Dimov, Franco Obersnel and Gino Tironi,
{\em On Tychonoff-type hypertopologies},
Proceedings of the Ninth Prague Topological Symposium, (Prague, 2001),
pp.~51--70, Topology Atlas, Toronto, 2002; {\tt arXiv:math.GN/0204121}}
\maketitle
\section{Introduction}
In 1975, M. M. Choban \cite{C} introduced a new topology on the set of all
closed subsets of a topological space for obtaining a generalization of
the famous Kolmogoroff Theorem on operations on sets.
This new topology is similar to the {\em Tychonoff topology} (known also
as {\em upper Vietoris topology}, or {\em upper semi-finite topology}
(\cite{M}), or {\em kappa-topology}) but is weaker than it.
In 1998, G. Dimov and D. Vakarelov \cite{DV} used a generalized version of
this new topology for proving an isomorphism theorem for the category of
all Tarski consequence systems.
This generalized version was called {\em Tychonoff-type topology}.
The present paper is devoted to a detailed study of Tychonoff-type
topologies on an arbitrary family $\MM$ of subsets of a set $X$.
When $\MM$ is a {\em natural family}, i.e.\ it contains all singletons, a
description of all Tychonoff-type topologies $\OO$ on $\MM$ is given (see
Proposition~\ref{char Tych}).
For doing this, the notion of {\em T-space} is introduced.
The natural morphisms for T-spaces are not enough to describe all
continuous maps between spaces of the type $(\MM,\OO)$, where $\MM$ is a
natural family and $\OO$ is a Tychonoff-type topology on it; we obtain a
characterization of those continuous maps which correspond to the
morphisms between T-spaces.
This is done by defining suitable categories and by proving that these
categories are isomorphic (see Theorem~\ref{isomorphism}).
In such a way we extend to any natural family $\MM$ on $X$ the
corresponding result obtained in \cite{DV} for the family $\FF in(X)$
of all finite subsets of $X$.
We investigate also the problem of {\em commutability between hyperspaces
and subspaces with respect to a Tychonoff-type topology}, i.e.\ when the
hyperspace of any subspace $A$ of a topological space $Y$ is canonically
representable as a subspace of the hyperspace of $Y$.
Such investigations were done previously by H.-J. Schmidt \cite{S} for the
lower Vietoris topology, by G. Dimov \cite{D1, D2} for the Tychonoff
topology and for the Vietoris topology, and by B. Karaivanov \cite{K} for
other hypertopologies.
We study also such a problem for a fixed subspace $A$ of $Y$.
Some results of \cite{D1, D2, Se} are generalized.
Finally, we study briefly some topological properties (separation axioms,
compactness, weight, density, isolated points, $P_\infty$) of the
hyperspaces $(\MM,\OO)$ with Tychonoff-type topologies $\OO$.
Some results of \cite{F, DV} are generalized.
Let us fix the notations.
\begin{notas}\label{initial notation}
We denote by $\omega$ the set of all {\em positive} natural numbers, by
$\RRRR$ --- the real line, and by $\Z$ --- the set of all integers.
We put $\NNNN=\omega\cup\{0\}$.
Let $X$ be a set.
We denote by $\PP(X)$ the set of all subsets of $X$.
Let $\MM, \AA \subseteq\PP(X)$ and $A\subseteq X$.
We will use the following notations:
\begin{itemize}
\item
$A_\MM^+:=\{M\in\MM : M\subseteq A\};$
\item
$\AA^+_\MM:=\{A^+_\MM : A\in\AA\};$
\item
$\FF in (X):=\{M\subseteq X : |M|<\aleph_0\}$;
\item
$\FF in_n (X):=\{M\subseteq X : |M|\le n\}$, where $n\in\omega$.
\end{itemize}
We will denote by $\AA^\cap$ (respectively by $\AA^\cup$) the closure
under finite intersections (unions) of the family $\AA$.
In other words,
\begin{itemize}
\item
$\AA^\cap:=\{\bigcap_{i=1}^k A_i : k\in \omega, A_i\in\AA\}$ and
\item
$\AA^\cup:=\{\bigcup_{i=1}^k A_i : k\in \omega, A_i\in\AA\}.$
\end{itemize}
Let $(X,\TT)$ be a topological space.
We put
\begin{itemize}
\item
$\CC L(X):=\{ M\subseteq X: M$ is closed in $X,\ M\not=\emptyset\}$
and
\item
$\CC omp(X):=\{ M\subseteq X: M { \rm \ is \ compact}\}$.
\end{itemize}
The closure of a subset $A$ of $X$ in $(X,\TT)$ will be denoted by $cl_X
A$ or $\ovl{A}^X$; as usual, for $U\sbe A\sbe X$, we put
\begin{itemize}
\item
$Ex_{A,X}U:=X\stm cl_X(A\stm U)$.
\end{itemize}
By a {\it base} of $(X,\TT)$ we will always mean an open base.
The weight (resp., the density) of $(X,\TT)$ will be denoted by $w(X,\TT)$
(resp., $d(X,\TT)$).
If $\CC$ denotes a category, we write $X\in |\CC|$ if $X$ is an object of
the category $\CC$.
For all undefined here notions and notations, see \cite{E} and \cite{J}.
\end{notas}
\section{Hypertopologies of Tychonoff-type}
\begin{fact}\label {intersection}
Let $X$ be a set and $\MM,\AA \subseteq\PP(X)$.
Then:
\begin{itemize}
\item[(a)]
$\bigcap \AA^+_\MM = \left(\bigcap \AA\right)^+_\MM;$
\item[(b)]
$A\subseteq B$ implies that $\Ap\subseteq B^+_\MM$ for all
$A$, $B \subseteq X$.
\end{itemize}
\end{fact}
\begin{defi}\label{Tych top}
Let $(X,\TT)$ be a topological space and let $\MM\subseteq \PP(X)$.
The topology $\OO_\TT$ on $\MM$, having as a base the family $\TT^+_\MM$,
will be called {\it Tychonoff topology on $\MM$ generated by
$(X,\TT)$}. When $\MM=\CC L(X)$, then $\OO_\TT$ is just the classical
Tychonoff topology on $\CC L(X)$.
Let $X$ be a set and $\MM\subseteq \PP(X)$.
A topology $\OO$ on $\MM$ is called a {\it Tychonoff topology on} $\MM$ if
there exists a topology $\TT$ on $X$ such that $\TT^+_\MM$ is a base of
$\OO$.
\end{defi}
\begin{defi}\label{Tych type top}
Let $X$ be a set and $\MM\subseteq\PP (X)$.
A topology $\OO$ on the set $\MM$ is called a {\it topology of
Tychonoff-type on $\MM$} if the family $\OO\cap\PP(X)^+_\MM$ is a base for
$\OO$.
\end{defi}
Clearly, a Tychonoff topology on $\MM$ is always a topology of
Tychonoff-type on $\MM$, but not viceversa (see Example~\ref{example}).
\begin{fact}\label{BO}
Let $X$ be a set, $\MM\subseteq\PP (X)$ and $\OO$ be a topology of
Tychonoff-type on $\MM$.
Then the family $\BB_\OO:=\{A\subseteq X : \Ap\in\OO\}$ is closed under
finite intersections, $X\in \BB_\OO$, and, hence, $\BB_\OO$ is a base for
a topology $\TT_\OO$ on $X$.
The family $\left(\BB_\OO\right)^+_\MM$ is a base of $\OO$.
\end{fact}
\begin{defi}\label{induced}
Let $X$ be a set, $\MM\subseteq\PP (X)$ and $\OO$ be a topology of
Tychonoff-type on $\MM$.
We will say that the topology $\TT_\OO$ on $X$, introduced in
Fact~\ref{BO}, is {\it induced by the topological space $(\MM,\OO)$.}
\end{defi}
\begin{pro}\label{char top Tych type}
Let $X$ be a set and $\MM\subseteq\PP(X)$.
A topology $\OO$ on $\MM$ is a topology of Tychonoff-type if and only if
there exists a topology $\TT$ on $X$ and a base $\BB$ for $\TT$ (which
contains $X$ and is closed under finite intersections) such that
$\BB^+_\MM$ is a base for $\OO$.
\end{pro}
\begin{proof}
Suppose $\OO$ is a topology of Tychonoff-type on $\MM$.
Then the topology $\TT_\OO$ induced by the topological space $(\MM,\OO)$
(see Fact~\ref{BO} and Definition~\ref{induced}) and the base $\BB_\OO$
have the required property.
Conversely, suppose $\TT$ and $\BB$ are given as in the statement.
Then $\BB^+_\MM$ is a base for $\OO$, and therefore also
$\OO\cap\PP(X)^+_\MM$ is a base for $\OO$.
\end{proof}
\begin{defi}\label{generates}
Let $X$ be a set and $\MM, \BB \subseteq\PP (X)$.
When $\BB_\MM^+$ is a base for a topology $\OO_\BB$ on $\MM$, we will say
that $\BB$ {\em generates a topology} on $\MM$.
(Obviously, the topology $\OO_\BB$ is of Tychonoff-type.
\end{defi}
\begin{pro}\label{char base}
Let $X$ be a set, $\MM\subseteq\PP(X)$ and $\BB\subseteq \PP(X)$.
The family $\BB$ generates a topology $\OO_\BB$ on $\MM$ if and only if
the family $\BB$ satisfies the following conditions:
\begin{itemize}
\item[(MB1)]
For any $M\in\MM$ there exists a $U\in \BB$ such that $M\subseteq U$;
\item[(MB2)]
For any $U_1$, $U_2 \in \BB$ and any $M\in\MM$ with
$M\subseteq U_1\cap U_2$ there exists a $U_3\in\BB$ such that
$M\subseteq U_3\subseteq U_1\cap U_2$.
\end{itemize}
\end{pro}
\begin{proof}
It follows from Proposition 1.2.1 \cite{E}.
\end{proof}
\begin{cor}\label{cord}
Let $X$ be a set and $\MM, \BB\sbe \PP(X)$.
If $\BB=\BB^\cap$ and $X\in \BB$, then $\BB$ generates a Tychonoff-type
topology on $\MM$.
\end{cor}
\begin{defi}\label{natural family}
Let $X$ be a set and $\MM\subseteq\PP(X)$.
We say that $\MM$ is a {\it natural family in $X$} if $\{x\}\in\MM$ for
all $x\in X$.
\end{defi}
\begin{cor}\label{base on X}
Let $X$ be a set and $\MM$ be a natural family in $X$.
If $\BB\subseteq\PP(X)$ generates a topology on $\MM$ (see Definition
\ref{generates}), then $\BB$ is a base for a topology on $X$.
\end{cor}
\begin{proof}
By Proposition~\ref{char base}, $\BB$ satisfies the conditions (MB1) and
(MB2).
Since $\MM$ is natural, this clearly implies that $\BB$ satisfies the
hypotesis of Proposition 1.2.1 \cite{E}.
So $\BB$ is a base for a topology on $X$.
\end{proof}
\begin{rem}
Trivial examples show that there exist sets $X$ and (non-natural) families
$\MM$,$\BB\subseteq\PP (X)$ such that $\BB_\MM^+$ is a base for a topology
on $\MM$ but
\begin{itemize}
\item[(a)]
$\bigcup\BB\not= X$, so that $\BB$ cannot serve even as subbase of a
topology on $X$ (take $X=\{0,1\}$, $\MM=\BB=\{\{0\}\}$);
\item[(b)]
$\BB$ is not a base of a topology of $X$, although $\bigcup\BB= X$
(take $X=\{0,1,2\}$, $\MM=\BB=\{\{0,1\},\{0,2\}\}$).
\end{itemize}
The example of (b) shows also that if we substitute in \ref{base on X}
naturality of $\MM$ with the condition ``$\bigcup\MM=X$'' then we cannot
prove that $\BB$ is a base of a topology on $X$; however, it is easy to
show that the condition ``$\bigcup\MM=X$'' implies that $\bigcup\BB=X$,
i.e.\ $\BB$ can serve as a subbase of a topology on $X$.
Of course, as it follows from Fact~\ref{intersection}, if $\BB^+_\MM$ is a
base of a topology $\OO$ on $\MM$, then $\tilde \BB=\BB^\cap \cup\{X\}$ is
a base for a topology on $X$ and $\tilde\BB_\MM^+$ is a base of $\OO$.
\end{rem}
\begin{cor}\label{Fin1}
Let $(X,\TT)$ be a topological space and let $\BB\subseteq\TT$ be a base of
$(X,\TT)$, closed under finite unions.
Then $\BB$ generates a topology of Tychonoff-type on $\FF in(X)$ and
$\CC omp(X)$.
\end{cor}
\begin{proof}
It follows easily from Proposition~\ref{char base}.
\end{proof}
\begin{pro}\label{O1=O2}
Let $X$ be a set, $\MM$,$\BB_1$,$\BB_2 \subseteq\PP(X)$, and suppose that
$\BB_1$ and $\BB_2$ generate, respectively, some topologies (of
Tychonoff-type) $\OO_{\BB_1}$ and $\OO_{\BB_2}$ on $\MM$.
Then $\OO_{\BB_1}=\OO_{\BB_2}$ if and only if the following conditions are
satisfied:
\begin{itemize}
\item[(CO1)]
For any $M\in\MM$ and any $U_1\in \BB_1$ such that $M\subseteq U_1$ there
exists $U_2\in \BB_2$ with $M\subseteq U_2\subseteq U_1$;
\item[(CO2)]
For any $M\in\MM$ and any $U_2\in \BB_2$ such that $M\subseteq U_2$ there
exists $U_1\in \BB_1$ with $M\subseteq U_1\subseteq U_2$.
\end{itemize}
\end{pro}
\begin{proof}
It follows from 1.2.B \cite{E}.
\end{proof}
\begin{cor}\label {Fin2}
Let $(X,\TT)$ be a topological space and $\BB_1$, $\BB_2$ be bases of
$(X,\TT)$, closed under finite unions.
Then they generate (see Corollary~\ref{Fin1}) equal topologies on
$\FF in(X)$ and $\CC omp(X)$.
In particular, every topology of Tychonoff-type on $\FF in(X)$ or on
$\CC omp(X)$, generated by a base of $(X,\TT)$ which is closed under
finite unions, coincides with the Tychonoff topology generated by
$(X,\TT)$ on the corresponding set.
\end{cor}
\begin{proof}
Check that conditions (CO1) and (CO2) of Proposition~\ref{O1=O2} are
satisfied.
\end{proof}
\begin{cor}\label{Mbase}
Let $(X,\TT)$ be a topological space, $\MM\subseteq\PP(X)$,
$\BB\subseteq\TT$, and suppose that $\BB$ generates a topology of
Tychonoff-type $\OO$ on $\MM$.
Then $\OO$ is the Tychonoff topology on $\MM$ generated by $(X,\TT)$ if
and only if for all $M\in\MM$ and for all $V\in\TT$ such that
$M\subseteq V$, there exists $U\in\BB$ with $M\subseteq U\subseteq V$.
In this case we will say that $\BB$ is an {\em $\MM$-base} for $(X,\TT)$.
Clearly, if $\MM$ is a natural family, then every $\MM$-base of $(X,\TT)$
is also a base of $(X,\TT)$.
\end{cor}
\begin{proof}
Put $\BB_1 :=\TT$ and $\BB_2 :=\BB$.
Then condition (CO2) of Proposition~\ref{O1=O2} is trivially satisfied.
The condition required in the statement is exactly condition (CO1).
\end{proof}
\begin{defi}\label{Mcover}
Let $X$ be a set, $\MM\subseteq\PP(X)$ and $A\subseteq X$.
A family $\UU\subseteq\PP(X)$ will be called an {\it $\MM$-cover} of $A$
if $A=\bigcup \UU$ and for all $M\in\MM$ with $M\subseteq A$ there exists
some $U\in\UU$ such that $M\subseteq U$.
\end{defi}
\begin{pro}\label{union2}
Let $X$ be a set and $\MM, \AA\subseteq \PP(X)$. Then the following
conditions are equivalent:
\begin{itemize}
\item[(U1)]
For all $U\in \AA$ and for all $x\in U$, there exists an $M\in\MM$ with
$x\in M\subseteq U$.
\item[(U2)]
For any $U\in\AA\cup\MM$ and for any subfamily
$\{U_\delta : \delta\in \Delta\}$ of $\AA\cup\MM$, the equality
$U^+_\MM = \bigcup_{\delta\in\Delta} \left( U_\delta\right)^+_\MM$
holds if and only if the family $\{U_\delta\}_{\delta\in \Delta}$ is an
$\MM$-cover of $U$.
\end{itemize}
\end{pro}
\begin{proof}
Observe that, trivially, in condition (U1) we can replace the requirement
`for all $U\in\AA$'' with ``for all $U\in\AA\cup\MM$''.
(U1)$\Rightarrow$(U2).
Let $U^+_\MM = \bigcup_{\delta\in\Delta} \left( U_\delta\right)^+_\MM$, with
$U$,$U_\delta\in\AA\cup\MM$ for all $\d\in\Delta$.
We will prove first that $\bigcup_{\delta\in\Delta}U_\delta = U$.
Let $x\in\bigcup_{\delta\in\Delta}U_\delta$.
Then there exists a $\delta\in\Delta$ such that $x\in U_\delta$.
By assumption, there exists an $M\in\MM$ with
$x\in M\subseteq U_\delta$.
Hence $M\in \left( U_\delta\right)^+_\MM$.
Since $U^+_\MM = \bigcup_{\delta\in\Delta} \left( U_\delta\right)^+_\MM$,
we obtain that $M\subseteq U$.
Thus $x\in U$.
Therefore, $\bigcup_{\delta\in\Delta}U_\delta \sbe U$.
Conversely, let $x\in U$.
By assumption, there exists an $M\in\MM$ such that
$x\in M\subseteq U$.
Hence
$M\in U^+_\MM = \bigcup_{\delta\in\Delta} \left( U_\delta\right)^+_\MM$.
Therefore there exists a $\delta\in\Delta$ such that
$M\in \left( U_\delta\right)^+_\MM$, i.e.\ $M\subseteq U_\delta$ and
$x\in U_\delta \subseteq \bigcup_{\delta\in\Delta} U_\delta$.
We have verified that $\bigcup_{\delta\in\Delta}U_\delta = U$.
Suppose $M\in\MM$ and $M\subseteq U$.
Then $M\in U^+_\MM $ and therefore there exists some $\gamma\in\Delta$
with $M\in \left( U_\gamma\right)^+_\MM$.
Hence $M\sbe U_\g$.
This shows that the family $\{U_\delta\}_{\delta\in \Delta}$ is an
$\MM$-cover of $U$.
The other implication can be easily proved.
(Let's remark that condition (U1) is not used in the proof of this last
implication.)
(U2)$\Rightarrow$(U1).
Suppose $U\in\AA$ and $x\in U$.
Clearly, we have
$$U^+_\MM=\bigcup\{ M^+_\MM: {M\in\MM,\ M\subseteq U}\}.$$
Then, by assumption, the family $\{M\in\MM:M\subseteq U\}$ is an
$\MM$-cover of $U$.
Therefore $U=\bigcup\{ M: {M\in\MM,\ M\subseteq U}\}$.
Hence there exists an $M\in\MM$ with $x\in M\subseteq U$.
\end{proof}
\begin{pro}\label{union}
Let $X$ be a set and $\MM\sbe\PP(X)$. Then the following
conditions are equivalent:
\begin{itemize}
\item[(a)]
$\MM$ is a natural family;
\item[(b)]
For any $U\sbe X$ and for any subfamily $\{U_\delta : \delta\in \Delta\}$
of $\PP(X)$, the equality
$U^+_\MM = \bigcup_{\delta\in\Delta} \left(U_\delta\right)^+_\MM$
holds if and only if the family $\{U_\delta\}_{\delta\in \Delta}$ is an
$\MM$-cover of $U$.
\end{itemize}
\end{pro}
\begin{proof}
Put $\AA=\PP(X)$ in Proposition~\ref{union2}.
\end{proof}
\begin{pro}\label{uniqueness}
Let $X$ be a set, $\MM\subseteq\PP(X)$, $\OO$ be a Tychonoff topology on
$\MM$ generated by a topology $\TT$ on $X$ and
$\MM$ be a network in the sense of Arhangel$'$ski\u{\i}\ for $\TT_\OO$.
Then
$\TT=\BB_\OO$ and
$\OO$
is generated by a unique topology on $X$, namely by $\TT_\OO$
(see Fact~\ref{BO} for the notation $\BB_\OO$ and $\TT_\OO$).
\end{pro}
\begin{proof}
We only need to show that $\BB_\OO\subseteq\TT$. Assume $A\in \BB_O$. Then
$A^+_\MM\in\OO$. Since
$\TT$ generates $\OO$, we have
$A^+_\MM=\bigcup_{\delta\in\Delta}\left(U_\delta\right)^+_\MM$, where
$U_\delta\in\TT$ for all $\delta\in\Delta$.
Clearly $\ U_\delta\in\BB_O$ for all
$\delta\in\Delta$. By Proposition~\ref{union2}, we obtain that
$A=\bigcup_{\delta\in\Delta} U_\delta$ and
therefore $A\in\TT$.
\end{proof}
\begin{rem}\label{rema}
Trivial examples show that there exist sets $X$, families
$\MM\sbe \PP(X)$ and Tychonoff topologies on $\MM$ which are generated
by more than one topology on $X$.
\end{rem}
\begin{cor}\label{char Tych top}
Let $X$ be a set, $\MM\subseteq\PP(X)$, $\OO$ be a
topology of Tychonoff-type on $\MM$ and
$\MM$ be a network in the sense of Arhangel$'$ski\u{\i}\ for $\TT_\OO$.
Then $\OO$ is a Tychonoff topology on $\MM$ if and only if $\BB_\OO=\TT_\OO$
(see
Fact~\ref{BO}
for the notations $\BB_\OO$ and $\TT_\OO$).
\end{cor}
\begin{proof}
Suppose $\BB_\OO=\TT_\OO$. Then the topology $\OO$ is generated by the
topology
$\TT_\OO$ on $X$ and hence, by
definition, $\OO$ is a Tychonoff topology on $\MM$.
Suppose $\OO$ is a Tychonoff topology on $\MM$.
Then $\OO$ is generated by some topology $\TT$ on $X$.
By Proposition~\ref{uniqueness}, we get $\TT=\BB_\OO$. Hence
$\TT_\OO=\BB_\OO$.
\end{proof}
\begin{cor}\label{unique}
Let $X$ be a set, $\MM$ be a natural family in $X$ and $\OO$ be a topology
of Tychonoff-type on
$\MM$. Then $\OO$ is a Tychonoff topology on $\MM$ if and only if
$\BB_\OO=\TT_\OO$.
\end{cor}
\begin{proof}
A natural family $\MM$ satisfies the hypothesis of
Corollary~\ref{char Tych top}.
\end{proof}
\begin{pro}\label {char BOB}
Let $(X,\TT)$ be a topological space, $\MM$ be a natural family in $X$,
$\BB\subseteq\TT$
and suppose that $\BB$ generates a topology $\OO_\BB$ on $\MM$.
Then
$$\BB_{\OO_\BB}=\{A\subseteq X : A \mbox{ \ is\ } \MM\mbox{-covered \ by\ a
\
subfamily \ of\ } \BB\},$$
$\BB_{\OO_\BB}\subseteq \TT$ and $\BB^\cap\subseteq \BB_{\OO_\BB}$ (see
Fact~\ref{BO} for the notation
$\BB_{\OO_\BB}$).
\end{pro}
\begin{proof}
It follows from Proposition~\ref{union} and Fact~\ref{BO}.
\end{proof}
\begin{pro}\label{TOB=T}
Let $(X,\TT)$ be a topological space, $\MM$ be a natural family in $X$,
$\BB$ be a base for
$(X,\TT)$
and suppose that $\BB$ generates a topology $\OO_\BB$ on $\MM$.
Let
$\TT_{\OO_\BB}$ be the topology on $X$ induced by $(\MM,\OO_\BB)$.
Then $\TT_{\OO_\BB}=\TT$.
\end{pro}
\begin{proof}
We have, by Proposition~\ref{char BOB},
that $\BB\subseteq \BB_{\OO_\BB}\subseteq \TT_{\OO_\BB}$.
Thus $\TT\subseteq\TT_{\OO_\BB}$.
As it is shown in Proposition~\ref{char BOB},
$\BB_{\OO_\BB}\subseteq \TT$ and
hence
$\TT_{\OO_\BB}\subseteq \TT$. So, $\TT=\TT_{\OO_\BB}$.
\end{proof}
\begin{exa}\label{ex1}
Let us show that
in Proposition~\ref{TOB=T}
the requirement
``$\MM$ is a natural family'' is essential.
Let $X=(0,1)\subset\RRRR$ be the open unit interval with the usual topology,
$\MM=\{[a,b] : 02$.
We consider now the natural family $\MM^\prime=\PP(X)\setminus\{\emptyset\}$
and we define $\OO$ as above.
$\OO$ is a Tychonoff-type topology on $\MM^\prime$ but it is not a Tychonoff
topology.
Again $\TT_\OO$ is the discrete topology on $X$. Hence $\MM^\prime = \CC
L((X,\TT_\OO))$.
We observe as before that $(\MM\ap,\OO)$ is not a T$_0$-space.
Note also that the family $\BB_\OO$ is $\MM^{\prime\prime}$-closed for every
natural family
$\MM^{\prime\prime}$ in $X$ which contains all two-points subsets of $X$ and
$\emptyset\not\in\MM^{\prime\prime}$.
\end{exa}
We will briefly discuss now some topological
properties of the hyperspaces $(\MM,\OO)$ with Tychonoff-type topologies
$\OO$.
\begin{fact}\label{sep axioms}
Let $X$ be a set, $\MM\subseteq\PP(X)$ and $\OO$ be a topology of
Tychonoff-type on $\MM$ generated by a subfamily $\BB$ of $\PP(X)$.
Then:
\begin{itemize}
\item[(a)]
the topological space $(\MM,\OO)$ is
a $T_0$-space (resp., $T_1$-space) if and only if for any
$F,G\in\MM$ with $F\not= G$, there exists a $B\in\BB$ such that either
$F\subseteq B$ and
$G\not\subseteq B$, or $G\subseteq B$ and $F\not\subseteq B$ (resp.,
$F\subseteq B$ and
$G\not\subseteq B$).
\item[(b)]
if for any $x\in X$ and for any $F\in\MM$ with $x\not\in F$,
there exists
a $B\in\BB$ such that $F\subseteq B$ and $x\not\in B$, then $(\MM,\OO)$
is a $T_0$-space.
\end{itemize}
\end{fact}
\begin{rem}\label{Choban}
Let us note that Fact~\ref{sep axioms}(b) implies the following assertion,
which was mentioned in \cite{C}, section 2 (after Lemma 3)
(the requirement that $X\in\Omega$ has
to be added there):
{\it if $(X,\TT)$ is a regular $T_1$-space, $\MM$ is a family consisting of
closed subsets of $(X,\TT)$ and $\BB$ is a base of $(X,\TT)$ such that
$\BB=\BB^\cap$ and
$U\in\BB$ implies that $X\stm \ovl{U}\in\BB$, then $(\MM,\OO_\BB)$ is
a $T_0$-space.}
\end{rem}
\begin{fact}\label{T_0}
Let $(X,\MM,\OO)\in\card{\HH\TT}$. Then the correspondence
$(X,\TT_\OO)\lra (\MM,\OO)$, $x\mapsto \{x\}$, is a homeomorphic embedding.
Hence, we have, in particular, that:
\begin{itemize}
\item[(a)]
$w(X,\TT_\OO)\le w(\MM,\OO)$;
\item[(b)]
if $(\MM,\OO)$ is a $T_0$-space then $(X,\TT_\OO)$ is a $T_0$-space.
\end{itemize}
\end{fact}
\begin{fact}\label{FsubsetG}
\mbox{}
\begin{itemize}
\item[(a)]
Let $X$ be a set, $\MM\subseteq\PP(X)$ be a family such that there exist
$F, G\in\MM$ with $F\subset G$ and $F\not= G$, and let $\OO$ be a topology
of Tychonoff-type on $\MM$. Then $(\MM,\OO)$ is not a $T_1$-space.
\item[(b)]
If $(X,\MM,\OO)\in\card{\HH\TT}$ then $(\MM,\OO)$ is a $T_1$-space
if and only if $(X,\TT_\OO)$ is a $T_1$-space and $\MM=\{\{x\}: x\in X\}$.
\end{itemize}
\end{fact}
\begin{fact}\label{compactness}
Let $(X,\MM,\OO)\in\card{\HH\TT}$.
Then $(\MM,\OO)$ is a compact space if and only if any $\MM$-cover of $X$,
consisting of elements of $\BB_\OO$, has a finite $\MM$-subcover.
\end{fact}
\begin{proof}
It follows from Proposition~\ref{union}.
\end{proof}
\begin{exas}
There are many examples of ``very nice'' spaces $X$ with
non-$T_0$-hyperspa\-ces $(\MM,\OO_\BB)$ (see Examples~\ref{example} and
\ref{bigexample}).
As an example of a non-$T_0$-space $(X,\TT)$ with a $T_0$-hyperspace
$(\MM,\OO_\TT)$, consider the two-points space $X=\{0,1\}$, with
$\TT=\MM=\{\emptyset,X\}$.
There exist non-compact spaces $X$ such that $(\CC L (X),\OO_\BB)$
is a compact non-$T_0$-space (e.g., the space $(\CC L (\RRRR),\OO_\BB)$,
described in Example~\ref{bigexample}).
To get an example of a non-compact space $X$ and a natural family $\MM$
such that $(\MM,\OO_\BB)$ is a compact $T_0$-space, consider $X:=\RRRR$
with its natural topology, $\MM:=\FF in_2(\RRRR) \cup\{\RRRR\}$ and take
$\BB$ as in Example~\ref{bigexample}.
As an example of a compact space $(X,\TT)$ with a non-compact hyperspace
$(\MM,\OO_\TT)$,
consider the unit interval $X=[0,1]$ with its natural topology and put
$\MM=\{\{x\}:x\in(0,1]\}$.
\end{exas}
The next three propositions are generalizations of, respectively,
Propositions 1, 2 and 3 of \cite{F}, and have proofs similar to those
given in \cite{F}.
(Let us note that in Proposition 2 of \cite{F} the requirement
``$\ems\not\in\CC$'' has to be added.)
\begin{pro}\label{F1}
Let $(X,\MM,\OO)\in\card{\HH\TT}$,
$w(\MM,\OO)=\aleph_0$,
$(X,\TT_\OO)$ be a $T_1$-space,
$\BB_\OO$ be closed under countable unions and
$\MM$ contain all infinite countable closed subsets of $(X,\TT_\OO)$.
Then $(X,\TT_\OO)$ is a compact space.
\end{pro}
\begin{pro}\label{F2}
Let $(X,\MM,\OO)\in\card{\HH\TT}$ and $\ems\not\in\MM$. Then
$d(\MM,\OO)=d(X,\TT_\OO)$.
\end{pro}
\begin{pro}\label{F3}
Let $(X,\MM,\OO)\in\card{\HH\TT}$ and $\ems\not\in\MM$. Then
$(\MM,\OO)$ has isolated points if and only if $(X,\TT_\OO)$
has isolated points.
\end{pro}
\section{On $\OO$-commutative spaces}
\begin{nist}\label{2-emb}
Let $(X,\TT)$ be a topological space and $A\subseteq X$. Recall that
$A$ is
said to be
{\em 2-combinatorially embedded in} $X$ (see \cite{CN}) if
the closures in $X$ of any two disjoint closed in $A$
subsets of $A$ are disjoint.
\end{nist}
\begin{defi}\label{2Bcomb}
Let $(X,\TT)$ be a topological space, $A\sbe X$
and $\BB\subseteq \PP(X)$.
We will
say that $A$ is {\em $2_\BB$-combinatorially embedded in $X$} if for any
$F\in \CC L(A)$ and for any $U\in\BB$
with $F\subseteq U$, there exists a $V\in\BB$ such that $\overline{F}^X
\subseteq V$ and $V\cap A\subseteq U$.
\end{defi}
\begin{pro}\label{2T}
Let $(X,\TT)$ be a topological space and $A\subseteq X$. Then $A$ is
2-com\-bi\-na\-to\-rially embedded in
$X$ if and only if $A$ is $2_\TT$-combinatorially embedded in $X$.
\end{pro}
\begin{proof}
($\Ra$)
Let $H\in \CC L(A)$, $V\in\TT$ and $H\sbe V$.
We put $U=V\cap A$ and
$F=A\setminus U$. Then $F$ and $H$ are two disjoint closed subsets of $A$.
Hence, by assumption, they have disjoint closures in $X$, i.e.\
$\overline{F}^X\cap\overline{H}^X=\emptyset$.
Let
$W=X\setminus\overline{F}^X$.
Then $W$ is open in $X$, $W\cap A=U=V\cap A$ and $\overline{H}^X\subseteq
W$.
($\La$)
Let $F$ and $G$ be two disjoint closed subsets of $A$. Put
$V=X\setminus\overline{G}^X$. Then $V$ is
open in $X$ and $F\subseteq V$. Hence, by assumption, there exists an open
set $W$ such that $\overline{F}^X\subseteq W$
and $W\cap A\subseteq V\cap A$. Let $U=A\setminus G$.
Then $Ex_{A,X}U=V$.
Hence $V\cap A=U$ and $W\subseteq V$.
We conclude that $\overline{F}^X\subseteq
W\subseteq
V=X\setminus\overline{G}^X$, i.e.
$\overline{F}^X\cap\overline{G}^X=\emptyset$.
\end{proof}
\begin{rem}\label{rem111}
In Example~\ref{exa111} below
we will show that there exist spaces $(X,\TT)$, subspaces
$A$ of $X$ and bases $\BB$ of $\TT$ such that $A$ is
$2_\BB$-combinatorially embedded in $X$
but $A$ is not
2-combinatorially embedded in $X$.
\end{rem}
\begin{pro}\label{iAX}
Let $(X,\TT)$ be a T$_1$-space, $\OO$ be a topology of Tychonoff-type on
$\CC L(X)$ and
$A\subseteq X$. Put $\BB_A=\{U\cap A : U\in\BB_\OO\}$
(see Fact \ref{BO} for the notation $\BB_\OO$).
The family $\BB_A$ generates a topology of Tychonoff-type $\OO_A$ on $\CC
L(A)$.
The function $i_{A,X}:(\CC L(A),\OO_A)\lra (\CC L(X),\OO)$, defined by
$i_{A,X}(F):=\overline{F}^X$, is
inversely continuous (i.e.\ it is injective and its inverse, defined on
$i_{A,X}\left(\CC L(A)\right)$, is a
continuous function) if and only if the set
$A$ is $2_{\BB_\OO}$-combinatorially embedded in $X$.
\end{pro}
\begin{proof}
The family $\BB:=\BB_\OO$ is closed under
finite intersections and $X\in\BB$ (see Fact \ref{BO}).
Hence
the family $\BB_A$ is closed under finite intersections and $A\in\BB_A$.
Therefore, by Corollary \ref{cord},
$\BB_A$ generates a topology of Tychonoff-type $\OO_A$ on $\CC L(A)$.
The function $i_{A,X}:(\CC L(A),\OO_A)\lra (\CC L(X),\OO)$ is clearly
injective.
Denote by $g$ its inverse defined on
$i_{A,X}(\CC L(A))$, i.e.
$g:i_{A,X}(\CC L(A))\lra \CC L(A)$.
($\Ra$)
Let $H\in\CC L(A)$, $U\in\BB$ and $H\subseteq U$. Then $H\in
\left(U\cap A\right)^+_{\CC L(A)}\in \OO_A$. Since
$g(\ovl{H}^X)=H$, the continuity of $g$ implies that
there exists a $V\in \BB$ such that
$$\ovl{H}^X\sbe V
\mbox{ and }
g(V^+_{\CC L(X)}\cap i_{A,X}(\CC L(A)))\sbe (U\cap A)^+_{\CC L(A)}.$$
Then
$V\cap A\subseteq U\cap A$.
Indeed, let $x\in V\cap A$.
Since $X$ is a $T_1$-space, we obtain that
$$
\{x\}\in V^+_{\CC L(X)}\cap i_{A,X}(\CC L(A))
\mbox{ and }
g(\{x\})=\{x\}.$$
Hence $x\in U\cap A$. So,
$A$ is $2_\BB$-combinatorially embedded in $X$.
($\La$)
Let $F\in i_{A,X}(\CC L(A))$
and $g(F)=H$. Then $F=\ovl{H}^X$ and
$H\in \CC L(A)$. Let $U\in\BB_A$ be such that $H\sbe U$. Then there
exists a $V\in \BB$ with $V\cap A=U$. Hence $H\sbe V$. Since
$A$ is $2_\BB$-combinatorially embedded in $X$,
there exists a $W\in\BB$ such that $F=\ovl{H}^X\sbe W$ and
$W\cap A\sbe V\cap A=U$.
Then $F\in W^+_{\CC L(X)}\in\OO$.
We will show that
$$g(W^+_{\CC L(X)}\cap i_{A,X}(\CC L(A)))\sbe U^+_{\CC L(A)}.$$
Indeed,
let $K\in \CC L(A)$,
$G=\ovl{K}^X$
and $G\sbe W$. Then $g(G)=K$
and
$$K=G\cap A\sbe W\cap A\sbe V\cap A=U,$$
i.e.\ $K\in U^+_{\CC L(A)}$, as we have to show. Hence, $g$ is a
continuous function.
\end{proof}
\begin{cor}[\cite{D2}, Theorem 2.1]\label{2.1D2}
If in Proposition \ref{iAX} we take $\OO$ to be the
Tychonoff topology on
$\CC L(X)$
generated by $(X,\TT)$
then
the function $i_{A,X}$ is inversely continuous if and only if
$A$ is 2-combinatorially embedded in $X$.
\end{cor}
\begin{proof}
It follows from Propositions~\ref{iAX}, \ref{uniqueness} and \ref{2T}.
\end{proof}
\begin{cor}\label{sequential}
Let $(X,\TT)$ be a T$_2$-space, $A\subseteq X$ and $\OO$ be a topology of
Tychonoff-type on $\CC L(X)$
generated by a subfamily of $\TT$.
Let $i_{A,X}$ be inversely continuous
(see Proposition~\ref{iAX} for the notation $i_{A,X}$).
Assume that the following condition is satisfied:
\begin{itemize}
\item[(*)] For any $U\in\TT$ and for
all countable $F\in\CC L(A)$ such that $\card{A\setminus
F}\geq\aleph_0$ and
$F\subseteq U$, there exists a $V\in\BB_\OO$
with $F\subseteq V\subseteq U$.
\end{itemize}
Then the set $A$ is sequentially closed.
\end{cor}
\begin{proof}Put $\BB :=\BB_\OO$.
Then, by Proposition~\ref{char BOB}, $\BB\sbe \TT$.
Assume that the set $A$ is not sequentially closed.
Then there exists a sequence
$(x_n)_{n\in\omega}$ in $A$ and a point $x\in X\setminus A$ such that
$\lim_{n\lra\infty}x_n=x$. Without loss
of generality we can assume $x_n\not=x_m$ for all $n\not=m$.
Let us consider the sets $F=\{x_{2n} : n\in\omega\}$ and
$G=\{x_{2n-1} : n\in\omega\}$.
Put $U=X\setminus \overline{G}^X$.
Then $F$ is a countable closed subset of $A$,
$\card{A\stm F}\ge\aleph_0$, $F\subseteq U$ and $U\in \TT$.
By (*), there exists a $V\in\BB$ such that $F\subseteq V\subseteq U$.
Since we are assuming that the function $i_{A,X}$ is inversely continuous,
we obtain, by Proposition~\ref{iAX}, that the set $A$ is
$2_\BB$-combinatorially embedded in $X$.
Hence there exists a $W\in\BB$ such that $\overline{F}^X\subseteq W$ and
$W\cap A\subseteq V\cap A$.
Then $x\in W$, because $x\in\ovl{F}^X$.
Since $W\in\TT$ and $x$ is a limit point of $G$, we have
$G\cap W\not=\emptyset$.
However this is a contradiction because
$$W\cap A\subseteq V\cap A\sbe U=X\setminus\overline{G}^X,$$
and hence $G\cap W=\ems$.
Therefore, $A$ is sequentially closed.
\end{proof}
\begin{rem}\label{rem222}
In Example~\ref{exa111} below we will show that condition (*) of
Corollary~\ref{sequential} is essential, i.e., if we omit it, then the set
$A$ could fail to be sequentially closed.
\end{rem}
\begin{cor}[\cite{D2}, Corollary 2.3]\label{2.3D2}
Let $(X,\TT)$ be a T$_2$-space, $A\subseteq X$, $\OO$ be the
Tychonoff topology on $\CC L(X)$
generated by $(X,\TT)$
and $i_{A,X}$ be inversely continuous
(see Proposition~\ref{iAX} for the notation $i_{A,X}$).
Then the set $A$ is sequentially closed.
\end{cor}
\begin{proof}
We have, by Proposition~\ref{uniqueness}, that $\BB_\OO=\TT$.
Then condition (*) of
Corollary \ref{sequential}
is trivially satisfied.
Hence, by
Corollary \ref{sequential},
$A$ is sequentially closed.
\end{proof}
\begin{cor}\label{eqv}
Let $(X,\TT)$ be a sequential T$_2$-space, $A\subseteq X$ and $\OO$ be
a topology of
Tychonoff-type on $\CC L(X)$
generated by a subfamily of $\TT$.
Assume that condition (*) of Corollary~\ref{sequential} is satisfied.
Then the following conditions are
equivalent
(see Proposition~\ref{iAX} for the notation $i_{A,X}$):
\begin{itemize}
\item[(a)] $i_{A,X}$ is a homeomorphic embedding;
\item[(b)] $i_{A,X}$ is inversely continuous;
\item[(c)] $A$ is closed in $X$.
\end{itemize}
\end{cor}
\begin{proof}
It is clear that (a) implies (b).
The implication (c)$\Ra$(a) is true for any $X$, because
if $A$ is a
closed subset of $X$ then $i_{A,X}$ is the inclusion map.
Let us show that (b) implies (c). By Corollary~\ref{sequential}, $A$ is
sequentially closed. Since $X$ is a
sequential space, we obtain that the set $A$ is closed.
\end{proof}
\begin{cor}[\cite{D2}, Corollary 2.4]\label{2.4D2}
Let $(X,\TT)$ be a sequential T$_2$-space, $A\subseteq X$ and $\OO$ be the
Tychonoff topology on $\CC L(X)$ generated by $(X,\TT)$.
Then the following conditions are equivalent
(see Proposition~\ref{iAX} for the notation $i_{A,X}$):
\begin{itemize}
\item[(a)] $i_{A,X}$ is a homeomorphic embedding;
\item[(b)] $i_{A,X}$ is inversely continuous;
\item[(c)] $A$ is closed in $X$.
\end{itemize}
\end{cor}
\begin{defi}\label{O-commutative}
Let $(X,\TT)$ be a topological space and let $\OO$ be a topology of
Tychonoff-type on $\CC L(X)$.
The space $(X,\TT)$ is called {\em $\OO$-commutative } if for any
$A\subseteq X$ the function $i_{A,X}$, defined in Proposition~\ref{iAX},
is a homeomorphic embedding.
When $\OO$ is the Tychonoff topology on $\CC L(X)$ generated by $(X,\TT)$,
the notion of ``$\OO$-commutative space'' coincides with the notion of
``{\em commutative space}'', introduced in \cite{D1, D2}.
\end{defi}
\begin{cor}\label{comm}
Let $(X,\TT)$ be a sequential T$_2$-space, $\OO$ be
a topology of
Tychonoff-type on $\CC L(X)$
generated by a subfamily of $\TT$ and
let condition (*) of Corollary~\ref{sequential} be satisfied
for every subspace $A$ of $X$.
Then $X$ is $\OO$-commutative
if and only if
$X$ is discrete.
\end{cor}
\begin{proof}
It follows from Corollary~\ref{eqv}.
\end{proof}
\begin{cor}[\cite{D2}, Corollary 2.5]
\label{2.5D2}
If $X$ is a sequential T$_2$-space then $X$ is commutative if and only if
$X$ is discrete.
\end{cor}
\begin{exa}\label{ex F}
Let us show that there exist spaces $X$ and
topologies $\OO$ of Tycho\-noff-type on $\CC L(X)$ that are
not Tychonoff topologies and that satisfy all hypothesis of Corollary
\ref{comm}.
Let $X=D(\aleph_1)$ be the discrete space of cardinality $\aleph_1$.
Let
$$\BB=\{A\subset X : |A|\leq\aleph_0\}\cup\{X\}$$
and $\MM=\CC L(X)$.
Then $\MM$ is a natural family on $X$, $\BB$ is $\MM$-closed,
$\BB^\cap=\BB$, $X\in\BB$ and
$\BB$ is a base for the discrete topology on $X$.
Let $\OO_\BB$ be the topology on $\MM$ generated
by $\BB$. Then $\OO_\BB$ is a
topology of Tychonoff-type on $\MM$,
however it is not a Tychonoff topology. In fact, by Proposition~\ref{BOB=B},
$\BB=\BB_{\OO_\BB}$. Hence
$\BB_{\OO_\BB}\not=\TT_{\OO_\BB}$ and,
by Corollary~\ref{unique}, $\OO$ cannot be a
Tychonoff topology.
Obviously, $\BB=\BB_{\OO_\BB}$ satisfies condition (*) of
Corollary \ref{sequential} for any subspace $A$ of $X$.
\end{exa}
\begin{defi}\label{O-HS-space}
Let $(X,\TT)$ be a topological space and $\OO$ be a topology of
Tycho\-noff-type on $\CC L(X)$.
The space $(X,\TT)$ is called {\em $\OO$-HS-space} if, for any
$A\subseteq X$, the function $i_{A,X}$,
defined in Proposition~\ref{iAX},
is continuous.
When
$\OO$ is the Tychonoff topology on $\CC L(X)$
generated by $(X,\TT)$,
the notion of ``$\OO$-HS-space'' coincides with the notion of
``{\em HS-space}'', introduced in \cite{BDN1, BDN2}.
\end{defi}
\begin{cor}\label{ocom}
Let $(X,\TT)$ be a T$_1$-space and $\OO$ be a topology of Tychonoff-type on
$\CC L(X)$.
Then $X$ is an $\OO$-commutative space if and only if $X$ is an
$\OO$-HS-space
and every subset $A$ of $X$ is $2_{\BB_\OO}$-combinatorially embedded in
$X$.
\end{cor}
\begin{proof}
It follows from Proposition~\ref{iAX}.
\end{proof}
\begin{cor}[\cite{D2},Corollary 2.2]\label{ocom1}
A $T_1$-space $X$ is commutative
if and only if $X$ is an HS-space
and every subspace of $X$ is 2-combinatorially embedded in $X$.
\end{cor}
\begin{proof}
It follows from Corollary~\ref{ocom} and Proposition~\ref{2T}.
\end{proof}
\begin{exa}\label{bigexample}
We will describe two Tychonoff-type, non Tychonoff topologies on two
different subfamilies of
$\PP(\RRRR)$ generated by the family $\BB$
of all open intervals of $\RRRR$. One of the resulting spaces
will be $T_0$ and
the other one will not.
Let $\TT$ be the natural topology on $X :=\RRRR$.
Then the family $\BB$ of all
open intervals in
$X$ is a base for $\TT$, it is closed under finite intersections
and $X\in\BB$.
Put $\MM :=\CC L(X,\TT)$ and
$\MM^\prime :=\FF in_2(X)$.
They are natural families.
The family $\BB$ is both $\MM^\prime$-closed and $\MM$-closed.
Indeed,
let $U\subseteq X$ be $\MM^\prime$-covered by a subfamily $\BB_U$ of $\BB$.
Then $U\in\TT$ and for every $x,y\in U$ there exists an open interval
$(\a,\b)\in\BB_U$ containing the points $x$ and $y$.
Hence $U$ is a connected open set in
$\RRRR$, i.e.\ $U\in\BB$.
Therefore, $\BB$ is an $\MM\ap$-closed family.
Since
$\MM^\prime\subset
\MM$, we obtain, by Proposition~\ref{M and M^prime}, that
$\BB$ is an $\MM$-closed family as well.
By Corollary~\ref{cord}, $\BB$ generates Tychonoff-type topologies
$\OO_\BB$ on $\MM$ and
$\OO^\prime_\BB$ on $\MM^\prime$.
As it follows from
Proposition~\ref{BOB=B},
$\BB_{\OO_\BB}=
\BB_{\OO^\prime_\BB}=
\BB\not=\TT$. Hence, by
Corollary~\ref{unique},
$\OO_\BB$
and $\OO^\prime_\BB$
are not Tychonoff topologies on $\MM$, respectively $\MM\ap$.
It is easy to see that $(\MM^\prime,\OO^\prime_\BB)$ is a
T$_0$-space. Indeed, let $\{x,y\}$ and $\{u,v\}$ be two distinct elements
in $\MM^\prime$.
We can assume $x0$.
Consider the interval $B=(x+\varepsilon,+\infty)$.
Then $\{u,v\}\in B^+_{\MM^\prime}$ but
$\{x,y\} \not\in B^+_{\MM^\prime}$.
Let's prove that $(\MM,\OO_\BB)$ is not a T$_0$-space.
Put $F=\{2k:k\in\Z\}$ and $G=\{2k+1:k\in\Z\}$. Then
$F,G\in\MM$ and $F\not=G$ but the only neighbourhood of
both $F$ and $G$ in
$\MM$ is $X^+_{\MM}=\MM$.
\end{exa}
\begin{exa}\label{bigexample1}
In the notations of Example~\ref{bigexample}, we will show that
$(\RRRR,\TT)$ {\em is an $\OO_\BB$-HS-space}.
We are working now with the space $(\MM,\OO_\BB)$ from
Example~\ref{bigexample}.
We will write simply $\OO$ instead of $\OO_\BB$.
Let $A\subseteq X$. We have to show that the function
$$i_{A,X}:(\CC L(A),\OO_A)\lra (\CC L(X),\OO),$$
where the topology $\OO_A$ on $\CC L(A)$ is generated by the family
$$\BB_A=\{A\cap U: U\in\BB\},$$
is continuous
(see Proposition~\ref{iAX} for the notation $i_{A,X}$).
Let
$B\in\BB$.
We will show that
$i_{A,X}^{-1}(B^+_\MM)$ is an open set.
Take an $F\in i_{A,X}^{-1}(B^+_\MM)$. Then $F\in\CC L(A)$ and
$\overline{F}^X\subseteq B$.
There exists an
$E\in\BB$ such that
$$\overline{F}^X\subseteq E\subseteq \overline{E}^X\subseteq B$$
(this is clear if $F$ is bounded, since in this case $\overline{F}^X$ is
compact; if $F$ is unbounded below, but is bounded above, then
$B=(-\infty,\beta)$, for some $\b\in\RRRR$, and we can pick
$E=(-\infty,\gamma)$ with $\sup F<\gamma<\beta$; similarly if $F$ is
unbounded above but not below; if $F$ is unbounded both above and below
then we have $B=\RRRR$ and we put $E := B$).
Then
$$F\in \left( E\cap A\right)^+_{\CC L(A)}\subseteq
i_{A,X}^{-1}(B^+_\MM).$$
Indeed, let $G\in \left( E\cap A\right)^+_{\CC L(A)}$.
Then
$$\overline{G}^X\subseteq \overline{E}^X\subseteq B,$$
i.e.\ $i_{A,X}(G)\in B^+_\MM$.
\end{exa}
\begin{rem}\label{rem112}
Let us note that
a similar proof shows that
{\em every subspace $Y$ of $(\RRRR,\TT)$
is an
$\OO_{\BB_Y}$-HS-space}
(see Examples~\ref{bigexample} and
\ref{bigexample1} for the notations).
More generally, let $Y$ be a topological space and $\DD$ be a base of $Y$.
We will say that $Y$ is {\it $\DD$-normal} if for every $F\in\CC L(Y)$
and for every $D\in\DD$ such that $F\sbe D$ there exists an $E\in\DD$ with
$F\sbe E\sbe \ovl{E}^Y\sbe D$.
Now, arguing as in Example~\ref{bigexample1},
we can prove that
{\em if $Y$ is a $\DD$-normal space, $\DD=\DD^\cap$ and $Y\in\DD$,
then $Y$ is an $\OO$-HS-space, where $\OO$ is the Tychonoff-type
topology on $\CC L(Y)$ generated by $\DD$.}
This generalizes the result of M. Sekanina \cite{Se} that any normal
space is a HS-space.
\end{rem}
\begin{exa}\label{exa111}
In the notations of Examples~\ref{bigexample} and \ref{bigexample1}, we
will show that {\em the function $i_{A,\RRRR}$ is a homeomorphic embedding
for any open interval $A$.}
We will argue for $A=(0,1)$;
the proof for any other open interval is similar.
We know, by Example~\ref{bigexample1}, that the function $i_{A,X}$ is
continuous.
Therefore we only need to prove that $i_{A,X}$ is inversely continuous.
By Proposition~\ref{iAX}, it is enough to show that the set $A$ is
$2_\BB$-combinatorially embedded in $X$.
So, let $H$ be a closed subset of $(0,1)$ and let $B=(\a,\b)\in\BB$ be
such that $H\subseteq B$.
We have to find a $D\in\BB$ such that $\overline{H}^X\subset D$ and
$D\cap (0,1)\subseteq B$.
Clearly, $\overline{H}^X\subseteq [0,1]$.
If $\overline{H}^X\subset (0,1)$, we can take $D=B$.
If $0\in\overline{H}^X$ but $1\not\in\overline{H}^X$ then $\a\le 0$ and we
can put $D=(-1,\b)$.
If $1\in\overline{H}^X$ but $0\not\in\overline{H}^X$ then $\b\ge 1$ and we
can put $D=(\a,2)$.
If $0,1\in\overline{H}^X$ then $\a\le 0$ and $\b\ge 1$ and we put
$D=(-1,2)$.
Therefore, {\em $A$ is $2_\BB$-combinatorially embedded in $(\RRRR,\TT)$.}
Note that $A$ {\em is not 2-combinatorially embedded in $(\RRRR,\TT)$.}
Observe that the triple $((\RRRR,\TT), A,\OO)$ satisfies all hypothesis of
Corollary~\ref{sequential} except for condition (*), but $A$
is not sequentially closed.
\end{exa}
\begin{exa}\label{prop111}
Let $Y\sbe \RRRR$. We will say, as usual, that a point $x\in Y$ is {\em
isolated from the right (left) (in $Y$)} if there exists an $\ep >0$ such
that if we put $U=(x,x+\varepsilon)$ ($U=(x-\varepsilon,x)$) then $U\cap
Y=\emptyset$. Now, in the notations of Examples~\ref{bigexample} and
\ref{bigexample1}, we have: {\em a subspace $Y$ of $(\RRRR,\TT)$ is
$\OO_{\BB_Y}$-commutative if and only if every point of $Y$ is either
isolated from the right or from the left.}
We first show that a space $Y$ that has a point $y_0$ which is
non-isolated both from the left and from the right cannot be
$\OO_{\BB_Y}$-commutative. Indeed, put $A=Y\setminus\{y_0\}$. We will
prove that $A$ is not $2_{\BB_Y}$-combinatorially embedded in $Y$. By
Proposition~\ref{iAX}, this will imply that the function $i_{A,Y}$ is not
inversely continuous and hence the space $Y$ will be not
$\OO_{\BB_Y}$-commutative. Let $H=\{y_n:n\in\omega\}$ be a decreasing
sequence in $Y$ converging to $y_0$. Then $H$ is a closed subset of $A$
and $H\subset (y_0,+\infty)\cap Y$. Suppose that there exists $B\in\BB$
such that $cl_Y H\subseteq B$ and $B\cap A\sbe (y_0,+\infty)$. Since
$y_0\in cl_Y H\sbe B$ and $y_0$ is not isolated from the left, we have
that $(B\cap A)\setminus (y_0,+\infty)\not =\ems$, which is a
contradiction. Hence, $A$ is not $2_{\BB_Y}$-combinatorially embedded in
$Y$.
Now we will show that a space $Y$ having only points which are isolated
either from the left or from the right is $\OO_{\BB_Y}$-commutative. Let
$A\subset Y$. We know, by Remark~\ref{rem112}, that the function $i_{A,Y}$
is continuous. Hence it is enough to show that it is inversely continuous,
i.e., according to Proposition~\ref{iAX}, that $A$ is
$2_{\BB_Y}$-combinatorially embedded in $Y$. So, let $H\in\CC L(A)$ and
let $H\subseteq B\cap Y$ for some $B=(\alpha,\beta)$. We have to find a
$D\in\BB$ such that $\overline{H}^Y\subset D\cap Y$ and $D\cap A\subseteq
B$. We have $ cl_Y H\subseteq \overline{B}^X=[\alpha,\beta]$. If $cl_Y
H\subseteq B$, we can take $D=B$ and we are done. If $\alpha\in cl_Y H$
and $\b\not\in cl_Y H$ then $\alpha\in Y$ and $\alpha$ is not isolated
from the right, being a limit point of $H$. Hence, by the assumption,
$\alpha$ is isolated from the left. Thus there exists a $\gamma<\alpha$
such that $(\gamma,\alpha)\cap Y=\emptyset$. Then $D=(\gamma,\b)$ is the
required interval. The other two possible cases are treated analogously.
\end{exa}
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\end{document}