Publication detail

# Problem 540 is (almost) solved

KOVÁR, M.

Original Title

Problem 540 is (almost) solved

English Title

Problem 540 is (almost) solved

Type

conference paper

Language

en

Original Abstract

Recall that a set is said to be saturated if it is the intersection of open sets. By the dual topology $\tau^d$ for a topological space $(X,\tau)$ we mean the topology on $X$ generated by taking the compact saturated sets of $X$ as a subbase for closed sets. The Problem 540 of J. D. Lawson and M. Mislove \cite{LM} in Open Problems in Topology (J. van Mill, G. M. Reed, eds.,1990) asks \medskip \roster \item which topologies can arise as dual topologies \smallskip and \smallskip \item whether the process of taking duals terminate after finitely many steps with the topologies that are duals of each other. \endroster \medskip For $T_1$ spaces, the solution of (2) simply follows from the fact that in $T_1$ spaces every set is saturated and hence the dual operator $d$ coincide with the compactness operator $\rho$ of J. de Groot, G. E. Strecker and E. Wattel \cite{GSW}. For more general spaces, the question (2) was partially answered by Bruce S. Burdick who found certain classes of (in general, non-$T_1$) spaces for which the process of taking duals of a topological space $(X,\tau)$ terminates by $\tau^{dd}=\tau^{dddd}$ -- the lower Vietoris topology on any hyperspace, the Scott topology for reverse inclusion on any hyperspace, and the upper Vietoris topology on the hyperspace of a regular space. B. Burdick presented his paper on The First Turkish International Conference on Topology in Istanbul 2000 \cite{Bu}. \medskip In this talk a general (and positive) solution of (2) with a short classification of topological spaces with respect to the number of distinct topologies generated by iterating duals will be presented. Our main result is the following theorem: \proclaim{Theorem} For every topological space $(X,\tau)$ it follows $\tau^{dd}= \tau^{dddd}$. \endproclaim On the other hand, we remark that this result cannot be improved since there exist a $T_1$ space $(X,\tau)$ generating four distinct topologies $\tau$, $\tau^d=\rho(\tau)$, $\tau^{dd}=\rho^2(\tau)$ and $\tau^{ddd}=\rho^3(\tau)$ (see e.g. Example 8 of \cite{GHSW} or Example 1 of \cite{Bu}).

English abstract

Recall that a set is said to be saturated if it is the intersection of open sets. By the dual topology $\tau^d$ for a topological space $(X,\tau)$ we mean the topology on $X$ generated by taking the compact saturated sets of $X$ as a subbase for closed sets. The Problem 540 of J. D. Lawson and M. Mislove \cite{LM} in Open Problems in Topology (J. van Mill, G. M. Reed, eds.,1990) asks \medskip \roster \item which topologies can arise as dual topologies \smallskip and \smallskip \item whether the process of taking duals terminate after finitely many steps with the topologies that are duals of each other. \endroster \medskip For $T_1$ spaces, the solution of (2) simply follows from the fact that in $T_1$ spaces every set is saturated and hence the dual operator $d$ coincide with the compactness operator $\rho$ of J. de Groot, G. E. Strecker and E. Wattel \cite{GSW}. For more general spaces, the question (2) was partially answered by Bruce S. Burdick who found certain classes of (in general, non-$T_1$) spaces for which the process of taking duals of a topological space $(X,\tau)$ terminates by $\tau^{dd}=\tau^{dddd}$ -- the lower Vietoris topology on any hyperspace, the Scott topology for reverse inclusion on any hyperspace, and the upper Vietoris topology on the hyperspace of a regular space. B. Burdick presented his paper on The First Turkish International Conference on Topology in Istanbul 2000 \cite{Bu}. \medskip In this talk a general (and positive) solution of (2) with a short classification of topological spaces with respect to the number of distinct topologies generated by iterating duals will be presented. Our main result is the following theorem: \proclaim{Theorem} For every topological space $(X,\tau)$ it follows $\tau^{dd}= \tau^{dddd}$. \endproclaim On the other hand, we remark that this result cannot be improved since there exist a $T_1$ space $(X,\tau)$ generating four distinct topologies $\tau$, $\tau^d=\rho(\tau)$, $\tau^{dd}=\rho^2(\tau)$ and $\tau^{ddd}=\rho^3(\tau)$ (see e.g. Example 8 of \cite{GHSW} or Example 1 of \cite{Bu}).

Keywords

saturated set, order of specialization, dual topology, compactness operator

RIV year

2001

Released

19.08.2001

Publisher

Matematicko-fyzikální fakulta Univerzity Karlovy

Pages from

45

Pages to

46

Pages count

2

Documents

BibTex


@inproceedings{BUT3562,
author="Martin {Kovár}",
title="Problem 540 is (almost) solved",
annote="Recall that a set is said to be saturated if it is the intersection of open sets. By the dual
topology $\tau^d$ for a topological space $(X,\tau)$ we mean the topology on $X$ generated by
taking the compact saturated sets of $X$ as a subbase for closed sets.
The Problem 540  of  J. D. Lawson and M. Mislove \cite{LM} in Open Problems  in
Topology (J. van Mill,   G. M. Reed, eds.,1990)

\medskip
\roster
\item  which topologies can arise as dual topologies

\smallskip
and
\smallskip

\item  whether the process of taking  duals  terminate after finitely many steps with
the  topologies that are duals of each other.
\endroster
\medskip

For $T_1$ spaces, the solution of (2) simply follows from the fact that in $T_1$ spaces
every set is saturated and hence the dual operator $d$ coincide
with the compactness operator  $\rho$ of J. de Groot, G. E. Strecker and E. Wattel \cite{GSW}.
For more general spaces, the question (2) was partially answered by Bruce S. Burdick who
found  certain classes of (in general, non-$T_1$) spaces for which the process of taking duals
of a topological space $(X,\tau)$
terminates by $\tau^{dd}=\tau^{dddd}$  -- the lower Vietoris topology on any hyperspace,
the Scott topology for reverse inclusion on any hyperspace, and the upper Vietoris topology
on the hyperspace of a regular space. B. Burdick presented his paper on The First Turkish
International Conference on Topology in Istanbul  2000 \cite{Bu}.

\medskip

In this talk a general (and positive) solution of  (2) with a short classification of topological spaces
with respect to the number of distinct topologies generated by iterating duals will be
presented.  Our main result is the following theorem:

\proclaim{Theorem} For every topological space $(X,\tau)$ it follows $\tau^{dd}= \tau^{dddd}$.
\endproclaim

On the other hand, we remark that this result cannot be improved since there exist a $T_1$ space
$(X,\tau)$ generating four distinct topologies $\tau$, $\tau^d=\rho(\tau)$, $\tau^{dd}=\rho^2(\tau)$ and
$\tau^{ddd}=\rho^3(\tau)$ (see e.g. Example 8 of \cite{GHSW}
or  Example 1 of \cite{Bu}).
",
}