Problem 540 is (almost) solved

článek ve sborníku ve WoS nebo Scopus

angličtina

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}).

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

KOVÁR, M.

2001

19. 8. 2001

Matematicko-fyzikální fakulta Univerzity Karlovy

45

46

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