The following is well-known: if $X$ is a topological spaces, then the union of compact subsets in it need not be compact. But, if $I$ is a compact set in the hyperspace $H(X)$ of all compact subsets of $X$ endowed with the Vietoris topology, then $\bigcup_{A\in I}A$ is compact. In other words, a compact union of compact sets is compact. Is there a similar result for closed sets? Namely, is a closed union of closed sets a closed set? If not, is it true for certain topological spaces? References are welcomed.

  • $\begingroup$ are you interested in the Vietoris monad? $\endgroup$ – Stu Kraji Apr 11 '16 at 17:12
  • $\begingroup$ Yes @StuKraji do you have a good reference? $\endgroup$ – Ittay Weiss Apr 12 '16 at 4:29
  • $\begingroup$ I will write a short answer for now (so that I can edit it later). May I ask you why are you interested in the vietoris monad? $\endgroup$ – Stu Kraji Apr 12 '16 at 7:06
  • $\begingroup$ I'm interested in a convenient formalism for upper/lower semicontinuous functions, primarily in the context of generalised inverse limits of compacta (and generalisations thereof). $\endgroup$ – Ittay Weiss Apr 12 '16 at 7:44

It could be useful to look in the literature for monads in the hyperspace of closed sets. Given the proximity with the powerset, you might find the type of result that you are looking for expressed by the multplication of the monad (I know, usually is the other way around :P).

If you are interested in a monad on $Top$ using the hyperspace of the closed sets, maybe a good starting point is the lower Vietoris monad - $V$ (the topology is generated by the sets of closed subsets that hit an open set). The multiplication of this monad gives you that if $I$ is a closed set in $VX$ then $\bigcup_{A\in I}A$ is a closed set in $X$.

Maybe you will find the following references useful:

Clementino and Tholen - A characterization of the Vietoris Topology

Michael - Topologies on space of subsets

Nachbin - Compact unions of closed subsets are compact and compact intersections of open subsets are open

  • $\begingroup$ Would anything go wrong if one takes the upper Vietoris topology? or the Vietoris topology? $\endgroup$ – Ittay Weiss Apr 12 '16 at 10:39
  • $\begingroup$ It does not work for the upper or the Vietoris topology (at least considering all topological spaces and the topology defined in terms of closed sets) $\endgroup$ – Stu Kraji Apr 13 '16 at 16:25

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