5

Any topology (family of open sets) on a given space determines a family of compact sets on the space (namely, all subsets of the space satisfying the definition of compactness).

Question: Can we go in the opposite direction, i.e. start with a certain family of subsets of the space which are said to be compact?

This would entail at least two sub-questions:

(a) Can a family of compact sets be described axiomatically without referencing a notion of open sets? (I.e. a family of compact sets "for some topology"?)

(b) Given a family of compact sets for some topology, is there only one topology such that this family of compact sets is induced by it? I.e. do we have a well-defined map from families of compact sets to families of open sets?

Motivation: Such a point of view, if possible, might be slightly useful

(1) for motivating proper maps as the analog of continuous maps
(2) in introductory functional analysis, in particular, for making explicit the duality that finer topologies/"more" open sets correspond to "fewer" compact sets, and "more" compact sets correspond to coarser topologies/"fewer" open sets.

In short I think it might be an interesting curiosity, but otherwise don't see any "real" applications.

Chill2Macht
  • 22,055
  • 10
  • 67
  • 178
  • 1
    You certainly don't have uniqueness : for example, the discrete topology and the cofinite topology have both finite sets as compact. But maybe starting from a family of compact (you should probably ask stability by intersections) you can create a topology. The naive way would be to take your topology to be the complement of the compact family. –  Apr 26 '17 at 19:22
  • 2
    This is related to the notion of a compactly generated space. – Nate Eldredge Apr 26 '17 at 19:23
  • 1
    @N.H.: In the cofinite topology, every set is compact, so that example doesn't work. – Nate Eldredge Apr 26 '17 at 19:24
  • 1
    Of course you're right. –  Apr 26 '17 at 19:27
  • 2
    On the other hand, we can take a topology on $\mathbb R$ where opens are the complements of countable set (plus the empty set). This is a topology on $\mathbb R$ and this time if I'm not mistaken the only compact set are the finite set. –  Apr 26 '17 at 19:29
  • 1
    You can find some more counterexamples at https://math.stackexchange.com/questions/517411/in-a-topological-space-only-finite-subsets-are-compact-sets?noredirect=1&lq=1 – Nate Eldredge Apr 26 '17 at 19:39
  • 1
    If S is finite then every subset of S is compact in any topology on S. But if S is finite and has more than 1 member, then there are at least 3 non-homeomorphic topologies on S. – DanielWainfleet Apr 26 '17 at 22:51

1 Answers1

5

No, you cannot uniquely generate a topology from a family of compact sets. Take, for instance, any finite space. No matter which topology you choose, you're bound to get the same family of compact sets. Thus you can't use the family of compact sets to differentiate between the different possible topologies.

As for the axiomatisation, at the very least you need the family to be closed under finite (but I believe arbitrary) intersections and finite unions, and it needs to contain all singletons of the space. Whether that's enough, I don't know.

Arthur
  • 204,511
  • For arbitrary intersections I think we need Hausdorff: https://math.stackexchange.com/questions/2184291/arbitrary-intersection-of-compact-sets-in-non-hausdorff-spaces Also the finite spaces is a very good example -- every open cover has a finite subcover because the power set itself is finite even. – Chill2Macht Apr 27 '17 at 05:15
  • For the axiomatization part, maybe we could use nets/filters/ultrafilters? https://en.wikipedia.org/wiki/Compact_space#Equivalent_definitions I am not sure though since I don't actually understand any of those concepts very well (they might implicitly depend on a notion of open set). – Chill2Macht Apr 27 '17 at 05:22