I was thinking to the following problem. Take a set $X$. If you take a compact topology T (non necessarily Hausdorff) you get the subposet $K_T$ of $\mathcal{P}(X)$ made of compact sets with respect to $T$.

One can list simple properties that $K_T $ respect: it is stable for union, it contains finite sets, it contains X and the empty set. But I don't think these are sufficient.

One possibly useful reformulation is that there exist a "core" $k_T \subset K_T$, which are the closed sets of the topology, with the following property.

$A \in K_T$ iff for every $C_i \in k_T$ with $\bigcap C_i \cap A = \emptyset$, then there exist a finite number of indices $S$ such that $\bigcap_{i \in S} C_i \cap A = \emptyset$.

If you pass to the complement, this is equivalent to the compactness. I was thinking on conditions that ensure that such a "core" exist.

Another observation is the following. If you take the cofinite topology and the indiscrete topology, they both make everything compact. So this core we are searching for could be not unique!

  • $\begingroup$ Maybe it would be simpler to ask the question for compact Hausdorff subsets of $X$ (not assuming $(X,T)$ itself Hausdorff, or even [quasi-]compact)? Compact Hausdorff often behaves better than quasi-compact, and specifically, in this case, they are automatically closed. So, a variant of your question: can we characterize families $\mathscr{K}\subseteq\mathscr{P}(X)$ which are the compact Hausdorff subspaces of $X$ for some topology on $X$ (with no assumptions on this topology)? $\endgroup$ – Gro-Tsen Dec 21 '19 at 11:58
  • $\begingroup$ It would be a starting point, but my initial motivation deals with the family of compact, non necessarily hausdorff, subsets! I don't know how to link, I am referring to "what are the algebras for the ultrafilter monad on topological spaces"? $\endgroup$ – Andrea Marino Dec 21 '19 at 12:03
  • $\begingroup$ Here's the question about algebras for the ultrafilter monad (to create a link, write [text](URL of link)). $\endgroup$ – Gro-Tsen Dec 21 '19 at 17:56
  • $\begingroup$ There is the class of anti-compact spaces $X$ where a subspace of $X$ is compact iff the subspace is finite (i.e. when it is forced to be, by cardinality). Discrete spaces are an example, and the cocountable topology on an uncountable set. It only shows that the finite sets are realisable in quite different ways.. $\endgroup$ – Henno Brandsma Dec 23 '19 at 22:06

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