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!

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). $\endgroup$ – Gro-Tsen Dec 21 '19 at 17:56forcedto 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