*[Cross-posted from MSE.](https://math.stackexchange.com/questions/4580731)*


Following [Bankston - The total negation of a topological property](https://doi.org/10.1215/ijm/1256048236), a topological space is called *anticompact* if all its compact subsets are finite. The linked MSE post above has two examples:


**Example 1:** The topology on $\mathbb{R}$ generated by the usual Euclidean topology together with the cocountable topology is an uncountable space that is connected anticompact and Hausdorff.


**Example 2:** Given a free ultrafilter $\mathfrak{F}$ on $\mathbb{N}$, take the ultrafilter topology $\tau = \mathfrak{F} \cup \{\emptyset\}$ on $\mathbb{N}$. This is a countably infinite, connected anticompact space. It is $T\_1$, but not $T\_2$.


Does there exist a countably infinite connected anticompact space that is Hausdorff?




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Note that such an example would have to be totally path disconnected. Also it cannot be sequential, as a consequence of [this answer](https://math.stackexchange.com/a/4567778) to [Is a space where only finite subsets are compact sets always discrete?](https://math.stackexchange.com/questions/517411/is-a-space-where-only-finite-subsets-are-compact-sets-always-discrete). In particular, it cannot be first countable.



Is there a connected Hausdorff anticompact space that is countably infinite?
Such an example has been constructed by [Banakh and Stelmakh](https://arxiv.org/abs/2211.12579).
More precisely, they constructed an anticompact countable connected Hausdorff space which is Brown and strongly rigid.


