# Is there an $\varepsilon$-space which is not $k$-Lindelöf?

Crossposted from https://math.stackexchange.com/questions/4717613

An $$\omega$$-cover $$\mathscr U$$ of a space $$X$$ is a collection of open sets so that $$X \not\in\mathscr U$$ and every finite subset of $$X$$ is contained in a member of $$\mathscr U$$. Similarly, a $$k$$-cover $$\mathscr U$$ of a space $$X$$ is a collection of open sets so that $$X \not\in\mathscr U$$ and every compact subset of $$X$$ is contained in a member of $$\mathscr U$$.

A space is an $$\varepsilon$$-space (aka $$\omega$$-Lindelöf) if every $$\omega$$-cover has a countable subset which is an $$\omega$$-cover. Likewise, a space is $$k$$-Lindelöf if every $$k$$-cover has a countable subset which is a $$k$$-cover.

It can be shown that $$k$$-Lindelöf $$\implies$$ $$\varepsilon$$-space $$\implies$$ Lindelöf. The Sorgenfrey line is an example of a space which is Lindelöf but not an $$\varepsilon$$-space. This leads us to the following question.

Is there an $$\varepsilon$$-space which is not $$k$$-Lindelöf? Can the example be Hausdorff? Metrizable?

If it can be of any help, it would suffice to find a non-compact $$\varepsilon$$-space $$X$$ so that $$\mathbb K(X)$$, the space of compact subsets of $$X$$ with the Vietoris topology, is not Lindelöf by Cor. 4.16 of https://doi.org/10.1016/j.topol.2021.107772.

No such example exists in the realm of separable metrizable spaces.

Claim. Every separable metrizable space is $$k$$-Lindelöf.

Proof. If $$X$$ is separable metrizable, then $$\mathbb K(X)$$ is also separable metrizable; hence, Lindelöf. If we let $$[U] = \{ K \in \mathbb K(X) : K \subseteq U \}$$, then, for a given $$k$$-cover $$\mathscr U$$ of $$X$$, $$\mathscr W := \{ [U] : U \in \mathscr U \}$$ is an open cover of $$\mathbb K(X)$$. Then a countable subcover of $$\mathscr W$$ corresponds to a countable subset of $$\mathscr U$$ which is a $$k$$-cover of $$X$$. $$\square$$