# Is $\mathbb R$ with cocountable topology star-$K$-compact?

1. A space $$X$$ is said to be starcompact if for every open cover $$\mathcal U$$ of $$X$$ there exists a finite subset $$\mathcal V\subseteq\mathcal U$$ such that $$St(\cup\mathcal V,\mathcal U)=X$$.

2. A space $$X$$ is said to be star-$$K$$-compact if for every open cover $$\mathcal U$$ of $$X$$ there exists a compact subset $$K$$ of $$X$$ such that $$St(K,\mathcal U)=X$$.

It is immediate that $$\mathbb R$$ with cocountable topology is starcompact. But we don't know whether $$\mathbb R$$ with cocountable topology is star-$$K$$-compact. It is clear from the definitions that every star-$$K$$-compact space is starcompact.

Yes, $$\mathbb R$$ with the cocountable topology is star-$$K$$-compact. If $$\mathcal U$$ is an open cover of $$\mathbb R$$ then, as any open set misses only countably many points, there is a countable $$\mathcal U'\subseteq\mathcal U$$ still covering $$\mathbb R$$. Now $$\mathbb R$$ is not the countable union of countable sets so that $$\bigcap \mathcal U'\neq\emptyset$$. It follows that there is a point $$x$$ with $$\mathbb R=St(\{x\},\mathcal U')=St(\{x\},\mathcal U)$$.