# Is there a metaLindelof nonLindelof space which has a dense hereditarily Lindelof subspace?

My question is as the title, i.e.,

Is there a metaLindelof nonLindelof space which has a dense hereditarily Lindelof subspace?

The question is related to the following result: Every separable metaLindelof space is Lindelof.

Thank you!

Suppose $$X$$ is metaLindelöf, and $$A \subseteq X$$ is hereditarily Lindelöf and dense.

Let $$\mathcal{U}$$ be an open cover of $$X$$ and let $$\mathcal{V}$$ be a point-countable refinement of $$\mathcal{U}$$. By a standard fact on covers, there is a discrete subset $$D \subseteq A$$ such that $$\operatorname{st}(D,\mathcal{V}_A) = A$$ (where $$\mathcal{V}_A$$ is the relativised version of $$\mathcal{V}$$). As $$A$$ is hereditarily Lindelöf, $$D$$ is countable and as $$A$$ is dense, $$\mathcal{V} = \bigcup_{d \in D} \{V \in \mathcal{V}: d \in V\}$$ and as $$\mathcal{V}$$ is point-countable, $$\mathcal{V}$$ is countable, and so $$\mathcal{U}$$ has a countable refinement which implies that $$X$$ is Lindelöf.

So there is no such example.

For similar ideas, see here, or one of some recent papers on star-properties like this survey,or e.g. this paper etc.

• Thanks, but how could one see that $\mathcal V$ covers $X$ (It is clear that $\mathcal V$ covers $A$)? – Joe Oct 6 '18 at 4:41
• @Joe It covers $X$ from the start. We use metaLindelöf there. – Henno Brandsma Oct 6 '18 at 4:45