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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!

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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.

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

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