# closed subset of weakly lindelof

A topological space $X$ is weakly Lindelof if every open cover has a countable subfamily $U$ such that $\bigcup \{ V: V\in U\}$ is dense in X.

Question: Are closed subsets of weakly Lindelof spaces necessarily weakly Lindelof?

The Niemytzki plane is weakly Lindelöf (the open upper half plane is a dense Lindelöf subspace); the $x$-axis is an uncountable closed and discrete subspace.
No. Consider the space whose points are all sequences of 0's and 1's of length $\leq\omega$. Visualize it as the binary tree plus "limits" for all paths through the tree, and topologize it accordingly. That is, each finite sequence is an isolated point, but a neighborhood of an infinite sequence $s$ must contain all sufficiently long finite initial segments of $s$. This space is weakly Lindelöf because the finite sequences constitute a countable dense set. But the infinite sequences constitute a closed, discrete, uncountable, and therefore not weakly Lindelöf subspace.
However, if $$X$$ is a normal weakly Lindelof space, $$F$$ is a closed subspace of $$X$$ and $$\mathcal{U}$$ is an open cover of $$F$$ then there is a countable subcollection $$\mathcal{V}$$ of $$\mathcal{U}$$ such that $$F \subseteq \overline{\bigcup \mathcal{V}}$$.
Indeed, if $$\mathcal{U}$$ covers $$X$$ then we're done. So we can assume that $$G:=X \setminus \bigcup \mathcal{U}$$ is non-empty. Noting that $$F$$ and $$G$$ are non-empty disjoint closed sets, use normality to find an open set $$O$$ such that $$G \subset O$$ and $$\overline{O} \cap F=\emptyset$$. Then $$\mathcal{U} \cup \{O\}$$ is an open cover of the weakly Lindelof space $$X$$ and hence it contains a countable $$\mathcal{C}$$ such that $$\bigcup \mathcal{C}$$ is dense in $$X$$. Since $$\overline{O} \cap F=\emptyset$$, the set $$\mathcal{V}:=\mathcal{C} \setminus \{O\}$$ is a countable subfamily of $$\mathcal{U}$$ such that $$F \subset \overline{\bigcup \mathcal{V}}$$.
This doesn't mean that $$F$$ is weakly Lindelof though. Indeed one can make the Niemytzki Plane normal by replacing the $$x$$-axis with a $$Q$$-set, that is an uncountable subset of the reals whose every subset is a relative $$G_\delta$$. It is consistent with ZFC that such a set exists. The $$Q$$-set would still be a closed non-weakly Lindelof subspace of the Niemytzki Plane but, in view of the above, it would at least be "weakly Lindelof" with respect to covers made up of open subsets of $$X$$.