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?
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Sign up to join this communityA 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 $X$.