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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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3 Answers 3

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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.

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Many thanks. Tick to the first answer (by a minute). Looking at the (non-Hausdorff) examples which motivated this question, I now see that one of them has a non-weakly-Lindelof closed set whose complement is a dense open compact set. Thus the whole space is weakly Lindelof. In fact it has the stronger property that every open cover has a finite subfamily with dense union. Presumably this stronger property cannot occur in a non-compact Hausdorff space? –  Douglas Somerset Apr 21 '12 at 20:04
It can, spaces in which this happens are called H-closed: en.wikipedia.org/wiki/H-closed_space –  KP Hart Apr 22 '12 at 8:08

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.

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However, closed subspaces of normal weakly Lindelof spaces are indeed weakly Lindelof.

Let $X$ be such a space, $F \subset X$ be closed and $\mathcal{U}$ be an open cover of $F$. If $\mathcal{U}$ covers $X$ then we're done. So we can assume that $G:=X \setminus \bigcup \mathcal{U}$ is a 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 $\mathcal{V}$ such that $\bigcup \mathcal{V}$ is dense in $X$. Since $\overline{O} \cap F=\emptyset$, the set $\mathcal{C}:=\mathcal{V} \setminus \{O\}$ is a countable subfamily of $\mathcal{U}$ such that $F \subset \overline{\bigcup \mathcal{C}}$.

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