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?
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.