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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?

flag	asked Apr 21 2012 at 18:05 Douglas Somerset 447●1●5

KP Hart · Accepted Answer · 2012-04-21 18:18:16Z UTC

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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.

answered Apr 21 2012 at 18:18

KP Hart
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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 2012 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 2012 at 8:08

Andreas Blass · Answer 2 · 2012-04-21 18:16:51Z UTC

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.

closed subset of weakly lindelof

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