closed subset of weakly lindelof - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:30:46Z http://mathoverflow.net/feeds/question/94757 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94757/closed-subset-of-weakly-lindelof closed subset of weakly lindelof Douglas Somerset 2012-04-21T18:05:09Z 2012-04-21T18:18:16Z <p>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.</p> <p>Question: Are closed subsets of weakly Lindelof spaces necessarily weakly Lindelof?</p> http://mathoverflow.net/questions/94757/closed-subset-of-weakly-lindelof/94759#94759 Answer by Andreas Blass for closed subset of weakly lindelof Andreas Blass 2012-04-21T18:16:51Z 2012-04-21T18:16:51Z <p>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.</p> http://mathoverflow.net/questions/94757/closed-subset-of-weakly-lindelof/94760#94760 Answer by KP Hart for closed subset of weakly lindelof KP Hart 2012-04-21T18:18:16Z 2012-04-21T18:18:16Z <p>The <a href="https://en.wikipedia.org/wiki/Moore_plane" rel="nofollow">Niemytzki plane</a> is weakly Lindel&ouml;f (the open upper half plane is a dense Lindel&ouml;f subspace); the $x$-axis is an uncountable closed and discrete subspace.</p>