4
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ – Douglas Somerset Apr 21 '12 at 20:04
  • $\begingroup$ It can, spaces in which this happens are called H-closed: en.wikipedia.org/wiki/H-closed_space $\endgroup$ – KP Hart Apr 22 '12 at 8:08
4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.