2
$\begingroup$

Let $X$ be a Paracompact Hausdroff space with a dense subset $A$ which is Lindelöf. Then, $X$ is Lindelof

I've written down my attenpt below -

As per the hint in the problem, as a paracompact $T_2$ space is regular, all I have to do is show that every open cover of $X$ has a countable subcollection whose closures cover.

So, for any open cover $\{U_\alpha\}$ of $X$, we get an open cover $\{V_\alpha\}$ of $A$, where $V_\alpha = A \cap U_\alpha$.

As $A$ is Lindelöf, we can thus get a countable subcollection $\{V_{\alpha_i}:i\in \mathbb{N}\}$, such that $\bigcup_\limits{i=1}^{\infty} \overline V_{\!\!\alpha_i} = A$.

So, I believe now that we will get $\bigcup_\limits{i=1}^{\infty} \overline U_{\!\!\alpha_i} = X$, thus showing $X$ is Lindelöf.

But, this is the part I'm stuck at. Somehow, we have to use the fact that $A$ is dense, but I just can't figure it out. Any help in solving this is appreciated!

$\endgroup$
  • 1
    $\begingroup$ It may help to first pick a locally finite refinement of the cover. $\endgroup$ – Daniel Fischer Aug 1 at 13:51
  • $\begingroup$ And observe that a locally finite family on a Lindelöf space is always at most countable. $\endgroup$ – Henno Brandsma Aug 1 at 14:15
3
$\begingroup$

First make a boring observation:

Lemma: any locally finite family of subsets on a Lindelöf space $X$ is at most countable.

Proof: for every $x \in X$, pick $O_x$ witnessing the local finiteness, and since this cover has a countable subcover, the original family of subsets is also at most countable.

Let $\mathcal{U}=\{U_i: i \in I\}$ be an open cover, and let $\mathcal{V}=\{V_i: i \in I\}$ be a locally finite open refinement of it, so that $\overline{V_i} \subseteq U_i$ for all $i$ (it is a standard fact that this can be done in paracompact Hausdorff spaces).

As $A$ is Lindelöf, $\{V_i \cap A: i \in I\}$ is at most countable. It is clear that $X$ is covered by the corresponding $U_i$ and we have a countable subcover.

| cite | improve this answer | |
$\endgroup$

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.