# Paracompact Hausdorff Space with Dense Lindelof subset is Lindelof

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!

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

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.