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!

  • 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

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.

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