show that every countable subcover is $\sigma$- locally finite refinement. (every regular Lindelöf space is paracompact)

Question

I was trying to show that the proposition above, if a topological space $X$ is regular (which I mean seperating the points and closed sets) and Lindelöf, then the space is paracompact. For any open cover $\cal{U}$ of $X$, for all $x\in X$ choose a $U_{x}\in\cal{U}$ and because $X$ is regular space, then choose $G_{x}\in\cal{U}$ for all $U_{x}$ such that $x\in G_{x}\subseteq cl(G_{x})\subseteq U_{x}$ and then $\mathcal{G}=\{G_{x} : x\in X\}$ is also an open cover of $X$, so it has a countable subcover too. But after that, I do not know what to do when I take countable subcover of $\cal{G}$. Is there any smooth solution? Any hint or help will be greatly appreciated.

Answer 1

Theorem: If $X$ is regular we have the following conditions are equivalent: Every open covering of $X$ has a refinement that is

• An open covering of $X$ and countably locally finite (A collection $\mathcal{B}$ said to be countably locally finite on $X$ if $\mathcal{B}=\cup_{n \in \mathbb{N}}B_n$ where each $B_n$ is locally finite on $X$).
• A covering of X and locally finite.
• A closed covering of $X$ and locally finite.
• An open covering of $X$ and locally finite.

As you mentioned in the question, one can find the collection $\mathcal{G}=\{G_x : x\in X\}$ such that each $G_x \subset U_x$. This collection is actully refinement of $\mathcal{U}$. Apply the theorem above we can conclude $X$ is paracompact.

Edit: To answer your question, just pick the countable subcover of $\mathcal{G}$ since $X$ is Lindelof and countable subcovering is automatically countably locally finite.

Answer 2

Lose the $G_x$, just take a countable subcover $\{U_n\mid n \in \omega\}$ of $\mathcal{U}$ that still covers $X$ (applying Lindelöfness), and as $\{U_n\mid n \in \omega\} = \bigcup_{n \in \omega} \{U_n\}$, which is a countable union of finite (so certainly locally-finite!) families, it is countably locally finite and a subcover is trivially a refinement.

So $\mathcal{U}$ has a $\sigma$-locally-finite refinement and as $X$ is regular, the standard theorem quoted by Huy Nguyen applies and so $X$ is paracompact.

That's the whole proof: two trivial observations, i.e. countable implies $\sigma$-locally finite and a subcover is a refinement.