This is an exercise of Munkres topology section 41.
Let $X$ be a regular space. If $X$ is a countable union of closed paracompact subspaces of $X$ whose interiors cover $X$, show $X$ is paracompact.
I Know this lemma:
Let $X$ be a regular space. Then TFAE (note: a refinement must itself also be a cover).
- Every open cover of $X$ has a locally finite open refinement. (i.e. $X$ is paracompact)
- Every open cover of $X$ has a $\sigma$-locally finite open refinement (where a family of sets is $\sigma$-locally finite iff it is a countable union of locally finite families)
- Every open cover of $X$ has a locally finite refinement (of any kind).
- Every open cover of $X$ has a locally finite closed refinement.
The following is what I tried.
Denote each closed parcompact subspace $U_{n}$. Think about an open covering $\mathscr{A}$ of $X$. To prove paracompactness, I should prove that $\mathscr{A}$ satisfies $1$ of the above lemma. However, because $X$ is a 'countable' union of $U_{n}$, I think that the way to the answer is proving that $\mathscr{A}$ satisfies $2$.
But I can't proceed any longer.
Best Answer
Let $\mathcal{U}$ be an open covering of a regular space $X=\bigcup_\mathbb{N}X_n$ which is a union of countably many closed paracompact subspaces $X_n\subseteq X$ whose interiors $(X_n)^\circ$ cover $X$.
For each $n\in\mathbb{N}$ the family $\mathcal{U}_n=\{U\cap X_n\}_{U\in\mathcal{U}}$ is an open covering of the paracompact $X_n$. Thus there is a family $\mathcal{V}_n'$ of open subsets of $X$ which has the property that $\{V\cap X_n\}_{V\in\mathcal{V}'_n}$ is a locally-finite open refinement (in $X_n)$ of $\mathcal{U}_n$. Write $\mathcal{V}_n=\{V\cap(X_n)^\circ\}_{V\in\mathcal{V}'_n}$ to obtain a locally-finite family of open subsets of $X$ which covers the interior $(X_n)^\circ$ and refines $\mathcal{U}_n$.
The collection $$\mathcal{V}=\bigcup_\mathbb{N}\mathcal{V}_n$$ is now a $\sigma$-locally-finite open refinement of $\mathcal{U}$.
Appeal to part $2$ of your theorem to complete.