Let \(X\) be a regular space.
If \(X\) is a finite union of closed paracompact subspaces of \(X\), then \(X\) is paracompact.
If \(X\) is a countable union of closed paracompact subspaces whose interiors cover \(X\), show \(X\) is paracompact.
Solution
We use the closed-refinement characterization: a regular space is paracompact iff every open cover has a locally finite closed refinement covering the space.
Suppose \[ X=A_1\cup\cdots\cup A_m \] with each \(A_i\) closed and paracompact. Let \(\mathcal U\) be an open cover of \(X\). Its restriction to \(A_i\) has a locally finite closed refinement \(\mathcal F_i\) covering \(A_i\). Because \(A_i\) is closed in \(X\), every member of \(\mathcal F_i\) is closed in \(X\). A finite union of locally finite families is locally finite, so \[ \mathcal F_1\cup\cdots\cup\mathcal F_m \] is a locally finite closed refinement of \(\mathcal U\) covering \(X\). Thus \(X\) is paracompact.
Suppose \[ X=\bigcup_{n\ge1}A_n \] where each \(A_n\) is closed and paracompact and the interiors \(A_n^\circ\) cover \(X\). Define \[ C_n=A_n-\bigcup_{i<n}A_i^\circ. \] Each \(C_n\) is closed in \(X\), is a closed subspace of the paracompact space \(A_n\), hence is paracompact, and the \(C_n\)’s cover \(X\). They are locally finite: if \(x\in A_m^\circ\), then the neighborhood \(A_m^\circ\) misses \(C_n\) for every \(n>m\).
Let \(\mathcal U\) be an open cover of \(X\). For each \(n\), choose a locally finite closed refinement \(\mathcal F_n\) of the restricted cover on \(C_n\), covering \(C_n\). Since \(C_n\) is closed in \(X\), all members of \(\mathcal F_n\) are closed in \(X\). The union \[ \mathcal F=\bigcup_{n\ge1}\mathcal F_n \] is locally finite: near a fixed point only finitely many \(C_n\)’s occur, and within each of those finitely many \(C_n\)’s the corresponding family \(\mathcal F_n\) is locally finite. Thus \(\mathcal F\) is a locally finite closed refinement of \(\mathcal U\) covering \(X\). Hence \(X\) is paracompact.