Paracompactness is weakly hereditary

Property-theoretic statement
The property of being a paracompact space is a weakly hereditary property of topological spaces.

Verbal statement
Any closed subset of a paracompact space is a paracompact space with the subspace topology.

Related facts

 * Compactness is weakly hereditary

Proof
Given: A paracompact space $$X$$, a closed subset $$A$$ of $$X$$.

To prove: Consider an open cover of $$A$$ by open sets $$U_i$$ with $$i \in I$$, an indexing set. The $$U_i$$ have a locally finite open refinement.

Proof:


 * 1) By the definition of subspace topology, we can find open sets $$V_i$$ of $$X$$ such that $$V_i \cap A = U_i$$, thus the union of the $$V_i$$s contains $$A$$.
 * 2) Since $$A$$ is closed, we can throw in the open set $$X \setminus A$$, and get an open cover of the whole space $$X$$.
 * 3) Since the whole space is compact, this open cover has a locally finite open refinement. In other words, there is a locally finite open refinement of the $$V_i$$s, that, possibly along with $$X \setminus A$$, covers the whole of $$X$$.
 * 4) By throwing out any member of this new refinement that do not intersect $$A$$, we get a locally finite open refinement of $$V_i$$s whose union contains $$A$$. The intersections of these with $$A$$ form a locally finite open refinement of the $$U_i$$s: The main point here is that if an open set in the refinement is contained in $$V_i$$, its intersection with $$A$$ is contained in $$U_i$$.