Paracompactness is weakly hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., paracompact space) satisfying a topological space metaproperty (i.e., weakly hereditary property of topological spaces)
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Statement

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 $U_{i}$.