Orthocompactness is weakly hereditary

Property-theoretic statement
The property of being an orthocompact 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
 * Paracompactness is weakly hereditary
 * Metacompactness is weakly hereditary

Proof in terms of open covers
Given: $$X$$ a orthocompact space, $$A$$ a closed subset of $$X$$ (with the subspace topology)

To prove: Consider an open cover of $$A$$ by open sets $$U_i$$ with $$i \in I$$, an indexing set. The $$U_i$$ have an open refinement with the property that at any point, the intersection of all members of the refinement is open.

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 orthocompact, this open cover has a refinement with the property that the intersection of all open subsets containing a point is open.
 * 4) By throwing out any member of this cover that is contained in $$X \setminus A$$, we get an open refinement of the $$V_i$$ with the property that the intersection of all open subsets containing any point is open.
 * 5) Consider the open cover of $$A$$ obtained by intersecting each member of this refinement with $$A$$. This is an open cover of $$A$$. Moreover, it is a refinement of the $$U_i$$ (the point here is that the intersection with $$A$$ of a subset contained in $$V_i$$ is contained in $$U_i$$). Finally, it continues to be true that the intersection of the open subsets containing any point is still open. This completes the proof.