Orthocompactness is weakly hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., orthocompact 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 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

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.