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)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about orthocompact space |Get facts that use property satisfaction of orthocompact space | Get facts that use property satisfaction of orthocompact space|Get more facts about weakly hereditary property of topological spaces

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

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:

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 $A$ .
Since $A$ is closed, we can throw in the open set $X ∖ A$ , and get an open cover of the whole space $X$ .
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.
By throwing out any member of this cover that is contained in $X ∖ A$ , we get an open refinement of the $V_{i}$ with the property that the intersection of all open subsets containing any point is open.
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.