Orthocompactness is weakly hereditary

From Topospaces

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

Proof

Proof in terms of open covers

Given: a orthocompact space, a closed subset of (with the subspace topology)

To prove: Consider an open cover of by open sets with , an indexing set. The 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 of such that , thus the union of the s contains .
  2. Since is closed, we can throw in the open set , and get an open cover of the whole space .
  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 , we get an open refinement of the with the property that the intersection of all open subsets containing any point is open.
  5. Consider the open cover of obtained by intersecting each member of this refinement with . This is an open cover of . Moreover, it is a refinement of the (the point here is that the intersection with of a subset contained in is contained in ). Finally, it continues to be true that the intersection of the open subsets containing any point is still open. This completes the proof.