Orthocompactness is weakly hereditary

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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: 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_is 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.