Paracompactness is weakly hereditary

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

Given: A paracompact space X, a closed subset A of X.

To prove: Consider an open cover of A by open sets U_i with i \in I, an indexing set. The U_i have a locally finite open refinement.

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 compact, this open cover has a locally finite open refinement. In other words, there is a locally finite open refinement of the V_is, that, possibly along with X \setminus A, covers the whole of X.
  4. By throwing out any member of this new refinement that do not intersect A, we get a locally finite open refinement of V_is whose union contains A. The intersections of these with A form a locally finite open refinement of the U_is: The main point here is that if an open set in the refinement is contained in V_i, its intersection with A is contained in U_i.