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
- Compactness is weakly hereditary
- Paracompactness is weakly hereditary
- Metacompactness is weakly hereditary
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.
- By the definition of subspace topology, we can find open sets of such that , thus the union of the s contains .
- Since is closed, we can throw in the open set , and get an open cover of the whole space .
- 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 , we get an open refinement of the with the property that the intersection of all open subsets containing any point is open.
- 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.