description/proof of that for Hausdorff topological space, if there is locally compact Hausdorff topological space of which space is locally closed subspace, space is locally compact
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of locally compact topological space.
- The reader knows a definition of locally closed topological subspace.
- The reader knows a definition of topological subspace.
- The reader admits the proposition that any topological subspace is locally closed if and only if it is the intersection of a closed subset and an open subset of the base space.
- The reader admits the proposition that any closed subspace of any locally compact topological space is locally compact.
- The reader admits the proposition that any subspace of any Hausdorff topological space is Hausdorff.
- The reader admits the proposition that any open subspace of any locally compact Hausdorff topological space is locally compact.
- The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any Hausdorff topological space, if there is a locally compact Hausdorff topological space of which the space is a locally closed subspace, the space is locally compact.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
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Statements:
\(\exists T' \in \{\text{ the locally compact Hausdorff topological spaces }\} (T \in \{\text{ the locally closed subspaces of } T'\})\)
\(\implies\)
\(T \in \{\text{ the locally compact topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(T = C' \cap U'\) for a closed \(C' \subseteq T'\) and an open \(U' \subseteq T'\); Step 2: see that \(C'\) is locally compact and that \(T\) is a locally compact subspace of \(C'\).
Step 1:
There are a closed \(C' \subseteq T'\) and an open \(U' \subseteq T'\) such that \(T = C' \cap U'\), by the proposition that any topological subspace is locally closed if and only if it is the intersection of a closed subset and an open subset of the base space.
Step 2:
\(C' \subseteq T'\) is a locally compact Hausdorff topological subspace, by the proposition that any closed subspace of any locally compact topological space is locally compact and the proposition that any subspace of any Hausdorff topological space is Hausdorff.
\(T = C' \cap U' \subseteq C'\), as the topological subspace of \(C'\), is an open subspace, by the definition of topological subspace, and is a locally compact topological subspace of \(C'\), by the proposition that any open subspace of any locally compact Hausdorff topological space is locally compact.
But \(T\) as the topological subspace of \(C'\) is \(T\) as the topological subspace of \(T'\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
As \(T\) we are talking about is the topological subspace of \(T'\), \(T\) is locally compact.
