description/proof of that locally compact Hausdorff topological space is open subspace of its \(1\)-point compactification
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally compact topological space.
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of topological subspace.
- The reader admits the proposition that for any locally compact Hausdorff topological space, the topology of the \(1\)-point compactification is the only topology that makes the \(1\)-point-augmented set compact Hausdorff with the original space as the subspace.
- The reader admits the proposition that for any Hausdorff topological space, any 1 point subset is closed.
Target Context
- The reader will have a description and a proof of the proposition that any locally compact Hausdorff topological space is an open subspace of its \(1\)-point compactification.
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:
\(T\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(T^+\): \(= \text{ the 1-point compactification }\)
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Statements:
\(T \in \{\text{ the open subspaces of } T^+\}\)
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2: Proof
Whole Strategy: Step 1: see that \(T\) is the subspace of the Hausdorff \(T^+\); Step 2: see that \(T = T^+ \setminus \{\infty\}\) is open on \(T^+\).
Step 1:
\(T^+\) is a Hausdorff topological space and \(T\) is the topological subspace of \(T^+\), by the proposition that for any locally compact Hausdorff topological space, the topology of the \(1\)-point compactification is the only topology that makes the \(1\)-point-augmented set compact Hausdorff with the original space as the subspace.
Step 2:
\(\{\infty\} \subseteq T^+\) is closed, by the proposition that for any Hausdorff topological space, any 1 point subset is closed.
So, \(T = T^+ \setminus \{\infty\}\) is open on \(T^+\).
So, \(T\) is an open topological subspace of \(T^+\).
