A description/proof of that for locally compact Hausdorff topological space, around point, there is open neighborhood whose closure is compact
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 neighborhood of point.
- The reader knows a definition of compact subset of topological space.
- The reader knows a definition of subspace topology.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any locally compact Hausdorff topological space, around any point, there is an open neighborhood whose closure is 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: Description
For any locally compact Hausdorff topological space, \(T\), and any point, \(p \in T\), there is an open neighborhood, \(U_p \subseteq T\), of \(p\) such that the closure, \(\overline{U_p}\), is compact on \(T\).
2: Proof
There is a compact neighborhood, \(N_p\), of \(p\). \(N_p\) is closed, by the proposition that any compact subset of any Hausdorff topological space is closed. There is an open neighborhood, \(U_p \subseteq N_p\). \(\overline{U_p} \subseteq N_p\) because \(\overline{U_p}\) is the smallest closed set that contains \(U_p\) while \(N_p\) is a closed set that contains \(U_p\). \(\overline{U_p}\) is closed on \(N_p\) by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset. \(N_p\) is a compact topological space, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace. \(\overline{U_p}\) is compact on \(N_p\), by the proposition that any closed subset of any compact topological space is compact. \(\overline{U_p}\) is compact on \(T\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
