A description/proof of that for locally compact Hausdorff topological space, in neighborhood around point, there is open neighborhood whose closure is compact and contained in neighborhood
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 for any locally compact Hausdorff topological space, in any neighborhood around any point, there is a compact neighborhood of the point.
- 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, in any neighborhood around any point, there is an open neighborhood of the point whose (the open neighborhood's) closure is compact and contained in the former neighborhood.
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, \(T\), any point, \(p \in T\), and any neighborhood, \(N_p \subseteq T\), of \(p\), there is an open neighborhood, \(U_p \subseteq T\), of \(p\) such that the closure, \(\overline{U_p}\), is compact on \(T\) and is contained in \(N_p\), which is \(U_p \subseteq \overline{U_p} \subseteq N_p\).
2: Proof
There is a compact neighborhood, \(C_p \subseteq T\), contained in \(N_p\), which is \(C_p \subseteq N_p\), by the proposition that for any locally compact Hausdorff topological space, in any neighborhood around any point, there is a compact neighborhood of the point. \(C_p\) is closed on \(T\), by the proposition that any compact subset of any Hausdorff topological space is closed.
There is an open neighborhood, \(U_p \subseteq T\), contained in \(C_p\), which is \(U_p \subseteq C_p\). \(\overline{U_p} \subseteq C_p \subseteq N_p\) because \(\overline{U_p}\) is the smallest closed set that contains \(U_p\) while \(C_p\) is such a closed set. \(\overline{U_p}\) is closed on \(C_p\) 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. \(C_p\) 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 \(C_p\), 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.
