A description/proof of that closed subspace of locally compact topological space is locally 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 closed set.
- The reader knows a definition of subspace topology.
- The reader admits the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any closed subspace of any locally compact topological 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: Description
For any locally compact topological space, \(T_1\), any closed subspace, \(T_2 \subseteq T_1\), is locally compact.
2: Proof
For any point, \(p \in T_2\), and any open neighborhood of \(p\), \(U_p \subseteq T_2\), is there a compact neighborhood of \(p\), \(N_p \subseteq T_2\), such that \(N_p \subseteq U_p\)?
\(U_p = U'_p \cap T_2\) where \(U'_p \subseteq T_1\) is open on \(T_1\), by the definition of subspace topology. \(p \in U'_p\), so, \(U'_p\) is an open neighborhood of \(p\) on \(T_1\). There is a compact neighborhood of \(p\), \(N'_p \subseteq T_1\), such that \(N'_p \subseteq U'_p\), by the definition of locally compact topological space.
Let us define \(N_p := N'_p \cap T_2 \subseteq T_2\). \(N_p\) is a neighborhood of \(p\) on \(T_2\), by the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace. \(N_p \subseteq U_p\), because \(N'_p \subseteq U'_p\) and \(N_p = N'_p \cap T_2 \subseteq U'_p \cap T_2 = U_p\).
Is \(N_p\) is compact on \(T_2\)? \(N'_p\) is a compact topological space, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace. \(N_p\) is closed on \(N'_p\), so, is compact on \(N'_p\), by the proposition that any closed subset of any compact topological space is compact. So, \(N_p\) is compact on \(T_1\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space. So, \(N_p\) is compact on \(T_2\), by the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.
