description/proof of that for locally compact Hausdorff topological space and 1st open subset whose closure is compact and contained in 2nd open subset, there is open subset that contains closure of 1st subset whose closure is compact and contained in 2nd subset
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 closure of subset of topological space.
- The reader knows a definition of countably compact subset of topological space.
- The reader admits 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.
- The reader admits the proposition that the closure of the union of any finite number of subsets is the union of the closures of the subsets.
- The reader admits the proposition that for any topological space, the union of any finite compact subsets is compact.
Target Context
- The reader will have a description and a proof of the proposition that for any locally compact Hausdorff topological space and any 1st open subset whose closure is compact and contained in any 2nd open subset, there is an open subset that contains the closure of the 1st subset whose (the 3rd subset's) closure is compact and contained in the 2nd subset.
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:
\(T\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(U_1\): \(\in \{\text{ the open subsets of } T\}\)
\(U_2\): \(\in \{\text{ the open subsets of } T\}\)
//
Statements:
\(\overline{U_1} \in \{\text{ the compact subsets of } T\} \land \overline{U_1} \subseteq U_2\)
\(\implies\)
\(\exists U_3 \in \{\text{ the open subsets of } T\} (\overline{U_3} \in \{\text{ the compact subsets of } T\} \land \overline{U_1} \subseteq U_3 \subseteq \overline{U_3} \subseteq U_2)\)
//
2: Note
The pair of \(U_1\) and \(U_2\) is typically taken as this: for any \(t \in T\) and any open neighborhood of \(t\), \(U_2\), there is an open neighborhood of \(t\), \(U_1\), such that \(\overline{U_1}\) is compact and \(\overline{U_1} \subseteq U_2\), by 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.
As the pair of \(U_1\) and \(U_3\) and the pair of \(U_3\) and \(U_2\) satisfy the conditions for this proposition, this proposition can be iteratively applied to the pairs, to the pairs derived from the pairs, and so on.
3: Proof
Whole Strategy: Step 1: for each \(t \in \overline{U_1}\), take an open neighborhood of \(t\), \(U_t\), such that \(\overline{U_t}\) is compact and \(\overline{U_t} \subseteq U_2\); Step 2: take a finite \(\{U_{t_j}\}\) such that \(\overline{U_1} \subseteq \cup_{j \in J} U_{t_j} := U_3\), and see that \(U_3\) satisfies the conditions for this proposition.
Step 1:
Let \(t \in \overline{U_1}\) be any.
As \(t \in \overline{U_1} \subseteq U_2\), there is an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(\overline{U_t} \subseteq T\) is compact and \(\overline{U_t} \subseteq U_2\), by 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.
Step 2:
\(\{U_t \vert t \in \overline{U_1}\}\) is an open cover of \(\overline{U_1}\).
As \(\overline{U_1}\) is compact, there is a finite subcover, \(\{U_{t_j} \vert j \in J\}\), where \(J\) is a finite index set.
Let us define \(U_3 := \cup_{j \in J} U_{t_j}\), open on \(T\).
\(\overline{U_1} \subseteq U_3\).
\(\overline{U_3} = \overline{\cup_{j \in J} U_{t_j}} = \cup_{j \in J} \overline{U_{t_j}}\), by the proposition that the closure of the union of any finite number of subsets is the union of the closures of the subsets.
\(\overline{U_3} = \cup_{j \in J} \overline{U_{t_j}} \subseteq T\) is compact, by the proposition that for any topological space, the union of any finite compact subsets is compact.
\(\overline{U_3} = \cup_{j \in J} \overline{U_{t_j}} \subseteq U_2\), because \(\overline{U_{t_j}} \subseteq U_2\) for each \(j \in J\).
So, \(U_3\) satisfies the conditions for this proposition.
