A description/proof of that for Hausdorff topological space and 2 disjoint compact subsets, there are disjoint open subsets each of which contains compact subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of compact subset.
Target Context
- The reader will have a description and a proof of the proposition that for any Hausdorff topological space and its any 2 disjoint compact subsets, there are some disjoint open subsets each of which contains one of the compact subsets.
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 Hausdorff topological space, \(T\), and any disjoint compact subsets, \(S_1, S_2 \subseteq T\), such that \(S_1 \cap S_2 = \emptyset\), there are some open subsets, \(U_1, U_2 \subseteq T\), such that \(S_i \subseteq U_i\) and \(U_1 \cap U_2 = \emptyset\).
2: Proof
For any point, \(p_{2, \alpha} \in S_2\), for each point, \(p_{1, \beta} \in S_1\), there are some disjoint open neighborhoods, \(U_{p_{1, \beta}} \cap U_{p_{2, \alpha, \beta}} = \emptyset\). \(\{U_{p_{1, \beta}}\}\) is an open cover of \(S_1\) and has a finite subcover, \(\{U_{p_{1, i}}\}\). There is the corresponding \(\{U_{p_{2, \alpha, i}}\}\). Let us define \(U_{p_{1, \alpha}} := \cup_{i} U_{p_{1, i}}\) and \(U_{p_{2, \alpha}} := \cap U_{p_{2, \alpha, i}}\), where \(U_{p_{1, \alpha}}\) is indexed with \(\alpha\), because it depends on \(p_{2, \alpha}\). \(U_{p_{i, \alpha}}\) is open and \(U_{p_{2, \alpha}}\) is nonempty (because \(p_{2, \alpha}\) is contained) open. \(S_1 \subseteq U_{p_{1, \alpha}}\) and \(U_{p_{1, \alpha}} \cap U_{p_{2, \alpha}} = \emptyset\), because for any \(p \in U_{p_{2, \alpha}}\), \(p \in U_{p_{2, \alpha, i}}\) for each \(i\) and \(p \notin U_{p_{1, i}}\) for each \(i\).
\(\{U_{p_{2, \alpha}}\}\) where \(\alpha\) is moved around is an open cover of \(S_2\) and has a finite subcover, \(\{U_{p_{2, i}}\}\), and there is the corresponding, \(U_{p_{1, i}}\). Let us define \(U_1 := \cap_{i} U_{p_{1, i}}\) and \(U_2 := \cup_{i} U_{p_{2, i}}\), both open. \(S_1 \subseteq U_1\) and \(S_2 \subseteq U_2\). \(U_1 \cap U_2 = \emptyset\), because for any \(p \in U_1\), \(p \in U_{p_{1, i}}\) for each \(i\) and \(p \notin U_{p_{2, i}}\) for each \(i\).
