description/proof of that countably-compactness of topological subset as subset equals countably-compactness as subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of countably compact subset of topological space.
- The reader knows a definition of topological subspace.
- The reader knows a definition of countably compact topological space.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
Target Context
- The reader will have a description and a proof of the proposition that the countably-compactness of any topological subset as the subset equals the countably-compactness as the subspace.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\), with the subspace topology
//
Statements:
\(S \in \{\text{ the countably compact subsets of topological spaces }\}\)
\(\iff\)
\(S \in \{\text{ the countably compact topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(S\) is a countably compact subset of \(T\); Step 2: see that \(S\) as the subspace is a countably compact topological space; Step 3: suppose that \(S\) as the subspace is a countably compact topological space; Step 4: see that \(S\) is a countably compact subset of \(T\).
Step 1:
Let us suppose that \(S\) is a countably compact subset of \(T\).
Step 2:
Let us see that \(S\) as the subspace is a countably compact topological space.
Let \(\{U_j \subseteq S \vert j \in J\}\), where \(J\) is any countable index set, be any countable open cover of \(S\) as the subspace.
\(U_j = U'_j \cap S\) where \(U'_j \subseteq T\) is an open subset of \(T\), by the definition of subspace topology.
\(\{U'_j \subseteq T \vert j \in J\}\) is a countable open cover of \(S\): as \(S = \cup_{j \in J} U_j\), \(S \subseteq \cup_{j \in J} U'_j\), because \(U_j \subseteq U'_j\).
As \(S\) is countably compact, there is a finite subcover, \(\{U'_j \subseteq T \vert j \in J^`\}\), where \(J^` \subset J\) is a finite index set.
Then, \(\{U_j \subseteq S \vert j \in J^`\}\) is a finite subcover of \(S\) as the topological subspace, because \(\cup_{j \in J^`} U_j = S\), because from \(S \subseteq \cup_{j \in J^`} U'_j\), \(S \cap S \subseteq (\cup_{j \in J^`} U'_j) \cap S = \cup_{j \in J^`} (U'_j \cap S)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(= \cup_{j \in J^`} U_j \subseteq S\), but the left hand side is \(S\), so, \(S \subseteq \cup_{j \in J^`} U_j \subseteq S\).
So, \(S\) as the subspace is a countably compact topological space.
Step 3:
Let us suppose that \(S\) as the subspace is a countably compact topological space.
Step 4:
Let us see that \(S\) is a countably compact subset of \(T\).
Let \(\{U'_j \subseteq T \vert j \in J\}\), where \(J\) is any countable index set, be any countable open cover of \(S\).
\(U_j := U'_j \cap S\) is an open subset of \(S\) as the topological subspace, by the definition of subspace topology.
\(\{U_j \subseteq S \vert j \in J\}\) is a countable open cover of \(S\) as the topological subspace, because \(\cup_{j \in J} U_j = S\), because from \(S \subseteq \cup_{j \in J} U'_j\), \(S \cap S \subseteq (\cup_{j \in J} U'_j) \cap S = \cup_{j \in J} (U'_j \cap S)\), as before, \(= \cup_{j \in J} U_j \subseteq S\), but the left hand side is \(S\), so, \(S \subseteq \cup_{j \in J} U_j \subseteq S\).
As \(S\) as the topological subspace is countably compact, there is a finite subcover, \(\{U_j \subseteq S \vert j \in J^`\}\), where \(J^` \subseteq J\) is a finite index set.
Then, \(\{U'_j \subseteq T \vert j \in J^`\}\) is a finite subcover of \(S\), because \(S \subseteq \cup_{j \in J^`} U'_j\), because \(S = \cup_{j \in J^`} U_j\) and \(U_j \subseteq U'_j\).
So, \(S\) is a countably compact subset of \(T\).
