description/proof of that closed subset of compact topological space is compact
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of compact subset of topological space.
- The reader knows a definition of closed subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that any closed subset of any compact topological space is 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the compact topological spaces }\}\)
\(C\): \(\in \{\text{ the closed subsets of } T\}\)
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Statements:
\(C\): \(\in \{\text{ the compact subsets of } T\}\)
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2: Proof
Whole Strategy: Step 1: take any open cover of \(C\), \(\{U_j \vert j \in J'\}\), take the open cover of \(T\), \(\{U_j \vert j \in J'\} \cup \{T \setminus C\}\), take a finite subcover of the cover of \(T\), \(\{U_j \vert j \in J\} \cup \{T \setminus C\}\), and see that \(\{U_j \vert j \in J\}\) is a finite subcover of the cover of \(C\).
Step 1:
Let \(\{U_j \vert j \in J'\}\) be any open cover of \(C\), where \(J'\) is a possibly uncountable index set.
\(\{U_j \vert j \in J'\} \cup \{T \setminus C\}\) is an open cover of \(T\).
As \(T\) is compact, there is a finite subcover of the cover of \(T\), \(\{U_j \vert j \in J\} \cup \{T \setminus C\}\), where \(J \subseteq J'\) is a finite index set: \(T \setminus C\) may not be necessary, but we can have it anyway, because it does no cause any harm: it is a finite subcover anyway.
Then, \(\{U_j \vert j \in J\}\) is a finite subcover of the cover of \(C\), because for each \(c \in C\), \(c \in U_j\) for a \(j \in J\), because \(c \notin T \setminus C\), so, \(C \subseteq \{U_j \vert j \in J\}\).
So, \(C\) is compact.
