description/proof of that union of finite compact subsets of 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.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the union of any finite compact subsets 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{S_j \subseteq T \vert j \in J\}\): such that \(S_j \in \{\text{ the compact subsets of } T\}\)
\(S\): \(= \cup_{j \in J} S_j\):
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Statements:
\(S \in \{\text{ the compact subsets of } T\}\)
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2: Proof
Whole Strategy: Step 1: take any open cover of \(S\), \(\{U_{j'} \vert j' \in J'\}\), and for each \(j \in J\), take a finite subcover for \(S_j\), \(\{U_{j'} \vert j' \in J'_j\}\), and see that \(\{U_{j'} \vert j' \in \cup_{j \in J} J'_j\}\) is a finite subcover for \(S\).
Let \(\{U_{j'} \subseteq T \vert j' \in J'\}\), where \(J'\) is any possibly uncountable index set, be any open cover of \(S\).
For each \(j \in J\), \(\{U_{j'} \vert j' \in J'\}\) is an open cover of \(S_j\), because \(S_j \subseteq S\).
As \(S_j\) is compact, there is a finite subcover, \(\{U_{j'} \vert j' \in J'_j\}\), where \(J'_j\) is a finite index set.
\(\{U_{j'} \vert j' \in \cup_{j \in J} J'_j\}\) is a finite subcover for \(S\), because \(S \subseteq \cup_{j' \in \cup_{j \in J} J'_j} U_{j'} = \cup_{j \in J} \cup_{j' \in J'_j} U_{j'}\), because for each \(s \in S\), \(s \in S_j\) for an \(S_j\), and \(s \in S_j \subseteq \cup_{j' \in J'_j} U_{j'} \subseteq \cup_{j \in J} \cup_{j' \in J'_j} U_{j'}\), and \(\cup_{j \in J} J'_j\) is finite.
So, \(S\) is compact.
