description/proof of that for topological space, compact subset of space that is contained in subspace is compact on subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of subspace topology of subset of topological space.
- 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, any compact subset of the space that is contained in any subspace is compact on 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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(K\): \(\in \{\text{ the compact subsets of } T'\}\), such that \(K \subseteq T\)
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Statements:
\(K \in \{\text{ the compact subsets of } T\}\)
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2: Note
When \(K\) is not contained in \(T\), \(K \cap T\) is not necessarily compact on \(T\), as is described in an article.
3: Proof
Whole Strategy: Step 1: take any open cover of \(K\) on \(T\), \(\{U_j \vert j \in J\}\), and take a finite subcover.
Step 1:
Let \(\{U_j \vert j \in J\}\) be any open cover of \(K\) on \(T\).
For each \(j \in J\), \(U_j\) is an open subset of \(T\), so, \(U_j = U'_j \cap T\) where \(U'_j \subseteq T'\) is an open subset, by the definition of subspace topology.
\(\{U'_j \vert j \in J\}\) is an open cover of \(K\) on \(T'\), because as \(\{U_j \vert j \in J\}\) covers \(K\), the possibly larger \(\{U'_j \vert j \in J\}\) covers \(K\).
As \(K\) is compact on \(T'\), there is a finite subcover, \(\{U'_j \vert j \in J^`\}\), where \(J^` \subseteq J\) is a finite subset.
\(\{U_j \vert j \in J^`\}\) is a finite subcover, because \(\cup_{j \in J^`} U_j = \cup_{j \in J^`} (U'_j \cap T)\), but each point, \(k \in K\), satisfies \(k \in U'_j\) for a \(j \in J^`\) and \(k \in T\), so, \(k \in U'_j \cap T = U_j\) for that \(j\).
So, \(K\) is compact on \(T\).
