description/proof of that for continuous map between topological spaces, image of compact subset of domain is compact subset of codomain
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of compact subset of topological space.
- The reader admits the proposition that for any map, the composition of the preimage after the map of any subset contains the argument set.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
- The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
- The reader admits the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(K_1\): \(\in \{\text{ the compact subsets of } T_1\}\)
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Statements:
\(f (K_1) \in \{\text{ the compact subsets of } T_2\}\)
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2: Proof
Whole Strategy: Step 1: take any open cover of \(f (K_1)\), \(\{U_j \vert j \in J'\}\); Step 2: see that \(K_1 \subseteq \cup_{j \in J'} f^{-1} (U_j)\); Step 3: take a finite subset, \(J \subseteq J'\), such that \(K_1 \subseteq \cup_{j \in J} f^{-1} (U_j)\); Step 4: see that \(f (K_1) \subseteq \cup_{j \in J} U_j\).
Step 1:
Let \(\{U_j \vert j \in J'\}\) be any open cover of \(f (K_1)\), where \(J'\) is any possibly uncountable index set, which means that \(f (K_1) \subseteq \cup_{j \in J'} U_j\).
Step 2:
\(f^{-1} (f (K_1)) \subseteq f^{-1} (\cup_{j \in J'} U_j)\).
\(K_1 \subseteq f^{-1} (f (K_1))\), by the proposition that for any map, the composition of the preimage after the map of any subset contains the argument set.
\(f^{-1} (\cup_{j \in J'} U_j) = \cup_{j \in J'} f^{-1} (U_j)\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
So, \(K_1 \subseteq \cup_{j \in J'} f^{-1} (U_j)\).
\(f^{-1} (U_j) \subseteq T_1\) is open, because \(f\) is continuous.
So, \(\{f^{-1} (U_j) \vert j \in J'\}\) is an open cover of \(K_1\).
Step 3:
There is a finite subcover of the open cover of \(K_1\), because \(K_1\) is compact, which means that there is a finite \(J \subseteq J'\), such that \(K_1 \subseteq \cup_{j \in J} f^{-1} (U_j)\).
Step 4:
\(f (K_1) \subseteq f (\cup_{j \in J} f^{-1} (U_j)) = \cup_{j \in J} f (f^{-1} (U_j))\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
\(f (f^{-1} (U_j)) \subseteq U_j\), by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
So, \(f (K_1) \subseteq \cup_{j \in J} U_j\).
That means that \(\{U_j \vert j \in J\}\) is a finite subcover of \(\{U_j \vert j \in J'\}\).
So, \(f (K_1)\) is compact.
