A description/proof of that continuous embedding between topological spaces with closed range is proper
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous map.
- The reader knows a definition of continuous embedding.
- The reader knows a definition of closed set.
- The reader knows a definition of range of map.
- The reader knows a definition of proper map.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
- The reader admits the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any continuous embedding between any topological spaces with any closed range is proper.
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: Description
For any topological spaces, \(T_1, T_2\), any continuous embedding with any closed range, \(f: T_1 \rightarrow T_2\), is proper.
2: Proof
The restriction of \(f\) on the range codomain, \(f': T_1 \rightarrow f (T_1)\), is a homeomorphism. Let \(S \subseteq T_2\) be any compact subset. \(S\) is compact as a subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace. Let \(O := \{U_\alpha \subseteq T_1\vert \alpha \in A\}\) where \(A\) is any possibly uncountable indices set be any open cover of \(f^{-1} (S)\). \(S \cap f (T_1)\) is closed on \(S\), so, is compact on \(S\), by the proposition that any closed subset of any compact topological space is compact, and is compact on \(T_2\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space, and is compact on \(f (T_1)\), by the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.
\(f (U_\alpha)\) is open on \(f (T_1)\), because \(f'\) is homeomorphic. \(\{f (U_\alpha) \vert \alpha \in A\}\) is an open cover of \(S \cap f (T_1) = S \cap f \circ f^{-1} (S)\) (that equal is because any point on \(T_1 \setminus f^{-1} (S)\) does not map into \(S\) anyway) on \(f (T_1)\), because for any point, \(p \in S \cap f \circ f^{-1} (S)\), \(f^{-1} (p) \in \cup_{\alpha \in A} U_\alpha\), \(p \in f (\cup_{\alpha \in A} U_\alpha) = \cup_{\alpha \in A} f (U_\alpha)\), 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. There is a finite subcover, \(\{f (U_i)\vert i \in I \subseteq A\}\) where \(I\) is a finite indices set. \(\{U_i\vert i \in I\}\) is a finite subcover of \(O\), because for any \(p \in f^{-1} (S)\), \(f (p) \in S \cap f \circ f^{-1} (S) \subseteq \cup_{i \in I} f (U_i)\), so, \(f (p) \in f (U_i)\) for an \(i\), which implies that \(p \in U_i\), because \(f\) is injective.
