396: Continuous Embedding Between Topological Spaces with Closed Range Is Proper

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A description/proof of that continuous embedding between topological spaces with closed range is proper

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Description

For any topological spaces, $T_1,T_2$, any continuous embedding with any closed range, $f:T_1\to T_2$, is proper.

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2: Proof

The restriction of $f$ on the range codomain, $f^{\prime }:T_1\to f(T_1)$, 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 $O:=\{U_{\alpha }\subseteq T_1|\alpha \in A\}$ 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 $f^{\prime }$ is homeomorphic. $\{f(U_{\alpha })|\alpha \in A\}$ 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, $\{f(U_i)|i\in I\subseteq A\}$ where $I$ is a finite indices set. $\{U_i|i\in I\}$ 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.

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References

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