description/proof of that closed continuous bijection is homeomorphism
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closed map.
- The reader knows a definition of continuous map.
- The reader knows a definition of bijection.
- The reader knows a definition of homeomorphism.
- The reader admits the proposition that for any injective map, the image of any subset minus any subset is the image of the 1st subset minus the image of the 2nd subset.
Target Context
- The reader will have a description and a proof of the proposition that any closed continuous bijection is a homeomorphism.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in\{\text{ the closed maps }\} \cap \{\text{ the continuous maps }\} \cap \{\text{ the bijections }\}\)
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Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(f^{-1}\) is continuous.
Step 1:
As \(f\) is a bijection, there is the inverse, \(f^{-1}: T_2 \to T_1\).
Let \(U \subseteq T_1\) be any open subset.
\({f^{-1}}^{-1} (U) = f (U) = f (T_1 \setminus C)\) where \(C \subseteq T_1\) is a closed subset.
\(= f (T_1) \setminus f (C)\), by the proposition that for any injective map, the image of any subset minus any subset is the image of the 1st subset minus the image of the 2nd subset, \(= T_2 \setminus f (C)\), because \(f\) is surjective.
\(f (C) \subseteq T_2\) is closed, because \(f\) is closed, and so, \(T_2 \setminus f (C) \subseteq T_2\) is open.
So, \(f^{-1}\) is continuous.
So, \(f\) is a homeomorphism.
