description/proof of that retract of Hausdorff topological space is closed
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of retract of topological space.
- The reader knows a definition of closed set.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that any retract on any Hausdorff topological space is closed on the space.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\in \{\text{ the retracts of } T\}\)
//
Statements:
\(S \in \{\text{ the closed subsets of } T\}\)
//
2: Natural Language Description
For any topological space, \(T\), any retract, \(S \subseteq T\), is closed on \(T\).
3: Proof
Whole Strategy: use the local criterion for openness for \(T \setminus S\); Step 1: take a retraction, \(r: T \to S\); Step 2: for each \(p \in T \setminus S\), take an open neighborhood of \(p\), \(U \subseteq T\), such that \(U \subseteq T \setminus S\).
Step 1:
There is a retraction, \(r: T \to S\), which is continuous, such that \(r \vert_S = id\).
Step 2:
Let \(p \in T \setminus S\) be any.
\(r (p) \neq p\), so, there are some open neighborhoods of \(p\) and \(r (p)\), \(U_p \subseteq T\) and \(U_{r (p)} \subseteq T\), such that \(U_p \cap U_{r (p)} = \emptyset\), because \(T\) is Hausdorff.
Let \(U := r^{-1} (U_{r (p)} \cap S) \cap U_p\).
\(p \in U\), because \(r (p) \in U_{r (p)} \cap S\), so, \(p \in r^{-1} (U_{r (p)} \cap S)\).
\(U\) is open on \(T\), because \(U_{r (p)} \cap S\) is open on \(S\) and \(r\) is continuous.
\(U \subseteq (T \setminus S)\), because \(r^{-1} (U_{r (p)} \cap S)\) consists of \(U_{r (p)} \cap S\) and some points outside \(S\), because as \(r\vert_S\) is the identity map, any point in \(S \setminus (U_{r (p)} \cap S)\) is not mapped into \(U_{r (p)} \cap S\), and as \(U_p \cap U_{r (p)} = \emptyset\), \(U\) consists only of points outside \(S\).
So, \(T \setminus S\) is open on \(T\), by the local criterion for openness. So, \(S\) is closed on \(T\).
