description/proof of that normal topological space is completely regular
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of normal topological space.
- The reader knows a definition of completely regular topological space.
- The reader admits the proposition that any normal topological space is regular.
- The reader admits the proposition that any regular topological space is Hausdorff.
- The reader admits the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma).
Target Context
- The reader will have a description and a proof of the proposition that any normal topological space is completely regular.
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:
\(T\): \(\in \{\text{ the normal topological spaces }\}\)
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Statements:
\(T \in \{\text{ the completely regular topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(T\) is Hausdorff; Step 2: for each \(t \in T\) and each closed \(C\) such that \(t \notin C\), take \(\{t\} \subseteq T \setminus C\) and apply the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma).
Step 1:
\(T\) is regular, by the proposition that any normal topological space is regular.
\(T\) is Hausdorff, by the proposition that any regular topological space is Hausdorff.
Step 2:
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
\(\{t\} \subseteq T\) is a closed subset, by the definition of normal topological space.
\(\{t\} \subseteq T \setminus C\), where \(T \setminus C \subseteq T\) is open.
By the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma), there is a continuous \(f: T \to [0, 1]\) such that \(f (\{t\}) = \{0\}\) and \(f (T \setminus (T \setminus C)) = \{1\}\).
But \(f (t) = 0\) and \(f (C) = f (T \setminus (T \setminus C)) = \{1\}\).
So, \(T\) is completely regular.
