description/proof of that completely regular topological space is regular
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of completely regular topological space.
- The reader knows a definition of regular topological space.
- The reader admits the proposition that for any Hausdorff topological space, any 1 point subset is closed.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
Target Context
- The reader will have a description and a proof of the proposition that any completely regular topological space is 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:
\(T\): \(\in \{\text{ the completely regular topological spaces }\}\)
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Statements:
\(T \in \{\text{ the regular topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(T\) satisfies the conditions to be regular.
Step 1:
While \(T\) is Hausdorff by definition, \(\forall t \in T (\{t\} \in \{\text{ the closed subsets of } T\})\), by the proposition that for any Hausdorff topological space, any 1 point subset is closed.
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
There is a continuous map, \(f: T \to [0, 1]\), such that \(f (t) = 0\) and \(f (C) = \{1\}\).
There are the open neighborhood of \(0\), \(B_{0, 1 / 4} \subseteq [0, 1]\) and the open neighborhood of \(1\), \(B_{1, 1 / 4} \subseteq [0, 1]\), such that \(B_{0, 1 / 4} \cap B_{1, 1 / 4} = \emptyset\).
\(f^{-1} (B_{0, 1 / 4}) \subseteq T\) is an open neighborhood of \(t\) and \(f^{-1} (B_{1, 1 / 4}) \subseteq T\) is an open neighborhood of \(C\), because \(f\) is continuous and \(f (t) = 0\) and \(f (C) = \{1\}\).
\(f^{-1} (B_{0, 1 / 4}) \cap f^{-1} (B_{1, 1 / 4}) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
So, \(T\) is regular.
