description/proof of that subspace of completely regular topological space is completely 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 topological subspace.
- The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any subspace of any completely regular 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:
\(T'\): \(\in \{\text{ the completely regular topological spaces }\}\)
\(T\): \(\subseteq T'\) with the subspace topology
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Statements:
\(T \in \{\text{ the completely regular topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(t \in T\) and each \(C \subseteq T\) such that \(t \notin C\), take \(C' \subseteq T'\) such that \(C = C' \cap T\), take \(f': T' \to [0, 1]\) such that \(f' (t) = 0\) and \(f' (C') = \{1\}\), and see that \(f := f' \vert_T\) will do.
Step 1:
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
\(C = C' \cap T\) for a closed \(C' \subseteq T'\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
While \(t \in T'\), \(t \notin C'\), because if \(t \in C'\), \(t \in C' \cap T = C\), a contradiction.
As \(T'\) is completely regular, there is a continuous \(f': T' \to [0, 1]\) such that \(f' (t) = 0\) and \(f' (C') = \{1\}\).
Let us take \(f := f' \vert_T = T \to [0, 1]\).
\(f\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
\(f (t) = 0\).
\(f (C) = \{1\}\), because \(C \subseteq C'\) and \(f (C) = f' (C)\).
So, \(T\) is completely regular.
