definition of completely regular topological space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of continuous, topological spaces map.
Target Context
- The reader will have a definition of completely regular topological 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:
\(*T\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
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Conditions:
\(\forall t \in T (\forall C \in \{\text{ the closed subsets of } T\} \text{ such that } t \notin C (\exists f: T \to [0, 1] \in \{\text{ the continuous maps }\} (f (t) = 0 \land f (C) = \{1\})))\)
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2: Note
The definition is equivalent if \(f (t) = 1 \land f (C) = \{0\}\) is required instead of \(f (t) = 0 \land f (C) = \{1\}\): if there is an \(f\) such that \(f (t) = 0 \land f (C) = \{1\}\), \(1 - f\) is \(: T \to [0, 1]\) and continuous, and \((1 - f) (t) = 1 - 0 = 1\) and \((1 - f) (C) = \{1 - 1\} = \{0\}\); if there is an \(f\) such that \(f (t) = 1 \land f (C) = \{0\}\), \(1 - f\) is \(: T \to [0, 1]\) and continuous, and \((1 - f) (t) = 1 - 1 = 0\) and \((1 - f) (C) = \{1 - 0\} = \{1\}\).
The definition is equivalent if a neighborhood of \(t\), \(N_t \subseteq T\), and a continuous \(f: T \to [0, 1]\) such that \(f (N_t) = \{0\}\) and \(f (C) = \{1\}\) are required: if there is an \(f\) such that \(f (t) = 0 \land f (C) = \{1\}\), there is an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(f (U_t) \subseteq [0, 1 / 2)\), and there is the \(f': [0, 1] \to [0, 1], [0, 1 / 2) \mapsto 0, t \in [1 / 2, 1] \mapsto 2 (t - 1 / 2)\), which is continuous, and \(f' \circ f\) is \(: T \to [0, 1]\) and continuous, and \(f' \circ f (U_t) = \{0\}\) and \(f' \circ f (C) = \{1\}\), so, \(U_t\) and \(f' \circ f\) will do; if there are a \(N_t\) and an \(f\) such that \(f (N_t) = \{0\}\) and \(f (C) = \{1\}\), \(f (t) = 0\) and \(f (C) = \{1\}\).
The definition is equivalent if a neighborhood of \(t\), \(N_t \subseteq T\), and a continuous \(f: T \to [0, 1]\) such that \(f (N_t) = \{1\}\) and \(f (C) = \{0\}\) are required: if there are an \(N_t\) and an \(f\) such that \(f (N_t) = \{0\}\) and \(f (C) = \{1\}\), \(1 - f\) is \(: T \to [0, 1]\) and continuous, and \((1 - f) (N_t) = \{1 - 0\} = \{1\}\) and \((1 - f) (C) = \{1 - 1\} = \{0\}\); if there are an \(N_t\) and an \(f\) such that \(f (N_t) = \{1\}\) and \(f (C) = \{0\}\), \(1 - f\) is \(: T \to [0, 1]\) and continuous, and \((1 - f) (N_t) = \{1 - 1\} = \{0\}\) and \((1 - f) (C) = \{1 - 0\} = \{1\}\).
