description/proof of that for 2nd-countable completely regular topological space, there is countable set of continuous maps into closed unit interval s.t. for each point and each closed subset that does not contain point, there is element of set that is \(0\) over open neighborhood of point and is \(1\) over closed subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of 2nd-countable topological space.
- The reader knows a definition of completely regular topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any 2nd-countable completely regular topological space, there is a countable set of some continuous maps into the closed unit interval such that for each point and each closed subset that does not contain the point, there is an element of the set that is \(0\) over an open neighborhood of the point and is \(1\) over the closed subset.
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:
\(T\): \(\in \{\text{ the 2nd-countable topological spaces }\} \cap \{\text{ the completely regular topological spaces }\}\)
\([0, 1]\): \(\subseteq \mathbb{R}\), with the subspace topology
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Statements:
\(\exists J \in \{\text{ the countable index sets }\}, \exists S = \{f_j: T \to [0, 1] \in \{\text{ the continuous maps }\} \vert j \in J\} (\forall t \in T (\forall C \in \{\text{ the closed subsets of } T\} \text{ such that } t \notin C (\exists f_j \in S (\exists U_t \in \{\text{ the open neighborhoods of } t\} (f (U_t) = \{0\} \land f (C) = \{1\})))))\)
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2: Proof
Whole Strategy: Step 1: take any countable basis, \(B\), take the set of the pairs of some elements of \(B\), \(J := \{(b_1, b_2)\}\), such that there is a continuous map, \(f: T \to [0, 1]\), such that \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\), choose such an \(f\) for each such pair, and take the set, \(S' := \{((b_1, b_2), f_{(b_1, b_2)}) \vert (b_1, b_2) \in J\}\); Step 2: take \(S := \{f_j \vert j \in J, (j, f_j) \in S'\}\); Step 3: see that \(S\) satisfies the requirements of this proposition.
Step 0:
When \(T = \emptyset\), the proposition vacuously holds, because \(J = \mathbb{N}\) and for each \(j \in J\), the only possible \(f_j: T \to [0, 1]\) will do, because there is no \(t \in T\).
Let us suppose otherwise, hereafter.
Step 1:
Let \(B\) be any countable basis for \(T\).
Let \(J \subseteq B \times B\) be the set of the pairs of some elements of \(B\) such that for each \((b_1, b_2) \in J\), there is a continuous map, \(f: T \to [0, 1]\), such that \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\).
\(J\) is countable, because it is a subset of the countable, \(B \times B\).
\(J\) is nonempty, in fact, for each nonempty \(b_1 \in B\), there is at least \(1\) \(b_2 \in B\) such that \((b_1, b_2) \in J\), because there is a \(t \notin T \setminus b_1\), so, there is an open neighborhood of \(t\), \(U_t \subseteq T\), and a continuous map, \(f: T \to [0, 1]\), such that \(f (U_t) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\), because \(T\) is completely regular (refer to Note for the definition of completely regular topological space), and there is a \(b_2 \in B\) such that \(t \in b_2 \in U_t\), so, \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\) hold.
Now, we have the function, \(F\), that the domain is \(J\) and for each \((b_1, b_2) \in J\), \(F ((b_1, b_2)) = \{f: T \to [0, 1] \in \{\text{ the continuous maps }\} \vert f (b_2) = \{0\} \land f (T \setminus b_1) = \{1\}\}\), which is nonempty.
By the axiom of choice, there is a function, \(F'\), that the domain is \(J\) and for each \((b_1, b_2) \in J\), \(F' ((b_1, b_2)) \in F ((b_1, b_2))\).
Let us take \(S' := \{((b_1, b_2), f_{(b_1, b_2)} := F' ((b_1, b_2))) \vert (b_1, b_2) \in J\}\).
Step 2:
Let us take \(S := \{f_j \vert j \in J, (j, f_j) \in S'\}\).
\(S\) is a countable set.
Step 3:
Let us see that \(S\) satisfies the requirements of this proposition.
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
\(t \in T \setminus C\), so, \(T \setminus C \subseteq T\) is an open neighborhood of \(t\).
So, there is a \(b_1 \in B\) such that \(t \in b_1 \subseteq T \setminus C\).
\(t \notin T \setminus b_1\).
There are an open neighborhood of \(t\), \(U_t \subseteq T\), and a continuous \(f: T \to [0, 1]\) such that \(f (U_t) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\), because \(T\) is completely regular.
There is a \(b_2 \in B\) such that \(t \in b_2 \subseteq U_t\).
So, \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\).
So, \((b_1, b_2) \in J\).
So, \(f_{(b_1, b_2)} \in S\) satisfies that \(f_{(b_1, b_2)} (b_2) = \{0\}\) and \(f_{(b_1, b_2)} (T \setminus b_1) = \{1\}\).
\(C \subseteq T \setminus b_1\), because for each \(c \in C\), \(c \notin T \setminus C\), so, \(c \notin b_1 \subseteq T \setminus C\), so, \(c \in T \setminus b_1\), so, \(f_{(b_1, b_2)} (C) = \{1\}\).
