description/proof of that on 2nd-countable topological space, open cover has countable subcover
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of 2nd-countable topological space.
Target Context
- The reader will have a description and a proof of the proposition that on any 2nd-countable topological space, any open cover of any subset has a countable subcover.
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:
\(T\): \(\in \{\text{ the 2nd-countable topological spaces }\}\)
\(S\): \(\subseteq T\)
\(J\): \(\{\text{ the possibly uncountable index sets }\}\)
\(O\): \(= \{U_j \subseteq T \vert j \in J\}\), \(\in \{\text{ the open covers of } S\}\)
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Statements:
\(\exists J^` \subseteq J \in \{\text{ the countable index sets }\} (\{U_j \subseteq T \vert j \in J^`\} \in \{\text{ the countable open covers of } S\})\)
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2: Proof
Whole Strategy: Step 1: take any countable basis, \(B = \{V_l \subseteq T \vert l \in L\}\), define \(L^` := \{l \in L \vert \exists j \in J (V_l \subseteq U_j)\}\) and \(f: L^` \to Pow (O), l \mapsto \{U_j \in O \vert V_l \subseteq U_j\}\), and take a choice function from the set of the nonempty subsets of \(O\) into \(O\), \(c\), and see that \(c \circ f (L^`)\) is a countable subcover.
Step 1:
Let \(B = \{V_l \subseteq T \vert l \in L\}\), where \(L\) is a countable index set, be any countable basis for \(T\).
Let us define the index set, \(L^` := \{l \in L \vert \exists j \in J (V_l \subseteq U_j)\}\), which is countable as a subset of countable \(L\).
Let us define the function, \(f: L^` \to Pow (O), l \mapsto \{U_j \in O \vert V_l \subseteq U_j\}\), which is not necessarily injective, but that is not any problem.
With respect to \(O\), there is a choice function from the set of the nonempty subsets into \(O\) that (the function) chooses an element from each nonempty subset, denoted by \(c\).
Each element of \(f (L^`)\) is a nonempty subset of \(O\), because \(L^`\) has been chosen such that each point image is nonempty, so, \(c \circ f: L^` \to O\) is well-defined.
\(c \circ f (L^`) \subseteq O\).
\(c \circ f (L^`)\) covers \(S\), because for each \(s \in S\), there is a \(U_j \in O\) such that \(s \in U_j\), there is an \(l \in L^`\) such that \(s \in V_l \subseteq U_j\), and there is the \(c \circ f (l)\), which may not be \(U_j\) but contains \(V_l\) anyway, so, \(s\) is contained in \(c \circ f (l)\).
\(c \circ f (L^`)\) is countable, because \(L^`\) is countable.
So, \(c \circ f (L^`)\) is a countable subcover of \(S\).
