description/proof of that for separable topological space induced by metric and subset, open cover of subset has countable subcover
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of separable topological space.
- The reader admits the proposition that any separable metric space is a 2nd-countable topological space.
- The reader admits the proposition that for any map, the cardinality of the range is equal to or smaller than the cardinality of the domain.
Target Context
- The reader will have a description and a proof of the proposition that for any separable topological space induced by any metric and any subset, any open cover of the 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:
\(T\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
\(S\): \(\subseteq T\)
\(\{U_j \in \{\text{ the open subsets of } T\} \vert j \in J'\}\): such that \(S \subseteq \cup_{j \in J'} U_j\)
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Statements:
\(T \in \{\text{ the separable topological spaces }\}\)
\(\implies\)
\(\exists J \subseteq J' (J \in \{\text{ the countable sets }\} \land S \subseteq \cup_{j \in J} U_j)\)
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2: Proof
Whole Strategy: Step 1: see that \(T\) has a countable basis, \(B = \{b_1, b_2, ...\}\); Step 2: for each \(s \in S\), take a \(U_j\) and a \(b_l \in B\) such that \(s \in b_l \subseteq U_j\), and take a map, \(f: S \to B\), and a map, \(g: f (S) \to \{U_j \vert j \in J'\}\), and see that \(S\) is covered by \(g \circ f (S)\); Step 3: see that \(g \circ f (S)\) is countable.
Step 1:
By the proposition that any separable metric space is a 2nd-countable topological space, \(T\) has a countable basis, \(B = \{b_1, b_2, ...\}\).
Step 2:
Let \(s \in S\) be any.
As \(S \subseteq \cup_{j \in J'} U_j\), there is a \(U_j\) such that \(s \in U_j\).
As \(U_j\) is an open neighborhood of \(s\), there is a \(b_l\) such that \(s \in b_l \subseteq U_j\).
So, we can take a map, \(f: S \to B, s \mapsto b_l\).
And we can take a map, \(g: f (S) \to \{U_j \vert j \in J'\}\), such that for each \(b_l \in f (S)\), \(b_l \subseteq g (b_l)\), which is possible because we have chosen \(b_l\) such that there is a \(U_j\) such that \(b_l \subseteq U_j\): there may be some \(s, s' \in S\) such that \(s \neq s'\) and \(s, s' \in b_l\) and \(b_l\) has been chosen as \(s \in b_l \subseteq U_j\) and \(s' \in b_l \subseteq U_{j'}\) with \(U_j \neq U_{j'}\), but that does not matter: the point is that there is at least 1 \(U_j\) such that \(b_l \subseteq U_j\) and we choose a \(U_j\) from some possibly multiple options.
\(S\) is covered by \(g \circ f (S) \subseteq \{U_j \vert j \in J'\}\), because for each \(s \in S\), \(s \in f (s) \subseteq g (f (s))\).
Step 3:
\(g \circ f (S) = g (f (S))\) is countable, because \(f (S) \subseteq B\) is countable, by the proposition that for any map, the cardinality of the range is equal to or smaller than the cardinality of the domain.
