description/proof of that separable metric space is 2nd-countable topological space
Topics
About: metric space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of metric space.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of separable topological space.
- 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 any separable metric space is a 2nd-countable 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:
\(T\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
//
Statements:
\(T \in \{\text{ the separable topological spaces }\}\)
\(\implies\)
\(T \in \{\text{ the 2nd-countable topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any countable dense subset, \(S = \{p_j \in T \vert j \in J\}\), and take the set of the open balls, \(B := \{B_{p_j, q} \subseteq T \vert j \in J, q \in \mathbb{Q}_+\}\); Step 2: see that \(B\) is a basis for \(T\).
Step 1:
There is a countable dense subset, \(S = \{p_j \in T \vert j \in J\} \subseteq T\), where \(J\) is a countable index set.
Let us think of the set of the open balls, \(B := \{B_{p_j, q} \subseteq T \vert j \in J, q \in \mathbb{Q}_+\}\), where \(\mathbb{Q}_+\) is the positive rational numbers set.
\(B\) is a countable set.
Step 2:
Let us see that \(B\) is a basis for \(T\).
Each \(B_{p_j, q} \subseteq T\) is open.
For each \(p_j \in S\) and each open neighborhood of \(p_j\), \(U_{p_j} \subseteq T\), there is a \(B_{p_j, q} \subseteq U_{p_j}\), because \(T\) has the topology induced by the metric. \(p_j \in B_{p_j, q} \subseteq U_{p_j}\).
For each \(p \in T \setminus S\) and each open neighborhood of \(p\), \(U_p \subseteq T\), there is an open ball around \(p\), \(B_{p, \epsilon} \subseteq T\), such that \(B_{p, \epsilon} \subseteq U_p\), because \(T\) has the topology induced by the metric.
There is a \(p_j \in S\) such that \(p_j \in B_{p, \epsilon / 4}\), because \(T\) is separable: \(p\) is an accumulation point of \(S\).
Let \(q \in \mathbb{Q}_+\) be any that satisfies \(\epsilon / 4 \lt q \lt \epsilon / 2\).
\(B_{p_j, q} \subseteq B_{p, \epsilon}\), because for each \(p' \in B_{p_j, q}\), \(dist (p', p) \le dist (p', p_j) + dist (p_j, p) \lt \epsilon / 2 + \epsilon / 4 = 3 / 4 \epsilon\). So, \(p \in B_{p_j, q} \subseteq B_{p, \epsilon} \subseteq U_p\).
So, \(B\) is a basis for \(T\), and \(T\) is 2nd-countable.
