description/proof of that for topological space induced by metric, space has countable basis iff space is separable
Topics
About: metric 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 knows a definition of basis of topological space.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space induced by any metric, the space has a countable basis if and only if the space is separable.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\) induced by any metric, \(dist: T \times T \to \mathbb{R}\)
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Statements:
\(\exists B \in \{\text{ the countable bases for } T\}\)
\(\iff\)
\(T \in \{\text{ the separable topological spaces }\}\)
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2: Note
It is crucial for \(T\) to be induced by a metric, as is seen in Proof.
3: Proof
Whole Strategy: Step 1: suppose that \(T\) is separable with any countable dense subset, \(S\); Step 2: see that the set of the 'rational-number'-open balls, \(\{B_{s_j, q} \vert s_j \in S, q \in \mathbb{Q}\}\), is a countable basis for \(T\); Step 3: suppose that \(T\) has a countable basis, \(\{B_j\}\); Step 4: for each \(B_j\), take any \(s_j \in B_j\), and take \(S := \{s_j\}\), and see that \(S\) is a countable dense subset of \(T\).
Step 1:
Let us suppose that \(T\) is separable with any countable dense subset, \(S = \{s_1, s_2, ...\} \subseteq T\).
Step 2:
Let us take the set of the 'rational-number'-open balls, \(\{B_{s_j, q} \vert s_j \in S, q \in \mathbb{Q}\}\).
It is countable because \(S\) and \(\mathbb{Q}\) are countable: let \(\mathbb{Q} = \{q_1, q_2, ...\}\) and count \(\{B_{s_j, q} \vert s_j \in S, q \in \mathbb{Q}\}\) as \(\{B_{s_1, q_1}, B_{s_1, q_2}, B_{s_2, q_1}, B_{s_1, q_3}, B_{s_2, q_2}, B_{s_3, q_1}, ...\}\): order \(B_{s_j, q_l}\) s by 1st \(j + l\) and then by \(j\).
Let us see that it is a basis for \(T\).
Let \(t \in T\) be any and let \(U_t \subseteq T\) be any open neighborhood of \(t\).
There are an open ball around \(t\), \(B_{t, \epsilon} \subseteq T\), such that \(B_{t, \epsilon} \subseteq U_t\), a 'rational-number'-open ball around \(t\), \(B_{t, q} \subseteq T\), such that \(B_{t, q} \subseteq B_{t, \epsilon}\), and the 'rational-number'-open ball around \(t\), \(B_{t, q / 2} \subseteq T\).
There is an \(s_j \in B_{t, q / 2} \cap S\), because \(\overline{S} = T\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Let us take \(B_{s_j, q / 2}\). \(t \in B_{s_j, q / 2}\), because \(dist (t, s_j) \lt q / 2\). \(B_{s_j, q / 2} \subseteq B_{t, q}\), because for each \(p \in B_{s_j, q / 2}\), \(dist (p, t) \le dist (p, s_j) + dist (s_j, t) \lt q / 2 + q / 2 = q\).
So, \(t \in B_{s_j, q / 2} \subseteq B_{t, q} \subseteq U_t\).
As each \(B_{s_j, q}\) is an open subset of \(T\), \(\{B_{s_j, q} \vert s_j \in S, q \in \mathbb{Q}\}\) is a basis for \(T\).
Step 3:
Let us suppose that \(T\) has a countable basis, \(\{B_j\}\).
Step 4:
For each \(B_j\), let us take any \(s_j \in B_j\) and take \(S := \{s_1, s_2, ...\}\).
Let us see that \(S\) is a countable dense subset of \(T\).
It is countable.
For each \(t \in S\), let \(U_t \subseteq T\) be any open neighborhood of \(t\).
As \(\{B_j\}\) is a basis, there is a \(B_j\) such that \(t \in B_j \subseteq U_t\).
As \(s_j \in B_j\), \(s_j \in B_j \subseteq U_t\), which means that \(U_t \cap S \neq \emptyset\).
So, \(t \in \overline{S}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, \(T \subseteq \overline{S}\), but as \(\overline{S} \subseteq T\), \(\overline{S} = T\).
