1100: For Topological Space Induced by Metric, Space Has Countable Basis iff Space Is Separable

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description/proof of that for topological space induced by metric, space has countable basis iff space is separable

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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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.

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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.

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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.

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Main Body

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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:
$\mathrm{\exists }B\in \{\text{ the countable bases for }T\}$
$⟺$
$T\in \{\text{ the separable topological spaces }\}$
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2: Note

It is crucial for $T$ to be induced by a metric, as is seen in Proof.

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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, $\{B_{s_j,q}|s_j\in S,q\in \mathbb{Q}\}$, 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, $\{B_{s_j,q}|s_j\in S,q\in \mathbb{Q}\}$.

It is countable because $S$ and $\mathbb{Q}$ are countable: let $\mathbb{Q}=\{q_1,q_2,...\}$ and count $\{B_{s_j,q}|s_j\in S,q\in \mathbb{Q}\}$ 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 $t$, $B_{t,ϵ}\subseteq T$, such that $B_{t,ϵ}\subseteq U_t$, a 'rational-number'-open ball around $t$, $B_{t,q}\subseteq T$, such that $B_{t,q}\subseteq B_{t,ϵ}$, 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 $\bar{S}=T$, 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 $dist(t,s_j)<q/2$. $B_{s_j,q/2}\subseteq B_{t,q}$, because for each $p\in B_{s_j,q/2}$, $dist(p,t)\le dist(p,s_j)+dist(s_j,t)<q/2+q/2=q$.

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 $T$, $\{B_{s_j,q}|s_j\in S,q\in \mathbb{Q}\}$ 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 $U_t\cap S\ne \mathrm{\varnothing }$.

So, $t\in \bar{S}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, $T\subseteq \bar{S}$, but as $\bar{S}\subseteq T$, $\bar{S}=T$.

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References

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