1070: Metric Subspace

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definition of metric subspace

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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

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Starting Context

• The reader knows a definition of metric space.

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Target Context

• The reader will have a definition of metric subspace.

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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:
$T^{\prime }$: $\in \{\text{ the metric spaces }\}$, with the metric, $dis{t}^{\prime }:T^{\prime }\times T^{\prime }\to \mathbb{R}$
$\ast T$: $\subseteq T^{\prime }$, $\in \{\text{ the metric spaces }\}$, with the metric, $dist:T\times T\to \mathbb{R}$, specified below
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Conditions:
$\mathrm{\forall }{t}_1,t_2\in T(dist(t_1,t_2)=dis{t}^{\prime }(t_1,t_2))$
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There is no restriction for $T$ as a subset.

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2: Note

Let us see that $dist$ is indeed a metric for $T$.

Let $t_1,t_2,t_3\in T$ be any.

$0\le dist(t_1,t_2)=dis{t}^{\prime }(t_1,t_2)$.

When $t_1=t_2$, $dist(t_1,t_2)=dis{t}^{\prime }(t_1,t_2)=0$.

When $dist(t_1,t_2)=0$, $dis{t}^{\prime }(t_1,t_2)=dist(t_1,t_2)=0$, so, $t_1=t_2$.

$dist(t_1,t_2)=dis{t}^{\prime }(t_1,t_2)=dis{t}^{\prime }(t_2,t_1)=dist(t_2,t_1)$.

$dist(t_1,t_3)=dis{t}^{\prime }(t_1,t_3)\le dis{t}^{\prime }(t_1,t_2)+dis{t}^{\prime }(t_2,t_3)=dist(t_1,t_2)+dist(t_2,t_3)$.

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References

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