437: Convergence of Sequence on Metric Space

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definition of convergence of sequence on metric space

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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 convergence of sequence on metric space.

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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:
$M$: $\in \{\text{ the metric spaces }\}$
$s$: $:\mathbb{N}\to M$, $\in \{\text{ the sequences of points on }M\}$
$\ast m$: $\in M$
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Statements:
$\mathrm{\forall }ϵ\in \mathbb{R}\text{ such that }0<ϵ(\mathrm{\exists }N\in \mathbb{N}(\mathrm{\forall }n\in \mathbb{N}\text{ such that }N<n(dist(m,s(n))<ϵ)))$
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2: Note

Sometimes, we may use $\mathbb{N}\setminus \{0\}$ instead of $\mathbb{N}$.

When $M$ is made the topological space with the canonical induced topology, $\mathbb{N}$ is a directed index set, $s$ is a net with the directed index set, and any convergence of $s$ as the sequence is a convergence of $s$ as the net with the directed index set, because for any open neighborhood, $U_m\subseteq M$, of $m$, there is an $ϵ$ open ball, $B_{m,ϵ}\subseteq U_m$, around $m$, and as $s(n)\in B_{m,ϵ}$, $s(n)\in U_m$.

The convergence, $m$, is inevitably unique: let us suppose that there was another convergence, $m^{\prime }\in M$, which implies that $0<dist(m,m^{\prime })$; $dist(m,m^{\prime })\le dist(m,s(j))+dist(s(j),m^{\prime })$, so, $dist(m,m^{\prime })-dist(m,s(j))\le dist(s(j),m^{\prime })$; for each $N<n$, $dist(m,s(j))<dist(m,m^{\prime })/2$; so, $dist(m,m^{\prime })/2=dist(m,m^{\prime })-dist(m,m^{\prime })/2\le dist(s(j),m^{\prime })$, which means that $s$ would not converge to $m^{\prime }$.

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References

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