1079: For Cauchy Sequence on Metric Space, if Subsequence Converges to Point, Sequence Converges to the Point

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description/proof of that for Cauchy sequence on metric space, if subsequence converges to point, sequence converges to the point

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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: Proof

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Starting Context

• The reader knows a definition of Cauchy sequence on metric space.

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Target Context

• The reader will have a description and a proof of the proposition that for any Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point.

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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:
$T$: $\in \{\text{ the metric spaces }\}$
$f$: $:\mathbb{N}\to T$, $\in \{\text{ the Cauchy sequences }\}$
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Statements:
$\mathrm{\exists }{f}^{\prime }:\mathbb{N}\to \mathbb{N}\text{ such that }\mathrm{\forall }j,k\in \mathbb{N}\text{ such that }j<k(f^{\prime }(j)<f^{\prime }(k))(f\circ f^{\prime }\to t\in T,\text{ which means that }f\circ f^{\prime }\text{ converges to }t)$
$⟹$
$f\to t\in T$, which means that $f$ converges to $t$
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2: Proof

Whole Strategy: Step 1: take the open ball around $t$, $B_{t,ϵ/2}$, and take an $N\in \mathbb{N}$ such that for each $j\in \mathbb{N}$ such that $N<j$, $f\circ f^{\prime }(j)\in B_{t,ϵ/2}$; Step 2: take an $N^{\prime }\in \mathbb{N}$ such that for each $j,k\in \mathbb{N}$ such that $N^{\prime }<j,k$, $dist(f(j),f(k))<ϵ/2$; Step 3: take $N^{″}:=max(N,N^{\prime })$, and see that for each $j\in \mathbb{N}$ such that $N^{″}<j$, $dist(t,f(j))<ϵ$.

Step 1:

Let us take the open ball around $t$, $B_{t,ϵ/2}$.

Let us take an $N\in \mathbb{N}$ such that for each $j\in \mathbb{N}$ such that $N<j$, $f\circ f^{\prime }(j)\in B_{t,ϵ/2}$, which is possible because $f\circ f^{\prime }$ converges to $t$.

Step 2:

Let us take an $N^{\prime }\in \mathbb{N}$ such that for each $j,k\in \mathbb{N}$ such that $N^{\prime }<j,k$, $dist(f(j),f(k))<ϵ/2$, which is possible because $f$ is a Cauchy sequence.

Step 3:

Let us take $N^{″}:=max(N,N^{\prime })$.

For each $j\in \mathbb{N}$ such that $N^{″}<j$, $dist(t,f(j))\le dist(t,f\circ f^{\prime }(j))+dist(f\circ f^{\prime }(j),f(j))<ϵ/2+ϵ/2=ϵ$, because while $f\circ f^{\prime }(j)=f(f^{\prime }(j))$, $j\le f^{\prime }(j)$ and $N^{\prime }\le N^{″}<f^{\prime }(j),j$.

So, $f$ converges to $t$.

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References

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