1081: For Metric Space, Convergent Sequence Is Cauchy Sequence, and N for Cauchy ϵ Condition Can Be Chosen to Be N for Convergence ϵ/2 Condition

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description/proof of that for metric space, convergent sequence is Cauchy sequence, and $N$ for Cauchy $ϵ$ condition can be chosen to be $N$ for convergence $ϵ/2$ condition

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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 metric space.
• 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 metric space, any convergent sequence is a Cauchy sequence, and $N$ for the Cauchy $ϵ$ condition can be chosen to be $N$ for the convergence $ϵ/2$ condition.

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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:
$M$: $\in \{\text{ the metric spaces }\}$
$f$: $:\mathbb{N}\to M$
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Statements:
$\mathrm{\exists }m\in M\text{ such that }\mathrm{\forall }ϵ\in \mathbb{R}\text{ such that }0<ϵ(\mathrm{\exists }{N}_{ϵ}\in \mathbb{N}(\mathrm{\forall }n\in \mathbb{N}\text{ such that }{N}_{ϵ}<n(dist(m,f(n))<ϵ/2)))$, which equals that $f$ converges to $m$
$⟹$
$\mathrm{\forall }ϵ\in \mathbb{R}\text{ such that }0<ϵ(\mathrm{\forall }n,o\in \mathbb{N}\text{ such that }{N}_{ϵ}<n,o(dist(f(n),f(o))<ϵ))$, which implies that $f$ is Cauchy
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The point of this proposition is that $N_{ϵ}$, which is a map from the positive real numbers set into the natural numbers set that (the map) is determined by the convergent condition, can be used for the Cauchy condition.

The reason why that is relevant is that if some sequences are uniformly convergent, we can conclude that also $N$ for the Cauchy $ϵ$ condition can be chosen uniformly: we can take $N$ for the convergent $ϵ/2$ condition, which ($N$) can be chosen uniformly.

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2: Proof

Whole Strategy: Step 1: take any $ϵ$ and find an $N_{ϵ}$ such that for each $n$ such that $N_{ϵ}<n$, $dist(m,f(n))<ϵ/2$; Step 2: see that for each $n,o$ such that $N_{ϵ}<n,o$, $dist(f(o),f(n))<ϵ$.

Step 1:

Let $ϵ\in \mathbb{R}$ be any such that $0<ϵ$.

By the supposition of this proposition, there is an $N_{ϵ}\in \mathbb{N}$ such that for each $n\in \mathbb{N}$ such that $N_{ϵ}<n$, $dist(m,f(n))<ϵ/2$.

That supposition equals $f$'s being convergent to $m$: if $f$ is convergent to $m$, for each $ϵ$, we can take $ϵ/2$ for the convergent condition, which means that there is an $N$ such that for each $n$ such that $N<n$, $dist(m,f(n))<ϵ/2$, and we can take $N_{ϵ}:=N$ for the supposition; if the supposition holds, for each $ϵ$, we can take $2ϵ$ for the supposition, which means that there is an $N_{2ϵ}$ such that for each $n$ such that $N_{2ϵ}<n$, $dist(m,f(n))<2ϵ/2=ϵ$, and we can take $N:=N_{2ϵ}$ for the convergent condition.

Step 2:

$N_{ϵ}$ is a map from the positive real numbers set into the natural numbers set, and let us see that $N_{ϵ}$ can be used for the Cauchy $ϵ$ condition.

Let $n,o\in \mathbb{N}$ be any such that $N_{ϵ}<n,o$.

$dist(f(o),f(n))\le dist(f(o),m)+dist(m,f(n))<ϵ/2+ϵ/2=ϵ$.

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References

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