1082: For Complete Metric Space and Sequence of Uniformly Convergent Sequences, if Sequence of Each Corresponding Terms of Sequences Converges, Sequence of Convergences Converges to Convergence of Sequence of Convergences of Sequences

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description/proof of that for complete metric space and sequence of uniformly convergent sequences, if sequence of each corresponding terms of sequences converges, sequence of convergences converges to convergence of sequence of convergences of sequences

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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 complete metric space.
• The reader admits 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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Target Context

• The reader will have a description and a proof of the proposition that for any complete metric space and any sequence of uniformly convergent sequences, if the sequence of each corresponding terms of the sequences converges, the sequence of the convergences of the uniformly-convergent sequences converges to the convergence of the sequence of the convergences of the corresponding-terms sequences.

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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:
$M$: $\in \{\text{ the complete metric spaces }\}$
$S$: $=\{f:\mathbb{N}\to M|f\in \{\text{ the convergent sequences }\}\}$
$g$: $:\mathbb{N}\to S$
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Statements:
(
$\mathrm{\forall }ϵ\in \mathbb{R}\text{ such that }0<ϵ(\mathrm{\exists }N\in \mathbb{N}(\mathrm{\forall }j\in \mathbb{N},\mathrm{\forall }n\in \mathbb{N}\text{ such that }N<n(dist(limg(j),g(j)(n))<ϵ)))$
$\wedge$
$\mathrm{\forall }n\in \mathbb{N}(li{m}_{j\to \mathrm{\infty }}g(j)(n)=h(n))$
)
$⟹$
$li{m}_{j\to \mathrm{\infty }}limg(j)=limh$
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Whenever "$\text{ lim }$" appears, it implies that the limit (which is nothing but the convergence) exists.

Just "$\text{ lim }$" means the convergence of the sequence: for example, $limg(j)$ means the convergence of the sequence, $g(j)$, not taking $j\to \mathrm{\infty }$. So, $limg(j):=li{m}_{n\to \mathrm{\infty }}g(j)(n)$ with $j$ fixed.

The 1st condition means uniform convergence for the sequence of the sequences: $N$ is chosen independent of $j$.

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2: Proof

Whole Strategy: Step 1: see that $h$ is a Cauchy sequence; Step 2: see that $li{m}_{j\to \mathrm{\infty }}limg(j)=limh$.

Step 1:

By the supposition of this proposition, the sequence, $h:\mathbb{N}\to M$, exists.

Let us see that $h$ is a Cauchy sequence.

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

By 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, each $g(j)$ is a Cauchy sequence, and $N$ for the Cauchy $ϵ/2$ condition can be chosen uniformly as $N$ for the convergence $ϵ/4$ condition.

So, let $N\in \mathbb{N}$ be such that for each $n,o\in \mathbb{N}$ such that $N<n,o$, $dist(g(j)(o),g(j)(n))<ϵ/2$, where $N$ does not depend on $j$.

For any $j$, $dist(h(o),h(n))\le dist(h(o),g(j)(o))+dist(g(j)(o),g(j)(n))+dist(g(j)(n),h(n))<dist(h(o),g(j)(o))+ϵ/2+dist(g(j)(n),h(n))$.

There are an $N_o\in \mathbb{N}$ such that for each $N_o<j$, $dist(h(o),g(j)(o))<ϵ/4$ and an $N_n\in \mathbb{N}$ such that for each $N_n<j$, $dist(h(n),g(j)(n))<ϵ/4$.

So, let us choose $j$ such that $N_o,N_n<j$, and then, $dist(h(o),h(n))<ϵ/4+ϵ/2+ϵ/4=ϵ$.

Although the choice of $j$ depends on $n$ and $o$, such an act does not restrict the choices of $n$ and $o$, and so, for each $n,o\in \mathbb{N}$ such that $N<n,o$, $dist(h(o),h(n))<ϵ$, which means that $h$ is a Cauchy sequence.

Step 2:

So, $limh$ exists, because $M$ is complete.

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

For any $n\in \mathbb{N}$, $dist(limh,limg(j))\le dist(limh,h(n))+dist(h(n),g(j)(n))+dist(g(j)(n),limg(j))$.

There are an $N_h\in \mathbb{N}$ such that for each $n\in \mathbb{N}$ such that $N_h<n$, $dist(limh,h(n))<ϵ/2$ and an $N\in \mathbb{N}$ such that for each $n\in \mathbb{N}$ such that $N<n$, $dist(g(j)(n),limg(j))<ϵ/4$, where $N$ does not depend on $j$ by the supposition of this proposition.

So, let us choose $n$ as $N_h,N<n$, and then, $dist(limh,limg(j))<ϵ/2+dist(h(n),g(j)(n))+ϵ/4$.

There is an $N^{\prime }\in \mathbb{N}$ such that for each $j\in \mathbb{N}$ such that $N^{\prime }<j$, $dist(h(n),g(j)(n))<ϵ/4$.

Although $N^{\prime }$ depends on $n$, that does not matter: the point is that $N^{\prime }$ exists anyway.

So, for each $j\in \mathbb{N}$ such that $N^{\prime }<j$, $dist(limh,limg(j))<ϵ/2+ϵ/4+ϵ/4=ϵ$.

That means that $li{m}_{j\to \mathrm{\infty }}limg(j)=limh$.

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References

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