1083: For Convergent Sequence on R, if Each Term Is Smaller or Larger Than Number, Convergence Is Equal to or Smaller or Larger Than Number, Respectively

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description/proof of that for convergent sequence on $\mathbb{R}$, if each term is smaller or larger than number, convergence is equal to or smaller or larger than number, respectively

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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
• 3: Proof

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Starting Context

• The reader knows a definition of Euclidean metric.

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Target Context

• The reader will have a description and a proof of the proposition that for any convergent sequence on $\mathbb{R}$, if each term is smaller or larger than a number, the convergence is equal to or smaller or larger than the number, respectively.

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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:
$\mathbb{R}$: with the Euclidean metric
$f$: $:\mathbb{N}\to \mathbb{R}$, $\in \{\text{ the convergent sequences }\}$
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Statements:
(
$\mathrm{\exists }M\in \mathbb{R}(\mathrm{\forall }j\in \mathbb{N}(f(j)<M))$
$⟹$
$limf\le M$
)
$\wedge$
(
$\mathrm{\exists }m\in \mathbb{R}(\mathrm{\forall }j\in \mathbb{N}(m<f(j)))$
$⟹$
$m\le limf$
)
//

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2: Note

As is easily guessed, $limf<M$ or $m<limf$ is not guaranteed.

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3: Proof

Whole Strategy: Step 1: suppose that $f(j)<M$; Step 2: see that $limf\le M$; Step 3: suppose that $m<f(j)$; Step 4: see that $m\le limf$.

Step 1:

Let us suppose that $\mathrm{\exists }M\in \mathbb{R}(\mathrm{\forall }j\in \mathbb{N}(f(j)<M))$.

Step 2:

Let us suppose that $M<limf$.

For any $j\in \mathbb{N}$, $limf=limf-f(j)+f(j)\le |limf-f(j)|+f(j)$, so, $limf-|limf-f(j)|\le f(j)$.

But $j$ can be chosen such that $|limf-f(j)|<limf-M$, and then, $limf-(limf-M)<f(j)$, which implies that $M<f(j)$, a contradiction.

So, $M<limf$ was wrong, which means that $limf\le M$.

Step 3:

Let us suppose that $\mathrm{\exists }m\in \mathbb{R}(\mathrm{\forall }j\in \mathbb{N}(m<f(j)))$.

Step 4:

Let us suppose that $limf<m$.

For any $j\in \mathbb{N}$, $f(j)=f(j)-limf+limf\le |f(j)-limf|+limf$.

But $j$ can be chosen such that $|f(j)-limf|<m-limf$, and then, $f(j)<m-limf+limf=m$, a contradiction.

So, $limf<m$ was wrong, which means that $m\le limf$.

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References

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