MHB Cauchy sequence in fixed point theory

[ozkan12]

İn some articles, I see something...

For example,

Let we define a sequence by ${x}_{n}=f{x}_{n}={f}^{n}{x}_{0}$$\left\{{x}_{n}\right\}$. To show that $\left\{{x}_{n}\right\}$ is Cauchy sequence, we suppose that $\left\{{x}_{n}\right\}$ is not a Cauchy sequence...For this reason, there exists a subsequence $\left\{{x}_{n\left(i\right)}\right\}$ of $\left\{{x}_{n}\right\}$ such that $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)}\right)\ge\varepsilon$...İn there, I didnt understant what is the ${x}_{n(i)}$ and ${x}_{n(i+1)}$...These terms are consecutive terms, that is distinction between these terms equal to 1...I didnt understand...Can you help me ? Thank you so much, best wishes

 

[Ackbach]

That's a "subsequence" notation. Imagine a sequence $\{1,2,3,4,5,\dots\}$. In this case, $x_1=1, x_2=2, x_3=3,\dots$. You can pick a subsequence out of this sequence, which preserves the order, but is a subset: $\{2,4,6,8,\dots\}$. Here, you could notate it as $x_{n(0)}=2, x_{n(1)}=4, x_{n(2)}=6,\dots$. This subsequence, of course, is "every other", but you could pick your subsequence however you like. I should point out that another very common notation for subsequence is $x_{n_i}$. So you'd have $x_{n_1}, x_{n_2}, x_{n_3},\dots$.

 

[ozkan12]

Dear Ackbach,

From this, can we say that the distance between two arbitrary points is bigger than epsilon?

For example, $d\left({x}_{5},{x}_{14}\right) > \varepsilon$

And why do we use $d\left({x}_{n(i)},{x}_{n(i+1)}\right)\ge \varepsilon$ ? What is mean of this ? I ask this...

 

[Ackbach]

Dear Ackbach,

From this, can we say that the distance between two arbitrary points is bigger than epsilon?

For example, $d\left({x}_{5},{x}_{14}\right) > \varepsilon$

In the context of the proof so far, you're assuming the sequence is NOT Cauchy. Thus, the statement above is an assumption, presumably leading to a contradiction as per the usual method of proof by contradiction.

And why do we use $d\left({x}_{n(i)},{x}_{n(i+1)}\right)\ge \varepsilon$ ? What is mean of this ? I ask this...

You're in the subsequence, and you're assuming a statement that is the direct negation of the sequence being Cauchy. In a Cauchy sequence, once you're beyond a certain term depending on a given epsilon, the distance between two elements of the sequence is smaller than epsilon. What would the negation of that be? That there is a subsequence where the distance between two elements is greater than epsilon.

 

[ozkan12]

n(i) and n(i+1) ...Why we examine distance between these terms...Very strange...I didnt understand...

 

[ozkan12]

Dear,

My question is not this...My question is "what is the n(i) and n(i+1)" can we say that $n(i) \le n(i+1)-1$

 

[Ackbach]

Dear,

My question is not this...My question is "what is the n(i) and n(i+1)" can we say that $n(i) \le n(i+1)-1$

Yes, I think you can say the last inequality. Basically, your last inequality says that the "index distance" between two elements in the subsequence is at least one, and that the sequence $n(i)$ is monotone increasing. That's certainly what you'd expect for indices.

 

[ozkan12]

Dear,

That is, we can say

for example n(i)=5 i n(i+1)=12 etc...that is, n(i) and n(i+1) are arbitrary points...We can't say that always these are consecutive terms, isn't it ?

 

[Evgeny.Makarov]

We can't say that always these are consecutive terms, isn't it ?

No, they don't have to be consecutive.

One of the definitions of Cauchy sequence is the following.
\[
\forall\varepsilon\,\exists m\,\forall n>m\;|x_m-x_n|<\varepsilon
\]
Its negation is
\[
\exists\varepsilon\,\forall m\,\exists n>m\;|x_m-x_n|\ge\varepsilon.\qquad(*)
\]
Take an $\varepsilon$ whose existence is guaranteed by (*). Take $m=1$ and define $n(1)=1$. Then there exists an $n>1$ such that $|x_m-x_n|\ge\varepsilon$. Take any such $n$ and call it $n(2)$. Then we have $|x_{n(1)}-x_{n(2)}|\ge\varepsilon$.

As a next step, take $m=n(2)$; then there exists some $n>m$ such that $|x_m-x_n|\ge\varepsilon$. Take one such $n$ and call it $n(3)$. Then we have $|x_{n(2)}-x_{n(3)}|\ge\varepsilon$, and so on.

 

[ozkan12]

Dear Evgeny Makarov,

What is the presentation of $d\left({x}_{n(i)},{x}_{n(İ+1)}\right)\ge \varepsilon$ ? Can you explain ? Why distance between them don't have to consecutive ? I didnt understand this ?

 

[Evgeny.Makarov]

What is the presentation of $d\left({x}_{n(i)},{x}_{n(İ+1)}\right)\ge \varepsilon$ ?

I am not sure what you mean. I wrote the definition of a Cauchy sequence of real numbers, where the distance between $x$ and $y$ is $|x-y|$. In a metric space the distance is $d(x,y)$.

Why distance between them don't have to consecutive ?

The negation of the definition says $\forall m\,\exists n>m\;d(x_m,x_n)\ge\varepsilon$. The claim is that there exists an $n$ and that it is strictly greater than $m$. It does not follow that $n=m+1$ necessarily, though it can be.

I didnt understand this ?

Are you asking or stating the fact that you don't understand? (Smile)

 

[ozkan12]

What I didnt understand is why we use $d\left({x}_{{n(i)}},{x}_{n(i+1)}\right)\ge\varepsilon$ ? This recall consecutive terms for me...:)

 

[Evgeny.Makarov]

What I didnt understand is why we use $d\left({x}_{{n(i)}},{x}_{n(i+1)}\right)\ge\varepsilon$ ?

This holds by definition of $n(i+1)$.

This recall consecutive terms for me.

$x_{n(i)}$ and $x_{n(i+1)}$ are indeed consecutive terms of the subsequence $x_{n(k)}$, $k=1,2,\dots$.

If you have further questions, please confirm that you understand my explanation in post #9.

 

[ozkan12]

Dear,

I understand this,but why we use $d\left({x}_{n\left(i\right)},{x}_{n(i+1)}\right)\ge \varepsilon$...I didnt understand this

 

[Evgeny.Makarov]

I understand this,but why we use $d\left({x}_{n\left(i\right)},{x}_{n(i+1)}\right)\ge \varepsilon$...I didnt understand this

Because it's a true proven statement under current assumptions (that the sequence is not Cauchy). In mathematics, we are allowed to write and use any statement that has been properly proved. The question: "Why we use it?" is not mathematical, strictly speaking. The only legitimate question is, "Why is it true?" And you seem to agree with my explanation that is it true.

 

[Ackbach]

Dear,

I understand this,but why we use $d\left({x}_{n\left(i\right)},{x}_{n(i+1)}\right)\ge \varepsilon$...I didnt understand this

If you look carefully at Evgeny's Post #9, the expression $d\left({x}_{n\left(i\right)},{x}_{n(i+1)}\right)\ge \varepsilon$ is the last portion of the negation of the Cauchy criterion. So that's why it is what it is: you're negating the Cauchy criterion as an assumption, and then presumably you're going to derive a contradiction somewhere, which would show that the sequence is Cauchy.

 

[ozkan12]

Look Dear :),

I understand these definitions but how always we say that n(i) and n(i+1) is not consecutive, how I get this from definitions...I didnt understand this...I am sorry...

 

[Evgeny.Makarov]

I understand these definitions but how always we say that n(i) and n(i+1) is not consecutive, how I get this from definitions...

I assume there is some statement whose proof you don't understand. Please write this statement without words, using only formulas.

 

[ozkan12]

Dear

İf you accept it, I can send article your email adress ? Are you okay ? İn this area, we can't talk this topic :)

 

[Evgeny.Makarov]

I am only prepared to discuss your questions on the forum. Feel free to take a photo or scan the relevant portion of the article page and post it here as a picture, though writing questions as text with LaTeX formulas is preferred.

 

[ozkan12]

Dear,

İndeed, I didnt understand only why w9e write $d\left({x}_{n(i)},{x}_{n(i+1)}\right)\ge\varepsilon$ ? İn post 9 we write n(1)=1 but can we write n(2)=9 or n(2)=15 or etc...?

 

[Evgeny.Makarov]

İn post 9 we write n(1)=1 but can we write n(2)=9 or n(2)=15 or etc...?

Then there exists an $n>1$ such that $|x_m-x_n|\ge\varepsilon$. Take any such $n$ and call it $n(2)$.

Yes, $n(2)$ could be 9 or 15, but not always.

 

[ozkan12]

That is, we can choose arbitrary points such that $d\left({x}_{n(i)},{x}_{n(i+1)}\right)\ge\varepsilon$...this is the answer of my question