Why Is the Existence of m and n Guaranteed in the Cauchy Sequence Proof?

[ozkan12]

to prove that $\left\{{x}_{n}\right\}$ is Cauchy seqeunce we use a method. I have some troubles related to this method. Please help me...
$\left\{{c}_{n}\right\}$=sup$\left\{d\left({x}_{j},{x}_{k}\right):j,k>n\right\}$.Then $\left\{{c}_{n}\right\}$ is decreasing. If ${c}_{n}$ goes to 0 as n goes to $\infty$, then we are done. Assume that ${c}_{n}$ goes to c>0. Choose $\varepsilon$ <$\frac{c}{8}$ small enough and select N such that for all n$\ge$N.$d\left({x}_{n},{x}_{n+1}\right)$<$\varepsilon$ and ${c}_{n}$<c+$\varepsilon$.

By definition of ${c}_{N+1}$ there exists m,n $\ge$ N+1 such that d(${x}_{m}$,${x}_{n}$)>${c}_{n}$-$\varepsilon$>c-$\varepsilon$.

why By definition of ${c}_{N+1}$ there exists m,n $\ge$ N+1 such that d(${x}_{m}$,${x}_{n}$)>${c}_{n}$-$\varepsilon$>c-$\varepsilon$. I didn't understand. Please explain to me...Thank you :)

 

[Euge]

Hi ozkan12,

Since $c_{N+1}$ is the least upper bound of the set $A_{N+1} := \{d(x_j,x_k): j, k > N+1\}$, $c_{N+1} - \varepsilon$ is not an upper bound of $A_{N+1}$. So there exists $m, n > N+1$ such that $c_{N+1} - \epsilon < d(x_m,x_n)$. This is what follows from definition. The inequalities $d(x_m,x_n) > c_n - \varepsilon > c - \epsilon$ are incorrect. Note that since $c - \varepsilon < c_n$ for all $n \ge N$, $c_N - \varepsilon > c - 2\varepsilon$, not $c_N - \varepsilon > c - \varepsilon$.

 

[ozkan12]

But this is written in article ? If you want, you send your email adress...I can send article to you

 

[Euge]

Let me get more to the point. The inequality $d(x_m,x_n) > c_n - \varepsilon > c - \varepsilon$ should be $d(x_m,x_n) > c_n - \varepsilon \ge c - \varepsilon$, so not all inequalities are strict. What we know is that $c_n$ decreases to $c$, so $c_n \ge c$ for all $n \in \Bbb N$. However, we do not know that $c_n > c$ for all $n \ge N$, which is what you'll need to establish the strict inequality $c_n - \varepsilon > c - \varepsilon$.

We can prove $d(x_m, x_n) > c_n - \varepsilon$ from the fact $d(x_m,x_n) > c_{N+1} - \varepsilon$. Indeed, since $n > N+1$ then $c_{N+1} \ge c_n$. Thus $d(x_m,x_n) > c_{N+1} - \varepsilon \ge c_n - \varepsilon$.

 

[ozkan12]

Yes...But I don't understand How we can prove d(xm,xn)>cn−ε from the fact d(xm,xn)>cN+1−ε and
why By definition of cN+1 there exists m,n ≥ N+1 such that d(xm,xn)>cn-ε>=c-ε

 

[Euge]

I did explain it in my last post, but let me try again. Since $\{c_k\}$ is decreasing and $n > N+1$, then $c_{N + 1} \ge c_n$. Hence $c_{N+1} - \varepsilon \ge c_n - \varepsilon$. Since $d(x_m, x_n) > c_{N+1} - \varepsilon$ and $c_{N+1}-\varepsilon \ge c_n - \varepsilon$, then $d(x_m, x_n) > c_n - \varepsilon$.

Now since $c_n \ge c$ (because the sequence $\{c_k\}$ decreases to $c$), $c_n - \varepsilon \ge c - \varepsilon$. Therefore $d(x_m, x_n) > c_n - \varepsilon \ge c - \varepsilon$.

 

[ozkan12]

I understood thank you so much :) But I have a little problem there exists m,n>N+1 such that cN+1−ϵ<d(xm,xn).

Why Hence cN+1>d(xm,xn)−ε. I didnt understand

 

[Euge]

I understood thank you so much :) But I have a little problem there exists m,n>N+1 such that cN+1−ϵ<d(xm,xn).

Why Hence cN+1>d(xm,xn)−ε. I didnt understand

The part "Hence $c_{N+1} > d(x_m,x_n) -\varepsilon" was accidental; it wasn't supposed to be there. (Tongueout) Sorry about that! I have removed that statement. Hopefully everything is clear now.

 

[ozkan12]

Thank you so much, it is very helpful for me :)