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The proof is about the statement: if $f$ is an uniformly continuous function and $(x_n)$ is a Cauchy sequence then $f(x_n)$ is a Cauchy sequence.

A sequence $(x_n)$ can be defined as Cauchy as

$$\forall\delta>0,\exists N\in\Bbb N:|x_i-x_j|<\delta\quad\quad\forall i,j>N\tag{1}$$

And a definition of uniformly convergent function on some set $A$ can be

$$\forall\varepsilon>0,\exists\delta>0,\forall x,y\in A: |x-y|<\delta\implies|f(x)-f(y)|<\varepsilon\tag{2}$$

Then if we combine both statement we can have a "proof" that uniformly continuous functions preserves Cauchy sequences. For some $\varepsilon>0$ and some Cauchy sequence $(x_n)$ we have that

$$\exists\delta>0:|x_i-x_j|<\delta\implies|f(x_i)-f(x_j)|<\varepsilon\tag{3}$$

I dont see why this last statement can proof the preservation of Cauchy sequences for some uniformly continuous function. My problem is that I cant deduce from $(3)$ that this implies the $\forall\delta>0$ required in the definition of a Cauchy sequence.

The statement $(3)$ just is saying that exist some $\delta>0$ on the domain for a Cauchy sequence $f(x_n)$ in the range (because I can choose statements for all $\varepsilon>0$).

So I get lost somewhere, there is something that Im not seeing, can you help me please? Thank you in advance.

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  • $\begingroup$ The confusion comes from deltas being used in two different definitions being used slightly differently. $\endgroup$ – Doug M Jun 2 '16 at 1:36
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You want to show that the sequence $(f(x_n))$ is a Cauchy sequence. What is the criteria for that? For every $\epsilon>0$, there exists $N>0$ such that $|f(x_i)-f(x_j)|<\epsilon.$

Now, combine the given definitions, for every $\epsilon>0$, there exists $\delta$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. So, choose $N$, this can be done as $x_n$ is a Cauchy sqeuence, such that $|x_i-x_j|<\delta$ for all $i,j>N$. Then, this same $N$ satisfies what you're looking for.

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  • $\begingroup$ Ah! I finally understand. Now my next step is to see why I dont wasnt seeing it before. I was focused on epsilon-delta but the important thing is the $N$. The $|x_i-x_j|$ is a Cauchy tail where you can choose eventually any $N$. Thank you very much. $\endgroup$ – Masacroso Jun 2 '16 at 2:01
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I don't know if you had follow, but you prove that (3) holds for any $\varepsilon >0$ That any $\varepsilon >0$ is the same that your delta in the definition of a Cauchy sequence. But... I would recommend you to see this prove in a more visual way. Do a drawing.

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  • $\begingroup$ I think I understand. I can see it visually but I cant understand the proof. $\endgroup$ – Masacroso Jun 2 '16 at 1:39
  • $\begingroup$ The problem here is that you can't use delta for 2 different things. So you need a new letter. $\endgroup$ – PrinceAndrew Jun 2 '16 at 1:42

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