# Back to cauchy sequences.

In this question I will write about a problem which seemingly crops up while proving that all Cauchy sequences are convergent.

Consider a Cauchy sequence. As it is a Cauchy sequence it must be a bounded one. As such by Bolzano Weierstrass theorem we can assert the existence of a sub sequence which is convergent. Let the sub sequence be converging to limit $l$. Now if the whole sequence is convergent it must converge to $l$.

For this to happen we must have a value of $n'$, corresponding to every positive value of $\varepsilon$, such that for all $n>n'$, $|x_n-l|<\varepsilon$.

Let $x_{n_k}$ denote the terms of the convergent sub sequence.

And as the sequence is Cauchy, for any given $\varepsilon$, we must have a value of $n_1$ such that for $m,n >n_1$

$$|x_m -x_n|<\varepsilon\tag1$$

$$\text{In particular we can replace x_m by x_{n_k} in (1)}\tag2$$

And hence we have

$$|x_n-l|=|x_n-x_{n_k}+x_{n_k}-l|<|x_n-x_{n_k}|+ |x_ {n_k}-l| <2\varepsilon.$$

Now put $\varepsilon= a/2$.

so we can guarantee the existence of a value of $n'$ such that for any given $a$ (changing the variable does not affect anything) $|x_n-l|<a$ for all $n>n'$.

So the sequence is convergent.

Now as we can see the main point of the proof is $(2)$. However, I feel a bit awkward to proceed through $(2)$ without having a proof that we can find $x_{n_k}$ beyond any given value of $n$, i.e. given any $n$ the set $P_{n_k} =\{ x_{n_k}\;\text{occurring after}\;x_n\}$ is non empty. I hope the members of this community would like to elaborate on this topic, throw some light on it and help me to resolve this problem. I may receive replies which may emphasize its obviousness but still then I don't feel it to be obvious and hence as per my views without the answer to this problem the proof lacks rigor.

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I edited the answer, but cannot understand where the $(2)$ goes, so please complete the editing yourself. –  enzotib Oct 5 '12 at 17:19
In particular we can replace x(m) by x(nk) in (1). This what I labelled as (2). –  Primeczar Oct 5 '12 at 17:37
do you mean all caucy sequences of reals are convergent? –  ryu jin Jan 10 '13 at 11:37

I'm not too sure exactly what you're asking here. Suppose you have a sequence $(a_n)_{n=1}^\infty$ with a subsequence $(a_{n_k})_{k=1}^\infty$. Then we must have for the subscripts $n_k \ge k$ (this shouldn't be too hard to see, if you want a fully rigorous explanation, consider an inductive proof). Because these are infinite sequences, both $n$ and $k$ are unbounded as indices. So given any $m$, there exists some $k > m$ which implies $n_k > m$.

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I hope you can prove that the terms of the sub sequence are well distributed throughout the sequence. –  Primeczar Oct 5 '12 at 17:12
What do you mean by "well distributed"? –  EuYu Oct 5 '12 at 17:13
that we can find X(nk) beyond any given value of n,i.e. given any n the set P(nk) ={ X(nk) occurring after X(n)} is non empty –  Primeczar Oct 5 '12 at 17:14
I have already shown that. There exists some $k$ larger than your $n$ by the Archimedean property of the naturals. And $n_k$ is larger than $k$. Therefore $x_{n_k}$ occurs after $x_n$. –  EuYu Oct 5 '12 at 17:16
If they were not "well distributed" they would all be below some $n$, i.e. it would only be a finite subsequence. –  Hagen von Eitzen Oct 5 '12 at 17:21

What you want is that $\{k_n:k_n \text{ occurring after } n\}$ is all of $\mathbb{N}$ and not $\{x_{k_n} \text{ occurring after } x_n\}$ is non empty. Also be careful that the subscript in $x_{k_n}$ is $n$ i.e. for each $n$ we get a natural number $k_n$ and an element of the sequence, namely $x_{k_n}$. Anyway, here is a proof.

First lets prove that if $k_n$ is a strictly increasing sequence of natural numbers then $k_n\geq n , \ \forall n \in \mathbb{N}.$
Indeed $k_1\geq 1$ and if $k_n\geq n$ then $k_{n+1}>k_n\geq n \Rightarrow k_{n+1}\geq n+1$.

Now let $\epsilon>0$.
Since $(x_n)_{n \in \mathbb{N}}$ is Cauchy exist $n_1$ such that $\forall m,n \geq n_1, \ |x_m-x_n|<\frac{\epsilon}{2}$.
Since $(x_{k_n})_{n \in \mathbb{N}}$ converges to $\ell$ exist $n_2$ such that $\forall n \geq n_2, \ |x_{k_n}-\ell|<\frac{\epsilon}{2}$.
If $n_0=\max\{n_1,n_2\}$ then $\forall n \geq n_0, \ |x_{k_n}-\ell|<\frac{\epsilon}{2} \text{ and } |x_{k_n}-x_n|<\frac{\epsilon}{2}$ (since $k_n\geq n \Rightarrow k_n,n \geq n_0\geq n_1$).

Therefore $\forall n \geq n_0, \ |x_{n}-\ell|=|x_{n}-x_{k_n}+x_{k_n}-\ell|\leq |x_{n}-x_{k_n}|+|x_{k_n}-\ell| < \epsilon$.
Hence $x_n \to \ell$.

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