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| 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.
Waiting for reply.