I am a little confused about the following concept from Real Analysis:
suppose $\{a_n\}$ is a sequence in $\mathbb{R}$ which is Cauchy. then it is convergent, and it is shown as follows:
first we show it is bounded (this I understand how to do). then, from the Bolzano-Weierstrass theorem(every bounded sequence of real numbers contains a convergent subsequence) we deduce that there exists a convergent subsequence $\{a_{n_k}\}$ with limit $a$ (say).
we now claim that $\lim_{n \to \infty} a_n = a$.
what we need to show then is that for any given $\varepsilon > 0$ we can find $N > 0$ such that $\tag{1} n \geq N \Rightarrow |a_n - a| < \varepsilon\,. $ by the triangle inequality we know that the following inequality is true, for all $n, n_k \in \mathbb{N}$: $\tag{2} |a_n - a| \leq |a_n - a_{n_k}| + |a_{n_k} - a| \,. $ i know that the usual argument is now to say we control the first term using the Cauchy property of $\{a_n\}$ and the second term using the convergence of $\{a_{n_k}\}$ to $a$.
and here is where I am struggeling: why are we allowed to constrain the $a_{n_k}$? this is not clear to me, given the statement ${(1)}$. it looks like we have to make a $second$ choice, namely for the right hand side of ${(2)}$ to be $< \varepsilon$, the $n_k$ must be large also.
in particular, this choice cannot (at least as far as i understand) be resolved in a straightforward way (such as simply taking the larger of two constants) because the subsequence is a sequecen indexed by the subscript $k$, not the subscript $n$ any more - so choosing $K$ so that $|a_{n_k} - a| < \varepsilon/2$ whenever $k \geq K$ cannot be combined with choosing $N$ so that $|a_n - a_{n_k}| < \varepsilon/2$ whenever $n \geq N$.
I added this last paragraph because I sense this is where my misunderstanding is coming from - what am I missing? Thanks a lot for your help!!