First the textbook proves that all Cauchy sequences are bounded, and so have a convergent subsequence, $\{a_{n_{k}}\}$ that converges to a limit, say $L$. Now we use this to prove that all Cauchy sequences are convergent.
So an $N_1$ exists such that $$\left|a_{n_{k}}-L\right|<\frac{\epsilon}{2}$$ for all $ k > N_1$,
and an $N_2$ exists such that $$\left|a_m-a_n\right|<\frac{\epsilon}{2}$$ for all $n,m > N_2 $.
Pick any $k > N_1$ such that $n_k > N_2$. Then for every $n > N_2$, $$\left| a_n – L \right | \leq \left| a_n – a_{n_{k}} \right| + \left| a_{n_{k}} – L \right| < \epsilon/2 + \epsilon/2 = \epsilon$$.
So $$\lim_{n \rightarrow \infty}a_n = L$$
I'm fine with this proof until the last part – I'm confused as to why we can pick an abitrary $k$ like we do? Should the limit not depend only on $n$? Now it appears like it also depends on $k$ and if we pick a $k < N_1$, the inequality isn't true. Can someone clarify this for me please?
Best Answer
At that point in the proof you’re trying to show that if $n>N_2$, then $|a_n-L|<\epsilon$. If you can find a real number $x$ such that $|a_n-x|<\frac{\epsilon}2$ and $|x-L|<\frac{\epsilon}2$, the triangle inequality will give you the desired result, so the proof boils down to finding such an $x$.
We take care of (1) first: there is an $N_1$ such that $|a_{n_k}-L|<\frac{\epsilon}2$ whenever $k>N_1$. This means that we can take our $x$ to be any $a_{n_k}$ with $k>N_1$, and we’ll have $|x-L|<\frac{\epsilon}2$.
Then we take care of (2): there is an $N_2$ such that $|a_n-a_m|<\frac{\epsilon}2$ whenever $m,n>N_2$. This means that since we’ve already specified that $n>N_2$, we can take our $x$ to be any $a_m$ with $m>N_2$, and we’ll have $|a_n-x|<\frac{\epsilon}2$.
Can we combine the two requirements? Is there an $a_{n_k}$ with $k>N_1$ that is also an $a_m$ with $m>N_2$? Equivalently, is there an $a_{n_k}$ with $k>N_1$ such that $n_k>N_2$? Sure: the sequence $\langle n_k:k\in\Bbb N\rangle$ is unbounded, so its tail $\langle n_k:k>N_1\rangle$ is also unbounded and contains a term $n_k>N_2$. Thus, we can set $x=a_{n_k}$ and satisfy both requirements, so that we have
$$|a_n-L|\le|a_n-x|+|x-L|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$
as desired. Note that this calculation doesn’t actually depend on a specific value of $k$: we could set $x=a_{n_\ell}$ for any $\ell\ge k$, the the calculation would be the same. As I said at the beginning, we’re really just trying to find one number $x$ that we can use to ‘tie’ $a_n$ to $L$ to within $\epsilon$; it turns out that there are lots of them, and it doesn’t matter which one we use.