How can I prove if $(x_n)_{n\in\mathbb N}$ is a cauchy sequence on $\mathbb{R}$, then it is convergent?

Question

Some blurry ideas in my mind push me to follow this: prove that if $(x_n)_{n\in\mathbb N}$ Cauchy, then bounded. If $(x_n)_{n\in\mathbb N}$ is bounded, then it has a convergent subsequence since there must be a monotone subsequence.

However, I don't know how to come up with the result suggesting that the mother sequence is also convergent.

Answer 1

Your ideas are good. So, you have this Cauchy sequence $(x_n)_{n\in\mathbb N}$ and you have already proved that it has a subsequnce $(x_{n_k})_{k\in\mathbb N}$ which converges to some limit $l$. And you want to prove that $\lim_{n\to\infty}x_n=l$. In order to do that, take $\varepsilon>0$. Now, take $p,q\in\mathbb N$ such that

1. $m,n\geqslant p\implies|x_m-x_n|<\frac\varepsilon2$;
2. $k\geqslant q\implies|x_{n_k}-l|<\frac\varepsilon2$.

Now, take $P\in\mathbb N$ such that $P\geqslant\max\{p, q\}$. Then $n_P\geqslant P\geqslant p$ and, if $n\geqslant P$,\begin{align}|x_n-l|&\leqslant|x_n-x_{n_P}|+|x_{n_P}-l|\\&<\frac\varepsilon2+\frac\varepsilon2\text{ (because $n,n_P\geqslant p$ and $P\geqslant q$)}\\&=\varepsilon.\end{align}