Cauchy sequences always have a largest or smallest element past an arbitrary index The problem is as follows:
Let $(x_n)_n$ be Cauchy. Show that either
$$
(1)\; \forall N\in\mathbb{N}, \exists\bar{n}\geq N \text{ s.t. } \forall n\geq N, x_n\leq x_{\bar{n}}
$$
or
$$
(2)\; \forall N\in\mathbb{N}, \exists\bar{n}\geq N \text{ s.t. } \forall n\geq N, x_n\geq x_{\bar{n}}
$$
Additionally, is it possible that $\forall N\in\mathbb{N}$, $\exists n_1,n_2\geq N$ such that $\forall n\geq N, x_{n_1}\leq x_n \leq x_{n_2}$?

I'm assuming this is a sequence of real numbers.
Now, here is the beginning of my attempt at a proof (I didn't get very far):
Let $(x_n)$ be Cauchy and suppose that (1) is false. The negation of (1) is
$\exists N\in\mathbb{N} \text{ s.t. } \forall\bar{n}\geq N, \exists n\geq N \text { s.t. } x_n > x_{\bar{n}}$.
Clearly we need to begin by letting $N\in\mathbb{N}$ be arbitrary since we are trying to show (2) is true. However, I'm not sure how to proceed from here. I know we need to find an $\bar{n}$ such that for $n\geq N$ we have $x_n\geq x_{\bar{n}}$. But I'm not sure how to pick the $\bar{n}$. I think the issue is that there are so many quantifiers running around I'm having trouble parsing the problem. Is the fact that Cauchy sequences are bounded necessary here? I even tested a couple of examples to get an intuition for why this is true, e.g.: take the sequence $x_n = \frac{(-1)^n}{n}$. Then when $N=1$, $\bar{n}=2$ works since for each index $n\geq 1$ we have $x_n < \frac{1}{2}$. When $N=2, \bar{n}=2$ works since for each index $n\geq 2$ we have $x_n \leq \frac{1}{2}$. However, I can't seem to be able to turn this process into a rigorous proof. Any help is greatly appreciated.
 A: To begin with, I will prove it for $N=1$. Suppose that what you wish to prove is false. Then, for each $n\in\mathbb N$, there will be $k,l\in\mathbb N$ such that $x_k<x_n<x_l$. Since your sequence is a Cauchy sequence, it converges to some real number $L$. Now, suppose that $x_1\geqslant L$. Let $n_1=1$. There is some natural number $n_2$ such that $x_{n_2}>x_{n_1}$. And there is some natural number $n_3$ such that $x_{n_3}>x_{n_2}$. Of course, it may happen that $n_3<n_2$, but in fact, there will always be a $n_3>n_2$ such that $x_{n_3}>x_{n_2}$. This is so because, if it turns out that your first choice for $n_3$ is smaller than $n_2$, you pick the $p\in\{1,2,\ldots,n_2-1\}$ such that $x_p$ is the largest. In particular, $x_p\geqslant x_{n_3}>x_{n_2}$ and you know that there is some natural number $m$ such that $x_m>x_p$. Then forget your first choice of $n_3$ and pick $n_3=m$ instead.
By this process, you get a strictly increasing sequence $(n_k)_{k\in\mathbb N}$. And the subsequence $(x_{n_k})_{k\in\mathbb N}$ of the sequence $(x_n)_{n\in\mathbb N}$ cannot possibly converge to $L$, since it is strictly increasing and its first term is already greater than or equal to $L$.
If $x_1\leqslant L$, then you do a similar construction, that will lead you to a strictly decreasing subsequence.
Now, let's do the general case. Take $N\in\mathbb N$ and apply the case $N=1$ to the Cauchy sequence $x_N,x_{N+1},x_{N+2},\ldots$
