Q:

Proving monotonocity and convergence for sequences $(s_n)$ and $(t_n)$

Let $X:=(x_n:n\in \mathbb N)$ be a bounded sequence, and for each $n\in \mathbb N$ let $s_n:=\sup\{x_k:k\geq n\}$ and $t_n:=\inf\{x_k:k\geq n\}$.
Prove that $(s_n)$ and $(t_n)$ are monotone and convergent.
My approach: Now as $X$ is bounded, it is evident that $\inf X\leq x_n\leq \sup X\ \forall\ n\in \mathbb N$
Let $X_n=(x_k:k\geq n)$ or $X_n$ be $\text{n-tail}$ of $X$
Thus $s_n=\sup X_n$ and $t_n=\inf X_n$
Now as $X_k$ is finite, $\sup X_k\in X_k$ and $\inf X_k\in X_k$
Also it is evident that $X_n\subset X_{n-1}\subset X_{n-2}\subset \ldots \subset X_2\subset X\ \forall\ n\in \mathbb N$
$\therefore \sup X_n\leq \sup X_{n-1}\leq \sup X_{n-2}\leq\ldots\leq\sup X_2\leq \sup X\ \forall\ n\in \mathbb N$
Therefore $s_n$ is decreasing and $s_n\leq \sup X\ \forall\ n\in \mathbb N$
Similarily it can be proved that $t_n$ is increasing and $t_n\geq \inf X\ \forall\ n\in \mathbb N$
Therefore both $s_n$ and $t_n$ are monotones and are convergent.
Please check this method for any mistakes.
Also $t_n\leq x_n\leq s_n\ \forall\ n\in \mathbb N$ means that $t_n$ and $s_n$ are lower and upper bounds for $X$ respectively, thus $t_n\leq \inf X$ and $s_n\geq \sup X$ which is in contradicition to what has been given before in the proof. I am doubtful of this statement as the upper and lower bounds are not fixed but change for every $n\in \mathbb N$ or are dependent on $n$. Is that allowed?
Please correct me wherever I have committed an error.
Thanks

A:

It’s not true that $X_k$ is necessarily finite. For instance, take $x_n=\left(-\frac12\right)^n$: this is clearly a bounded sequence, and all of its points are distinct, so each tail contains infinitely many distinct points.
But you have observed the key point, which is that if $X_n=\{x_k:k\ge n\}$, then $X_n\supseteq X_{n+1}$ for each $n\in\Bbb N$: from this it is immediate that $t_n\le t_{n+1}$ and $s_n\ge s_{n+1}$ and hence that $\langle s_n:n\in\Bbb N\rangle$ and $\langle t_n:n\in\Bbb N\rangle$ are monotone. If you already know that a bounded, monotone sequence converges, you’re practically done at that point.
No, in general it’s not true $t_n$ and $s_n$ are bounds on the whole sequence. Again you can look at my example at the beginning of the answer: for instance, $t_3=-\frac18$, which is bigger than $x_1=-\frac12$ and therefore not a lower bound for the whole sequence.

