# 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

• $X_k$ is not in general finite. E.g., take $x_n=\left(-\frac12\right)^n$. – Brian M. Scott Jul 18 '20 at 20:46
• Here, finite means that a bijection can be defined on $X_n$ from $\mathbb N_n$, which in fact is possible here. – Devansh Kamra Jul 18 '20 at 20:48
• No, it isn’t necessarily possible; I just gave you a counterexample. Every tail of my sequence contains infinitely many different real numbers. – Brian M. Scott Jul 18 '20 at 20:49
• No. The members of $\Bbb N_n$ are the natural numbers less than $n$ (or $\le n$, depending on your notational conventions). The indices of points of $X_n$ are natural numbers greater than or equal to $n$. – Brian M. Scott Jul 18 '20 at 20:54
• 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. – Brian M. Scott Jul 18 '20 at 20:59

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.