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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

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  • $\begingroup$ $X_k$ is not in general finite. E.g., take $x_n=\left(-\frac12\right)^n$. $\endgroup$ – Brian M. Scott Jul 18 '20 at 20:46
  • $\begingroup$ Here, finite means that a bijection can be defined on $X_n$ from $\mathbb N_n$, which in fact is possible here. $\endgroup$ – Devansh Kamra Jul 18 '20 at 20:48
  • $\begingroup$ No, it isn’t necessarily possible; I just gave you a counterexample. Every tail of my sequence contains infinitely many different real numbers. $\endgroup$ – Brian M. Scott Jul 18 '20 at 20:49
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    $\begingroup$ 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 $n$. $\endgroup$ – Brian M. Scott Jul 18 '20 at 20:54
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    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jul 18 '20 at 20:59
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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.

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  • $\begingroup$ Thanks a lot for the help. $\endgroup$ – Devansh Kamra Jul 18 '20 at 21:20
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    $\begingroup$ @DevanshKamra: You’re very welcome. $\endgroup$ – Brian M. Scott Jul 18 '20 at 21:20

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