The sequence ${u_n}$ is a bounded sequence and $x_r$ = min $[u_{r},u_{r+1},u_{r+2},…..]$, $y_r$ = max $[u_r,u_{r+1},u_{r+2},u_{r+3},……]$ , for $r\ge1$. Prove that ${x_n}$ and ${y_n}$ are both monotone convergent sequences. If lim $x_n$ = lim $y_n$ = $l$, prove that the sequence ${u_n}$ converges to $l$.
What I tried is that $x_1$ = min $[u_1, u_2, u_3,….]$ =inf ${u_n}$,
$y_1$ = max $[u_1,u_2,u_3,…….]$ = sup ${u_n}$
So $x_1\leq$ $u_1\leq$ $y_1$ and by mathematical induction we can prove that $x_n\leq$ $y_n\leq$ $u_n$.
I am not sure if it is correct, Please correct me if I'm wrong. Also, I could not prove that the sequence ${u_n}$ converges to $1$.
Best Answer
Actually, the question doesn't make sense, because that $\max$ and that $\min$ may well not to exist. But it makes sense if we work with $\sup$ and $\inf$ instead.
Since $x_n=\inf\{u_n,u_{n+1},u_{n+2},\ldots\}$, since $x_{n+1}=\inf\{u_{n+1},u_{n+2},u_{n+3},\ldots\}$, and since$$\{u_n,u_{n+1},u_{n+2},\ldots\}\supset\{u_{n+1},u_{n+2},u_{n+3},\ldots\},$$it is clear that $x_n\leqslant x_{n+1}$. By the same argument, $y_n\geqslant y_{n+1}$.
Furthermore, if $a,b\in\Bbb R$ are such that $(\forall n\in\Bbb N):u_n\in[a,b]$, then it follows from the definitions of $x_n$ and $y_n$ and from the fact that $[a,b]$ is a closed interval that $x_n,y_n\in[a,b]$. So, both sequence $(x_n)_{n\in\Bbb N}$ and $(y_n)_{n\in\Bbb N}$ are monotonic and bounded and therefore both of them converge.
Finally, if $\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=l$, since you always have $x_n\leqslant u_n\leqslant y_n$, it follows from the squeeze theorem that $\lim_{n\to\infty}u_n=l$ too.