Show $\sup\{x_k : k \ge n\}$ and $\inf\{x_k : k \ge n\}$ are monotone, where $\{x_k\}$ is bounded. [closed]

Question

Let $\{x_n\}$ be a bounded sequence of real numbers. Let us define $y_n = \sup\{x_k : k \ge n\}$ and $z_n = \inf\{x_k : k \ge n\}$. Show that the sequence $\{y_n\}$ is decreasing and $\{z_n\}$ is increasing.

Answer 1

Partial answer:

$(x_n)$ is bounded;

$y_n:=\sup \{x_k: k\ge n \}$ exists and is decreasing .

$y_{n+1}=\sup \{x_k: k\ge (n+1)\};$

1) $x_n \le x_{n+1};$

Then

$ y_{n}= \sup \{x_k: k\ge n\} = \sup \{x_k: k\ge n+1\} =y_{n+1};$

2) $x_n > x_{n+1};$

Then

$y_{n+1}=\sup \{ x_k : k \ge n+1\} \le \sup \{x_k: k \ge n\}=y_n;$

Answer 2

It is obvious that $ \inf(a, b)\leqslant b $, so for all $ k\in\mathbb{N} $ we have $$ \inf(x_{k+1}, x_{k+2}, \dots)=\inf(x_{k+1}, \inf(x_{k+2}, x_{k+3}, \dots))\leqslant \inf(x_{k+2}, x_{k+3}, \dots), $$ which implies $ \inf\{x_{k}:k\geqslant n\} $ is increasing. Using this method we can have that $ \sup\{x_{k}:k\geqslant n\} $ is decreasing.