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

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.

• Please, show us what you tried and where you are stuck Apr 21, 2020 at 13:09

$$(x_n)$$ 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

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

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.

• please can you tell how to use this method to prove that yn is decreasing?
– yuko
Apr 23, 2020 at 8:45