1. ## Limit Superior

Suppose that $\displaystyle \displaystyle{x_n}$ is a bounded sequence of real numbers.Show that $\displaystyle \displaystyle{y_n=\sup_{m \geq n}x_m}$ is a bounded decreasing sequence and deduce that $\displaystyle \displaystyle{y_n \longrightarrow y}$,ie $\displaystyle \displaystyle{\limsup_{n\to\infty}x_n=y}$.

2. Originally Posted by Markeur
Suppose that $\displaystyle \displaystyle{x_n}$ is a bounded sequence of real numbers.Show that $\displaystyle \displaystyle{y_n=\sup_{m \geq n}x_m}$ is a bounded decreasing sequence and deduce that $\displaystyle \displaystyle{y_n \longrightarrow y}$,ie $\displaystyle \displaystyle{\limsup_{n\to\infty}x_n=y}$.
If $\displaystyle N_1< N_2$, the $\displaystyle \{x_m| m\ge N_2\}$ is a subset of $\displaystyle \{x_m| m\ge N_2\}$. Any upper bound for the second is an upper bound for the first.

3. Originally Posted by HallsofIvy
If $\displaystyle N_1< N_2$, the $\displaystyle \{x_m| m\ge N_2\}$ is a subset of $\displaystyle \{x_m| m\ge N_2\}$. Any upper bound for the second is an upper bound for the first.
Hmm you mean $\displaystyle \{x_m| m\ge N_2\}$ is a subset of $\displaystyle \{x_m| m\ge N_1\}$?And how to prove that the sequence is decreasing as well?