# Limit Superior

• Mar 7th 2010, 02:49 AM
Markeur
Limit Superior
Suppose that $\displaystyle{x_n}$ is a bounded sequence of real numbers.Show that $\displaystyle{y_n=\sup_{m \geq n}x_m}$ is a bounded decreasing sequence and deduce that $\displaystyle{y_n \longrightarrow y}$,ie $\displaystyle{\limsup_{n\to\infty}x_n=y}$.
• Mar 7th 2010, 04:37 AM
HallsofIvy
Quote:

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

If $N_1< N_2$, the $\{x_m| m\ge N_2\}$ is a subset of $\{x_m| m\ge N_2\}$. Any upper bound for the second is an upper bound for the first.
• Mar 7th 2010, 04:47 AM
Markeur
Quote:

Originally Posted by HallsofIvy
If $N_1< N_2$, the $\{x_m| m\ge N_2\}$ is a subset of $\{x_m| m\ge N_2\}$. Any upper bound for the second is an upper bound for the first.

Hmm you mean $\{x_m| m\ge N_2\}$ is a subset of $\{x_m| m\ge N_1\}$?And how to prove that the sequence is decreasing as well?