Limit Superior

**Markeur** · March 7th 2010, 02:49 AM

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 $\displaystyle{\limsup_{n\to\infty}x_n=y}$ .

**HallsofIvy** · March 7th 2010, 04:37 AM

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.

**Markeur** · March 7th 2010, 04:47 AM

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?

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