Supremum of a series

Question

Let $( x_n )$ be a bounded sequence, and define for each $m \in \Bbb{N}$, the number $u_m := \sup{x_n : n \le m}$. Show that $(u_m)$ is a bounded, decreasing sequence. I'm having trouble even conceptualizing this question, specifically, how to define $(x_n)$ in terms of each ($m \in \Bbb N$). Any insight would be much appreciated.

Answer 1

As $(x_n)$ is bounded, there exist $C < \infty$ such that $|x_n|<C$. Then $u_m<C$. Notice that $$ \{ x_n : n\le m\} \subset \{x_n : n \le m+1\}.$$ Then, $u_m\le u_{m+1}$.