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.
Supremum of a series
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sequences-and-series
1 Answers
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As $(x_n)$ is bounded, there exist $C < \infty$ such that $|x_n|
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0than$k$ yo$u$, much less complicated than I was ma$k$ing it out to be. – 2012-10-25