University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 489: 128

Answer

Proof given below.

Work Step by Step

Every least upper bound is an upper bound. Let $M_{1}$ and $M_{2}$ be least upper bounds. Taking $M_{1}$ to be a least upper bound and $M_{2}$ to be just an upper bound, we can write $M_{1}\leq M_{2}$ Now, taking $M_{2}$ to be a least upper bound and $M_{1}$ to be just an upper bound, we can write $M_{2}\leq M_{1}$ By properties of real numbers, $(M_{1}\leq M_{2})$ and $(M_{2}\leq M_{1})\quad \Leftrightarrow\quad M_{1}=M_{2}$. There can only be one.
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