Let A CR be non-empty and bounded from above. Put s = sup A. Provethat for every n EN there exists xn € A such that8-1n-

Answered: Let ACR be non-empty and bounded from…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Let A CR be non-empty and bounded from above. Put s = sup A. Prove
that for every n EN there exists xn € A such that
8-
1
n
-<Xn≤ 8.
In

Expert Solution

Step 1: using by completeness property

To prove this statement, will use the definition of the supremum and properties of real numbers.

Recall that $s = sup (A)$ means that:

1. For every $x \in A$ , $x \leq s$ .
2. For every $ϵ > 0$ , there exists $x \in A$ such that $s - ϵ < x$ .

Given $n \in N$ , let's consider the set $B = {x \in A ∣ s - \frac{1}{n} < x}$ .

By property (2) of the supremum, we know that for $ϵ = \frac{1}{n}$ , there exists $x_{0} \in A$ such that $s - \frac{1}{n} < x_{0}$ . This implies that $x_{0} \in B$ , so $B$ is non-empty.

Since $A$ is bounded from above, there exists a real number $M$ such that $a \leq M$ for all $a \in A$ . In particular, this means $s - \frac{1}{n} < M$ .

This shows that $B$ is bounded above by $M$ , and by the completeness property of real numbers, it has a supremum. Let $y = sup (B)$ .

It claims that $y = s$ .

1. First, will show that $y \leq s$ . By the definition of $B$ , we have $s - \frac{1}{n} < x$ for all $x \in B$ . This implies that $s - \frac{1}{n}$ is an upper bound for $B$ , and by the completeness property, $y \leq s - \frac{1}{n}$ . Taking the limit as $n$ approaches infinity, we get $y \leq s$ .