**Lemma 1.5.1**An upper bound $ u $ is the supremum of $ S $ in $ \mathbb{R} $ if and only if $ \forall \, \varepsilon > 0, \exists \, a \in S $ such that $ u - \varepsilon < a $.**Diagram Explanation:**The diagram is a number line representing the concept of the supremum $ u $ of a set $ S $ within the real numbers $ \mathbb{R} $. - The area marked $ S $ indicates elements within the set.- The number 0 is labeled for reference on the line.- The point $ u - \varepsilon $ is depicted as a mark on the number line left of $ a $.- The element $ a $ from the set $ S $ is shown as being greater than $ u - \varepsilon $.- The supremum $ u = \sup S $ indicates the least upper bound, positioned to show its relation to the set $ S $.This visualization emphasizes that for every positive $ \varepsilon $, you can find an element $ a $ in $ S $ that is greater than $ u - \varepsilon $, affirming $ u $ as the supremum of $ S $. Let $(x_n)$ be a bounded sequence and let $s = \sup\{x_n: n \in \mathbb{N}\}$. Show that if $s \not\in \{x_n: n \in \mathbb{N}\}$, then there is a subsequence of $(x_n)$ that converges to $s$. (Hint: Use Lemma 1.5.1 and use $s - 1 < s, \, s - 1/2 < s, \, s - 1/3 < s, \ldots$)

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Description: **Lemma 1.5.1**An upper bound $ u $ is the supremum of $ S $ in $ \mathbb{R} $ if and only if $ \forall \, \varepsilon > 0, \exists \, a \in S $ such that $ u - \varepsilon < a $.**Diagram Explanation:**The diagram is a number line represe

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Math

Advanced Math

Let (xn) be a bounded sequence and let s = sup{xn : n e N}. Show that if s ¢ {xn : n E N}, then there is a subsequence of (xn) that converges to s. (Hint: Use Lemma 1.5.1 and use s - 1< s, s- 1/2 < s, s – 1/3 < s, ...)

Let (xn) be a bounded sequence and let s = sup{xn : n e N}. Show that if s ¢ {xn : n E N}, then there is a subsequence of (xn) that converges to s. (Hint: Use Lemma 1.5.1 and use s - 1< s, s- 1/2 < s, s – 1/3 < s, ...)

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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Concept explainers

Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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**Lemma 1.5.1**

An upper bound $ u $ is the supremum of $ S $ in $ \mathbb{R} $ if and only if $ \forall \, \varepsilon > 0, \exists \, a \in S $ such that $ u - \varepsilon < a $.

**Diagram Explanation:**

The diagram is a number line representing the concept of the supremum $ u $ of a set $ S $ within the real numbers $ \mathbb{R} $.

- The area marked $ S $ indicates elements within the set.
- The number 0 is labeled for reference on the line.
- The point $ u - \varepsilon $ is depicted as a mark on the number line left of $ a $.
- The element $ a $ from the set $ S $ is shown as being greater than $ u - \varepsilon $.
- The supremum $ u = \sup S $ indicates the least upper bound, positioned to show its relation to the set $ S $.

This visualization emphasizes that for every positive $ \varepsilon $, you can find an element $ a $ in $ S $ that is greater than $ u - \varepsilon $, affirming $ u $ as the supremum of $ S $.

Let $(x_n)$ be a bounded sequence and let $s = \sup\{x_n: n \in \mathbb{N}\}$. Show that if $s \not\in \{x_n: n \in \mathbb{N}\}$, then there is a subsequence of $(x_n)$ that converges to $s$. (Hint: Use Lemma 1.5.1 and use $s - 1 < s, \, s - 1/2 < s, \, s - 1/3 < s, \ldots$)

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