**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 \(s - 1 < s, \, s - 1/2 < s, \, s - 1/3 < s, \ldots\))

Name: **Lemma 1.5.1**An upper bound \( u \) is the supremum of \( S \) in \
Uploaded: 2024-03-20T06:38:58.000Z
Channel: Bartleby
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