**Axiom 5.2.4.** The set of real numbers \(\mathbb{R}\) is *complete* under the linear order \(\leq\). That is, for every subset \(A \subseteq \mathbb{R}\) that is bounded above, there exists a least upper bound \(m\), in the sense that, if \(M\) is any other upper bound of \(A\), then \(m \leq M\).Our set \(S = \{r \in \mathbb{Q} : r \leq \sqrt{2}\}\) is a subset of \(\mathbb{Q}\) and has many upper bounds that are elements of \(\mathbb{Q}\). However, there does not exist a least upper bound of \(S\) in \(\mathbb{Q}\). For this reason, \(\mathbb{Q}\) is not complete.*Construct another subset of \(\mathbb{Q}\) that shows \(\mathbb{Q}\) is not complete.*

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Axiom 5.2.4. The set of real numbers R is complete under the linear order s. That is, for every every subset ASR that is bounded above, there exists a least upper bound m, in the sense that, if M is any other upper bound of A, then ms M. Our set S= {re Q:rs v2) is a subset of Q and has many upper bounds that are elements of Q. However, there does not exist a least upper bound of S in Q. For this reason, Q is not complete. Construct another subset of Q that shows Q is not complete

Axiom 5.2.4. The set of real numbers R is complete under the linear order s. That is, for every every subset ASR that is bounded above, there exists a least upper bound m, in the sense that, if M is any other upper bound of A, then ms M. Our set S= {re Q:rs v2) is a subset of Q and has many upper bounds that are elements of Q. However, there does not exist a least upper bound of S in Q. For this reason, Q is not complete. Construct another subset of Q that shows Q is not complete

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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