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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