The discussion centers on the identification of a bounded subset of rational numbers (Q) that lacks a least upper bound (lub) within Q. A proposed example, consisting of the sequence [1/8, 1/4, 3/8, 1/2, 5/8, 3/4,...], is deemed incorrect as it is not bounded. The correct approach involves recognizing that irrational numbers are absent from Q, and thus, a bounded subset such as [1/8, 1/4, 3/8, 1/2, 5/8, 3/4] approaches an irrational limit, specifically √2, which serves as its least upper bound.
PREREQUISITESStudents of mathematics, particularly those studying real analysis, educators teaching concepts of rational and irrational numbers, and anyone interested in the properties of bounded sets and least upper bounds.
ciarax said:Homework Statement
Give an example of a bounded subset of Q which has no least upper
bound in Q. Explain why your answer has this property.
Homework Equations
The Attempt at a Solution
[1/8, 1/4, 3/8, 1/2, 5/8, 3/4...infinity]
is this correct?
Your set appears to be integer multiples of 1/8. This set is not bounded, so doesn't qualify as an example in this problem.ciarax said:Give an example of a bounded subset of Q which has no least upper
bound in Q. Explain why your answer has this property.
[1/8, 1/4, 3/8, 1/2, 5/8, 3/4...infinity]