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]
Rational numbers are numbers that can be expressed as a ratio of two integers, where the denominator is not equal to zero. They can be written in the form of a/b, where a and b are integers.
A bounded subset is a set of numbers that has a finite upper and lower limit. This means that all the numbers in the subset fall within a specific range.
This means that the subset does not have a maximum value. While there is an upper limit to the numbers in the subset, there is no single number that is greater than all the others.
Yes, a rational number can be a bounded subset with no least upper bound. This can happen when the subset contains numbers that are infinitely close to the upper limit, but not equal to it.
One example is the set of rational numbers between 0 and 1, where the upper limit is 1 but there is no rational number that is greater than all the others. Another example is the set of rational numbers between 1 and 2, where 2 is the upper limit but there is no single rational number that is greater than all the others.