Can You Solve a Problem Using the Definition of Supremum?

In summary, the article explores the concept of supremum, which is the least upper bound of a set in mathematics. It presents a problem-solving approach that involves applying the definition of supremum to determine whether a given value is indeed the supremum of a particular set. Through examples and explanations, the article emphasizes the importance of understanding the properties and implications of supremum in various mathematical contexts.
  • #1
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Homework Statement
I have solved the problem below; however, I am unsure whether it is correct (I list the step below which I am not sure whether I am allowed to do).
Relevant Equations
##M = sup S##
##s > M - \epsilon##
For this problem,
1710202583981.png

My solution:
Using definition of Supremum,
(a) ##M ≥ s## for all s
(b) ## K ≥ s## for all s implying ##K ≥ M##

##M ≥ s##
##M + \epsilon ≥ s + \epsilon##
##K ≥ s + \epsilon## (Defintion of upper bound)
##K ≥ M ≥ s + \epsilon## (b) in definition of Supremum
##M ≥ s + \epsilon##

Now we have two cases for the inequality:

(1) ##M = s + \epsilon##
(2) ##M > s + \epsilon##

For case (1), ##M ≠ s + \epsilon## from definition of Supremum so,

##M > s + \epsilon##
##M - \epsilon > s##

I am unsure why I get the wrong inequality direction. However, apart for that, it seems correct. I would appreciate any help.

Thanks for any help - Chiral.
 

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  • #2
I wonder if you could use a proof by contradiction., i.e.,

Assume there exists an ##\epsilon>0## such that there does not exist a ##s\in S## such that##s>M-\epsilon##, then it must be that ##M \neq \text{sup}(S)##.
 
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  • #3
ChiralSuperfields said:
Homework Statement: I have solved the problem below; however, I am unsure whether it is correct (I list the step below which I am not sure whether I am allowed to do).
Relevant Equations: ##M = sup S##
##s > M - \epsilon##

For this problem,
View attachment 341649
My solution:
Using definition of Supremum,
(a) ##M ≥ s## for all s
(b) ## K ≥ s## for all s implying ##K ≥ M##

##M ≥ s##
Is this any particular s? Suppose it is thousands below M. What will that prove?
ChiralSuperfields said:
##M + \epsilon ≥ s + \epsilon##
Is this any particular ##\epsilon##? What if it is huge?
ChiralSuperfields said:
##K ≥ s + \epsilon## (Defintion of upper bound)
No. If s is close enough to M and ##\epsilon## is huge, this is not true.
I think this approach is doomed to fail. You need to rethink it.
 
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FAQ: Can You Solve a Problem Using the Definition of Supremum?

What is the definition of supremum?

The supremum (or least upper bound) of a set S is the smallest number that is greater than or equal to every element in S. Formally, a number s is the supremum of S if: 1) s is an upper bound of S, and 2) for any other upper bound u of S, s ≤ u.

How do you prove that a number is the supremum of a set?

To prove that a number s is the supremum of a set S, you need to show two things: 1) s is an upper bound of S, meaning for all x in S, x ≤ s, and 2) for any ε > 0, there exists an element y in S such that s - ε < y. This ensures that s is the least upper bound.

What is the difference between supremum and maximum?

The supremum of a set S is the smallest upper bound of S, which may or may not be an element of S. The maximum of S, if it exists, is the largest element of S. Every maximum is a supremum, but not every supremum is a maximum.

Can the supremum be an element of the set?

Yes, the supremum can be an element of the set. In this case, the supremum is also the maximum of the set. For example, in the set {1, 2, 3}, the supremum is 3, which is also the maximum.

How do you find the supremum of an unbounded set?

If a set is unbounded above, it does not have a finite supremum. For example, the set of all real numbers or the set of all positive integers are unbounded above, and thus their supremum is considered to be infinity.

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