Real Analysis - Simple supremum/infimum problem

In summary, The supremum of the set S is 1 and the infimum is -1. This can be shown by using the property that the supremum and infimum of a set can be determined by adding the supremums and infimums of its subsets. Since the subsets {1/n} and {-1/m} both have supremum and infimum values of 1 and -1 respectively, the set S must also have a supremum of 1 and an infimum of -1.
  • #1
rolandc
1
0

Homework Statement


If S = { 1/n - 1/m | n, m [tex]\in[/tex] N}, find inf(S) and sup(S)

I'm having a really hard time wrapping my head around the proper way to tackle sumpremum and infimum problems. I've included the little that I've done thus far, I just need a nudge in the right direction. Correct me if I'm wrong but I gather that there are basically 2 ways to tackle these problems.

1. An "epsilon" argument using the definition of supremum/infimum
2. An argument using some variation of the archimedean property


Homework Equations



N/A

The Attempt at a Solution



S is bounded above by 1. Therefore, S has a supremum. Let u = sup(s). We know that u <= 1. We will show that u = 1 by proving that u cannot be less than 1.

Assume u < 1.

1/n - 1/m <= u < 1

Where do I go from here? Thanks.
 
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  • #2
Intuitively you should see that the candidates for inf and sup are -1 and +1, respectively. You've got the right idea for the proof.

By definition, you have to show any upper bound u for S has 1 <= u. To do this, try to exhibit an element of the set that is strictly bigger than u if you assume u < 1.
 
  • #3
rolandc said:

Homework Statement


If S = { 1/n - 1/m | n, m [tex]\in[/tex] N}, find inf(S) and sup(S)

I'm having a really hard time wrapping my head around the proper way to tackle sumpremum and infimum problems. I've included the little that I've done thus far, I just need a nudge in the right direction. Correct me if I'm wrong but I gather that there are basically 2 ways to tackle these problems.

1. An "epsilon" argument using the definition of supremum/infimum
2. An argument using some variation of the archimedean property


Homework Equations



N/A

The Attempt at a Solution



S is bounded above by 1. Therefore, S has a supremum. Let u = sup(s). We know that u <= 1. We will show that u = 1 by proving that u cannot be less than 1.

Assume u < 1.

1/n - 1/m <= u < 1

Where do I go from here? Thanks.

You could first prove that if [itex]S_1 \subseteq \mathbb{R}[/itex] and [itex]S_2 \subseteq \mathbb{R}[/itex] and you define [itex]S = S_1 + S_2 = \{ x+y : x\in S_1, y\in S_2 \}[/itex] then [itex]\text{sup}(S) = \text{sup}(S_1) + \text{sup}(S_2)[/itex] .

Then you can read off the supremum of [itex]\{1/n - 1/m : n, m \in \mathbb{N} \} = \{1/n : n \in \mathbb{N} \} + \{-1/m : m \in \mathbb{N} \}[/itex] easily.

A similar proof would yield [itex]\text{inf}(S) = \text{inf}(S_1) + \text{inf}(S_2)[/itex] and again you can read off the infimum of the set you want very easily after deducing this result.

[by the way the sup and inf's of [itex]\{1/n : n \in \mathbb{N} \} ,\{-1/m : m \in \mathbb{N} \}[/itex] are not [itex]\pm 1[/itex] ]
 
Last edited:

1. What is the definition of supremum and infimum in real analysis?

In real analysis, supremum and infimum are two important concepts used to describe the behavior of a set of numbers. The supremum of a set is the smallest number that is greater than or equal to all the numbers in the set. The infimum, on the other hand, is the largest number that is less than or equal to all the numbers in the set.

2. How do you find the supremum and infimum of a given set of numbers?

To find the supremum and infimum of a set of numbers, you need to first arrange the numbers in ascending order. The supremum will be the last number in the set, while the infimum will be the first number. If the set is unbounded, the supremum or infimum may not exist.

3. Can a set have multiple supremums or infimums?

No, a set can have only one supremum and one infimum. If a set has multiple supremums (or infimums), all these numbers will be equal to each other.

4. How are supremum and infimum related to the concepts of maximum and minimum?

The supremum of a set is always greater than or equal to the maximum of the set, while the infimum is always less than or equal to the minimum. However, the maximum or minimum of a set may not always exist, while the supremum and infimum always exist for a set.

5. In what real-life situations are the concepts of supremum and infimum useful?

The concepts of supremum and infimum are useful in various real-life situations, such as optimizing a function, finding the highest or lowest possible value in a set, and analyzing the behavior of a system. They are also commonly used in fields such as economics, physics, and engineering.

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