Real Analysis - Simple supremum/infimum problem

[rolandc]

Homework Statement

If S = { 1/n - 1/m | n, m \in 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.

 

[michael.wes]

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.

 

[Ted123]

Homework Statement

If S = { 1/n - 1/m | n, m \in 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 S_1 \subseteq \mathbb{R} and S_2 \subseteq \mathbb{R} and you define S = S_1 + S_2 = \{ x+y : x\in S_1, y\in S_2 \} then \text{sup}(S) = \text{sup}(S_1) + \text{sup}(S_2) .

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

A similar proof would yield \text{inf}(S) = \text{inf}(S_1) + \text{inf}(S_2) 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 \{1/n : n \in \mathbb{N} \} ,\{-1/m : m \in \mathbb{N} \} are not \pm 1 ]