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Real Analysis - Simple supremum/infimum problem

  • Thread starter rolandc
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  • #1
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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.
 

Answers and Replies

  • #2
michael.wes
Gold Member
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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
446
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.
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:

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