Supremum & Infimum Homework Statement

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In summary, we are trying to prove that the supremum of the set S, which is defined as the set of all fractions n/(n+m) where n and m are natural numbers, is equal to 1, and the infimum is equal to 0. To do this, we need to find values for n and m such that n/(n+m) is always greater than 1-ε for any given ε. Using the hint provided, we can rewrite n/(n+m) as 1 - m/(n+m) and choose m=1. This results in ε>1/(n+1), which we can then solve for n.
  • #1
drawar
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Homework Statement


Let $$S = \left\{ {\frac{n}{{n + m}}:n,m \in N} \right\}$$. Prove that sup S =1 and inf S = 0

Homework Equations


The Attempt at a Solution



So I was given the fact that for an upper bound u to become the supremum of a set S, for every ε>0 there is $$x \in S$$ such that x>u-ε. In this case, I'm supposed to find n and m such that $${\frac{n}{{n + m}} > 1 - \varepsilon }$$ for every ε given. However, I cannot express n and m in terms of ε explicitly. Any hints or comments will be very appreciated, thanks!
 
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  • #2
hi drawar! :smile:
drawar said:
I'm supposed to find n and m such that $${\frac{n}{{n + m}} > 1 - \varepsilon }$$ for every ε given.

hint: n/(n+m) = 1 - m/(n+m) :wink:
 
  • #3
tiny-tim said:
hi drawar! :smile:hint: n/(n+m) = 1 - m/(n+m) :wink:

Hi tiny-tim, thanks for the hint. Do you mean:

$${1 - \varepsilon < \frac{n}{{n + m}} = 1 - \frac{m}{{n + m}}}$$
Choosing m=1:
$${\varepsilon > \frac{m}{{n + m}} > \frac{1}{{n + 1}}}$$
and then solve for n?
 
  • #4
yup! :smile:

except, that's ##\frac{1}{\frac{n}{m}+1}## :wink:
 

FAQ: Supremum & Infimum Homework Statement

What is the definition of Supremum and Infimum?

Supremum and Infimum are two important concepts in mathematical analysis. The Supremum (or least upper bound) of a set is the smallest number that is greater than or equal to all elements in the set. The Infimum (or greatest lower bound) of a set is the largest number that is less than or equal to all elements in the set.

How are Supremum and Infimum related to Maximum and Minimum?

The Supremum and Infimum are related to the Maximum and Minimum in that the Maximum and Minimum are just the largest and smallest elements in a set, respectively. However, the Supremum and Infimum may not necessarily be actual elements of the set, but are defined as the smallest upper bound and largest lower bound, respectively.

How do you find the Supremum and Infimum of a set?

To find the Supremum and Infimum of a set, you can use the following steps:
1. Arrange the elements of the set in ascending order.
2. If the set has a maximum element, then the maximum is the Supremum.
3. If the set has a minimum element, then the minimum is the Infimum.
4. If the set does not have a maximum or minimum element, then the Supremum and Infimum can be found by observing the pattern of the set and using logical reasoning.

What is the importance of Supremum and Infimum in mathematical analysis?

The concept of Supremum and Infimum is important in mathematical analysis as it helps us define limits, continuity, and the behavior of functions. They also allow us to determine whether a set has a maximum or minimum element, which is crucial in optimization problems and finding solutions to equations.

Can a set have multiple Supremum or Infimum?

No, a set can only have one Supremum and one Infimum. This is because the Supremum and Infimum are unique and are defined as the smallest upper bound and largest lower bound, respectively. If a set has multiple Supremum or Infimum, then they would not be unique and would not accurately represent the set.

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