Supremum & Infimum Homework Statement

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Homework Help Overview

The problem involves the set \( S = \left\{ \frac{n}{n + m} : n, m \in \mathbb{N} \right\} \) and requires proving that the supremum of \( S \) is 1 and the infimum is 0. The discussion centers around understanding the conditions for supremum and infimum in the context of this set.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the requirement for an upper bound to be the supremum, specifically focusing on finding \( n \) and \( m \) such that \( \frac{n}{n + m} > 1 - \varepsilon \) for any given \( \varepsilon > 0 \). There is an exploration of expressing \( n \) and \( m \) in terms of \( \varepsilon \) and the implications of rearranging the equation.

Discussion Status

The discussion is active, with participants providing hints and engaging in clarifying the mathematical expressions involved. There is a collaborative effort to explore the implications of the hints provided, particularly regarding the relationship between \( n \) and \( m \) in the context of the supremum.

Contextual Notes

Participants are navigating the challenge of expressing variables in terms of \( \varepsilon \) and ensuring that the conditions for supremum are met. There is an acknowledgment of the need for further exploration of the relationships within the set \( S \).

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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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:
 
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?
 
yup! :smile:

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

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