Supremum & Infimum Homework Statement

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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

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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:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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