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Supremum and infimum

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Let $$S = \left\{ {\frac{n}{{n + m}}:n,m \in N} \right\}$$. Prove that sup S =1 and inf S = 0



    2. Relevant equations



    3. 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!
     
  2. jcsd
  3. Jan 30, 2013 #2

    tiny-tim

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    hi drawar! :smile:
    hint: n/(n+m) = 1 - m/(n+m) :wink:
     
  4. Jan 30, 2013 #3
    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?
     
  5. Jan 30, 2013 #4

    tiny-tim

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    yup! :smile:

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