How can I proof the infimum and supremum of this set?

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Solution 1

Assume that the $\sup E$ exists, thus there is some number $b \in R_{>0}$ s.t $\forall(x+y), x,y \in R_{>0}$ then $b \geq (x+y)$, but then $b+1$ is a sum of two numbers in $R_{>0}$ and $b+1>b$, contradiction.

Solution 2

$\bullet$ The supremum of $E$ does not exist because $E$ is not bounded above (for every $R>0$ the element $R+R$ belongs to $E$ and is striclty greater than $R$).

$\bullet$ The infimum of $E$ is $0$ because (as every element of $E$ is positive) is it a lower bound, and for every $\varepsilon>0$ the element $\varepsilon/3+\varepsilon/3$ belongs to $E$ and is strictly smaller than $\varepsilon$ (so there can't be a greater lower bound of $E$).

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Viktor Raspberry
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Updated on August 01, 2022

Comments

  • Viktor Raspberry
    Viktor Raspberry over 1 year

    $E = \{{x+y : x,y \in\Bbb R_{>0}}$}

    I was able to figure out that this set does not have a supremum, but I am not able to prove it. Also, how can I prove the infimum of this set ?

    This is my logic: This set does not have a supremum, because x,y can not be negative real numbers, and the the definition of the set is x+y, which will produce a bigger real number, and this set will grow without bound. When it comes to infimum, I think that it should be zero. For this set, the infimum of A is not an element of the set, because x,y are strictly greater than zero.

    Can someone provide a sufficent proof for the supremum and infimum of this set ?

    • Stefan Mesken
      Stefan Mesken almost 7 years
      You pretty much proved it already. Just sit down and formalize what you've already observed.
    • copper.hat
      copper.hat almost 7 years
      Note that $\mathbb{N} \subset E$.
  • Viktor Raspberry
    Viktor Raspberry almost 7 years
    How can I prove that the infimum is zero ?