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Proof of points arbitrarily close to supremum

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Let [itex] S \subset \mathbb{R} [/itex] be bounded above. Prove that [itex] s \in \mathbb{R} [/itex] is the supremum of [itex] S [/itex] iff. [itex] s [/itex] is an upper bound of [itex] S [/itex] and for all [itex] \epsilon > 0 [/itex], there exists [itex] x \in S [/itex] such that [itex] |s - x| < \epsilon [/itex].

    2. Relevant equations
    **Assume I have only the basic proof methods, some properties of inequalities, and the Completeness Axiom at my disposal**

    Assume I have already proved that if [itex] s \in \mathbb{R} [/itex] is the supremum of [itex] S [/itex], then for all [itex] \epsilon > 0 [/itex], there exists [itex] x \in S [/itex] such that [itex] s - x < \epsilon [/itex].

    3. The attempt at a solution
    The [itex] \Rightarrow [/itex] direction:
    By the definition of the supremum, [itex] s [/itex] is an upper bound.

    Thus [itex] s \geq x [/itex] for all [itex] x \in S [/itex], so [itex] s - x \geq 0 [/itex]. Then [itex] |s - x| = s - x < \epsilon [/itex], by the result mentioned above.

    The [itex] \Leftarrow [/itex] direction:
    Suppose that [itex] s \in \mathbb{R} [/itex] is an upper bound of [itex] S [/itex] and that for all [itex] \epsilon > 0 [/itex], there exists [itex] x \in S [/itex] such that [itex] |s - x| < \epsilon [/itex]. Now suppose for a contradiction that [itex] s [/itex] is not the supremum of [itex] S [/itex]. Then there exists an upper bound [itex] u [/itex] of [itex] S [/itex] with [itex] u < s [/itex].

    But since [itex] S [/itex] is bounded above, it does have a supremum. Call the supremum [itex] t [/itex]. We have that [itex] t \leq s [/itex], and for every [itex] \epsilon > 0 [/itex], there exists [itex] y \in S [/itex] such that [itex] |t - y| < \epsilon [/itex].


    I'm wondering whether there are any logical errors so far, and whether someone could give me a tiny hint as to whether I'm on the right track, and what I might think about next. I'm posting here rather than Math.SE because I would prefer "Socratic" help rather than an answer. Thanks!
     
  2. jcsd
  3. Mar 9, 2016 #2

    Samy_A

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    Homework Helper

    The ⇒ direction seems correct.

    For the ⇐ direction:
    You have ##s## and the supremum ##t##.
    If ##s=t##, you are done.
    ##s<t## is impossible, as ##t## is the supremum, the least upper bound.
    That leaves the case ##s>t##.
    Now try to prove that then there must exist an ##x \in S## satisfying ##x>t##, which contradicts ##t## being a supremum.
    Hint: set ##\epsilon=\frac{s-t}{3}##.
     
    Last edited: Mar 9, 2016
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