1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted