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Proof about bounds

  1. Oct 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the sets below. For each one, decide whether the set is bounded above. If it is, give the supremum in ##\mathbb{R}##. Then decide whether or not the set is bounded below. If it is, give the infimum. Finally, decide whether or not the supremum is a maximum, and whether to not the infimum is a minimum.

    c) the natural numbers ##\mathbb{N}##
    2. Relevant equations
    In my class ##\mathbb{N} = {0, 1, 2, ... }##

    3. The attempt at a solution
    Proof: ##\mathbb{N}## is not bounded above. We will show this with a contradiction. Suppose M is an upper bound on ##\mathbb{N}##. Then (##\lceil M \rceil + 1) \space \epsilon \space \mathbb{N}## and ##\lceil M \rceil + 1 > M##, a contradiction. Therefore ##\mathbb{N}## does not have an upper bound and is not bounded above.

    We will now show ##\mathbb{N}## is bounded below. Let ##U = 0## and let ##x \space \epsilon \space \mathbb{N}##. Then ##U = 0 \le x##. Therefore ##0## is a lower bound on ##\mathbb{N}##.

    In order to show inf##\mathbb{N} = 0## we must show 0 is the greatest lower bound. We proceed by contradiction. Suppose ##m## is a lower bound on ##\mathbb{N}## such that ##m > 0##. But ##0 \space \epsilon \space \mathbb{N}## so m is not a lower bound, a contradiction. We conclude that there does not exist a lower bound ##m## such that ##m > 0## and so inf##\mathbb{N} = 0##.

    In order to show min##\mathbb{N} = 0##. we must show 0 is a lower bound on ##\mathbb{N}## and ##0 \space \epsilon \space \mathbb{N}##. We've already shown 0 is a lower bound on ##\mathbb{N}## and ##0 \space \epsilon \space \mathbb{N}## is a true statement. Therefore min##\mathbb{N} = 0##. []

    My question: Is this proof clear and easy to follow?
     
  2. jcsd
  3. Oct 14, 2017 #2

    fresh_42

    Staff: Mentor

    Yes, this is correct.
     
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