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!

Max/Min proof I can't follow

  1. Oct 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Give an example of a bounded set that has neither a maximum nor a minimum. (The proof below is given by the book).

    We claim that the set ##(0,2)## is bounded and has neither a maximum nor a minimum.

    Proof: For each ##x \epsilon (0,2)##, we know that ##0 < x < 2##. Therefore 0 is a lower bound of the set and 2 is an upper bound. Thus, (0,2) is bounded. To see that it has no maximum, suppose to the contrary that ##s## is a maximum of the set ##(0,2)##. Then, by definition of maximum, s must be in the set ##(0,2)##. But
    ##0 < s < \frac {2+s}{2} < 2## and therefore ##\frac {2+s}{2}## is in the set (0,2) and larger than s, a contradiction. In a similar fashion, you can check that there is no minimum.
    2. Relevant equations


    3. The attempt at a solution
    I don't get where ##\frac {2+s}{2}## comes from. I know that since ##s < 2##, then
    ##s + 2 < 2 + 2## so ##s + 2 < 4## so ##\frac {s+2}{2} < 2##. But how do we know ## s < \frac {s+2}{2} ##
     
  2. jcsd
  3. Oct 9, 2017 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##s=\frac{s+s}2<\frac{s+2}2## because ##0<s<2##.
     
  4. Oct 9, 2017 #3
    Thank you, now I get it
     
  5. Oct 9, 2017 #4
    Ok, and to show minimum we would do this:

    Suppose ##h## is a minimum of ##(0,2)##. Then ##0 < h < 2## by definition of minimum. But ##0 < \frac {h}{2} < h < 2##. Thus h is not a minimum, a contradiction. We conclude that ##(0,2)## does not have a minimum. []

    note: for the minimum, we could have divided ##h## by any ##n > 0## and would have found another minimum.
     
  6. Oct 10, 2017 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Sometimes visualization is helpful.

    Draw a number line and mark the points x = 0 and x = 2 on it. The segment between these two marked points shows the region ##I = (0,2).## Now for any ##a \in I## the point ##b=(a+2)/2## is the mid-point of the segment from ##a## to ##2##, so lies between that segment's endpoints ##a## and ##2##.
     
  7. Oct 10, 2017 #6
    I will keep this in mind while going through this chapter, Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Max/Min proof I can't follow
  1. Max/ min (Replies: 3)

  2. Min and max value of r (Replies: 17)

Loading...