Max/Min proof I can't follow

  • #1
334
44

Homework Statement


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.

Homework Equations




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} ##
 

Answers and Replies

  • #3
334
44
##s=\frac{s+s}2<\frac{s+2}2## because ##0<s<2##.
Thank you, now I get it
 
  • #4
334
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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.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement


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.

Homework Equations




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} ##

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##.
 
  • #6
334
44
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##.
I will keep this in mind while going through this chapter, Thank you.
 

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