# Max/Min proof I can't follow

1. Oct 9, 2017

### fishturtle1

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. Oct 9, 2017

### andrewkirk

$s=\frac{s+s}2<\frac{s+2}2$ because $0<s<2$.

3. Oct 9, 2017

### fishturtle1

Thank you, now I get it

4. Oct 9, 2017

### fishturtle1

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. Oct 10, 2017

### Ray Vickson

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 $2$.