In class I was given this example, $E=\{r \in \mathbb{Q} : -2 \leq r \leq 2\}$ , as a counter example for the proposition that
If max E exists for $E \ ordered \subset S$, then Sup E exists and Max E = Sup E.
The proof was something like this:
$ Let \ B_0 \in \ E, \ then \ \frac{B_0 +2}{2} \geq B_0 \ and \frac{B_0 +2}{2} \ \in E \ so \ B_0 \ is \ not \ an \ upper \ bound \ and \ is \ not \ max.$
But later my professor said that there was a suprema and infima with Sup E = 2 and Inf E = -2. I'm honestly really confused and I would appreciate it if someone could help me out. I don't understand where the fraction $\frac{B_0 +2}{2}$ came from or how we somehow inferred that the suprema is 2 and the infima is -2.
I have no idea what you are saying. The set $\{r\in Q| -2\le r\le 2\}$ certainly does have both max and min. The inf= min is -2 and the sup= max is 2. I wonder if you do not something like $\{r\in Q| -2\le r^2\le 2\}$ (although the "-2" would be peculiar since a square is never negative) It is true that there is no rational r such that $r^2= 2$ so the set $\{r\in Q| 0\le r^2\le 2\}$ has no max or min.