S has no upper bound
Proof by contradiction: Suppose S has an upperbound, a. Certainly, 0 is in S so we must have 0< a. Then a< a+ 1 so a+ 1 is not in S. But S contains all numbers larger than -2 so that is a contradiction
S has no smallest member.
Proof by contradicton: Suppose S has smallest member b. Since b is in S, -2< b. Now, -2< (b- 2)/2< b is larger than -2 so in S but less than b which contradicts the assumption that b is the smallest member of S.