Neither $\max S$ nor $\min S$ exists for $S=(-1,1)$ The set $S=(-1,1)$ is bounded. However, neither $\max S$ nor $\min S$ exists.
I think I understand this question intuitively, but how do I actually prove it mathematically? I know it is bounded because I can find a number above and below the set.
 A: Assume by contradiction that $x = \max(S) \in S$. Then $x < 1$ and so $$x < \frac{1+x}{2} < 1$$ but $(1+x)/2 \in S$, giving a contradiction to the maximality of $x$. The proof for $\min(S)$ is similar.
A: Suppose $x=\max (-1,1)$. Then $x<x+\frac{1-x}{2}\in (-1,1)$, a contradiction.
Suppose $y=\min (-1,1)$. Then $y>y-\frac{y+1}{2}\in (-1,1)$, a contradiction.
A: If $S$ is a bounded non empty set then the supremum and the infimum exist but the maximum and the minimun may not exist.
A: By definition, for example:
$$\min S:=\{s\in S\;;\;s\le t\;\;\forall\;t\in S\}\,$$
Can you find an element within S that fulfills the above? No...why? Suppose there is one such element
$$s_0=\min S\implies s-(-1)=\epsilon>0$$
Now show there's an element in $\,S\,$ between $\,-1\,$ and $\,s_0\,$ ...
A: If you know how to prove that $\sup S = 1$ (and $\inf S = -1$) then this follows from the fact that a set of real numbers has a maximum if and only if its supremum is a set element (and a  minimum if and only if its infimum is a set element).
