Proof that the $\sup \left(a,b\right) = b$. I'm trying to prove that the least upper bound of the interval, $(a,b)$, is $b$, but am struggling with establishing that $b$ is the least such upper bound. 
So, using Rudin's definition, I want to prove both that $b$ is an upper bound of the interval and that it's the lowest such upper bound. 
First, by definition, $(a,b) = \{x \in \mathbb{R} \mid a < x < b\}$. So, $\forall x \in (a,b), x < b$, so $b$ is clearly an upper bound. 
Second -- where I find myself confused, though this is my best attempt -- let's assume, for a contradiction, that $\gamma \neq b$ is the least upper bound of $(a,b)$. So, $x \leq \gamma$ for all $x \in (a,b)$. Then, since $b$ is clearly an upper bound of $(a,b)$, $b \geq \gamma$. But, if $b > \gamma$, then $\gamma \in (a,b)$ by definition. But, if this is the case, then we can find some larger value in $(a,b)$, i.e., by taking the average of $b$ and $\gamma$, producing some value, $\alpha$, such that $\gamma < \alpha < b$. Thus, $\gamma$ is not an upper bound of this set, so it must be the case that $b = \gamma$. Thus, $b$ is the least upper bound of $(a,b)$. 
How does this look?
 A: Your proof is correct!
However you can prove it directly via definition of least upper bound. As you have noted $b$ is an upper bound. Let's take any $c<b$ and show that $c$ is not a upper bound. Clearly any number between $\max(a, c) $ and $b$ lies in $(a, b) $ and is greater than $c$ and thus $c$ is not an upper bound. Thus $\sup(a, b) =b$. 
A: Assuming $a<b.$
$b$ IS an upper bound for $(a,b)$ by the definition of $(a,b)$ and the definition of "upper bound". But if  $c<b$ then $c<\frac {b+\max (a,c)}{2}\in (a,b)$ so $c$ is NOT an upper bound for $(a,b).$
If $b$ belongs to a set $U$... (e.g.if $U$is the set of upper bounds for $(a,b)\;$),... and if no member of $U$ is less than $b$,  then $b$ is the least member of $U.$
A: You're on the right track, but you can simplify the proof. Let $c$ be an upper bound for $(a,b)$
If $c<b$, then
$$
a<\frac{a+b}{2}\le c<b
$$
and so
$$
a<\frac{a+b}{2}\le c<\frac{b+c}{2}<b
$$
This is a contradiction to $c$ being an upper bound for $(a,b)$, because $(b+c)/2\in(a,b)$.
Therefore $c\ge b$.
