Unique upper bound I'm self studying real analysis. 
I know I can find this proof easily on the internet, but I want to learn from my mistake. Can anyone tell me if my proof is correct or point me where is incorrect. thank you.
Proove that if $b^0$ = supremum E, then $b^0$ is unique
Suppose that $b^0$ is the supremum of E, we will show that $b^0$ is unique. 
Let x $\in E$, 
if  $b^0 - x > 0$ then $x<b^0$ and $b^0$ is an unique upper bound of E. 
if $x-b^0=0$ then $x=b^0$ and $b^0$ is and unique upper bound of E
if $b^0 - x <0$ then $b^0$ is not an upper bound or x $\notin E$

UPDATE
Suppose that $b_0$ and $b_1$  are both  an upper bound of E, we will show that the least upper bound is unique.
if $b_0 - b_1 >0 $ then $b_0 >  b_1$ . Since $b_0 >  b_1$  then $b_0$ is the least upper bound and unique.
if $b_0 - b_1 < 0 $ then $b_0 <  b_1$. Since $b_0 <  b_1$  then $b_1$ is the least upper bound and unique.
if $b^0 - b^1  = 0 $ then $b^0 =  b^1$ . Since $b^0 =  b^1$  then the least upper bound is unique.
 A: There's no reason to pick an $x \in E$. Since $b^0$ is a supremum of $E$, we already know that $x \leq b^0$.
What you want to do is consider two supremums $b^0$ and $b^1$ and show that they must be equal. By definition, since $b^1$ is an upper bound of $E$ and $b^0$ is a least upper bound, we must have $b^0 \leq b^1$. The same argument with $b^0$ and $b^1$ reversed shows that $b^1 \leq b^0$. Hence we must have the desired uniqueness, $b^0 = b^1$.
A: Unfortunately, your proof is not at all correct. Nowhere does it address the issue of uniqueness, and your conclusions don't follow. For example, the line

if $b^0 - x > 0$ then $x < b^0$ and $b^0$ is a unique upper bound of $E$

doesn't really make any sense - you can conclude that $x < b^0$, but what does that have to do with the conclusion? At best, it seems that you're assuming the conclusion here.

For an approach to the problem, however, try the usual setup for a uniqueness proof. Take two candidates, called $s$ and $t$. If $s < t$, show there's a big problem that arises. For the same reason, $t > s$ leads to a problem. Then conclude $s = t$.
