Unique upper bound

Question

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 $\notin E$

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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.

Answer 1

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.

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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 $s = t$.

Answer 2

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, $b^0 = b^1$.