I have attempted to prove that an upper bound $u$ is the supremum of $\mathit{A}$ if and only if for all $\epsilon>0$ there exists an $a\in\mathit{A}$ such that $u-\epsilon<a$.
Here is my attempted proof.
Let $u$ be an upper bound of non-empty set $\mathit{A}$ in $\mathbb{R}$. We shall first prove the if-part of the statement than the only-if-part of the statement.
We shall use proof by contradiction to prove that $u$ is the supremum of $\mathit{A}$ if for all $\epsilon>0$ there is an $a\in\mathit{A}$ such that $u-\epsilon<a$. Let $u$ be an upper bound of $\mathit{A}$.
Suppose that for all $\epsilon>0$ there is an $a\in\mathit{A}$ such that $u-\epsilon<a$.
Also suppose that $u$ is not the least upper bound of $\mathit{A}$.
Then there is a $\beta$ such that $\beta<u$ and $\beta$ is an upper bound.
Now let $\epsilon$ be $u-\beta$.
We know that $\epsilon$ is positive since we assumed $\beta$ is greater than $u$.
So we can replace $u-\beta$ with the inequality that we first assumed and write $\beta<a$.
But this is a contradiction since we assumed $\beta$ is an upper bound of $\mathit{A}$.
Therefore $u$ is the least upper bound of $\mathit{A}$ if for all $\epsilon>0$ there exists an $a\in\mathit{A}$ such that $u-\epsilon<a$.
Now lets prove the only-if-part, which is for all $\epsilon>0$ there exists an $a\in\mathit{A}$ such that $u-\epsilon<a$ if $u$ is the supremum of $\mathit{A}$.
We shall prove this statement by proof by contradiction. Let $u$ be an upper bound of $\mathit{A}$ Suppose that u is the supremum of $\mathit{A}$.
Also suppose that there exists an $\epsilon>0$ such that for all $a\in\mathit{A}$ we have $a<u-\epsilon$.
Then $u-\epsilon$ is also an upper bound of $\mathit{A}$.
Since $\epsilon>0$ it is obvious that $u-\epsilon<u$.
But this is a contradiction since we know that $u$ is the least upper bound of $\mathit{A}$ and $\mathit{A}$ can not have a smaller upper bound than $u$.
Therefore for all $\epsilon>0$ there exists an $a\in\mathit{A}$ such that $u-\epsilon<a$ if $u$ is the supremum of $\mathit{A}$.
Q.E.D
Is my proof attempt correct? I could not be sure about the second contradiction that I have found. Also is this proof suitable for formal mathematical writing? Thanks!
Best Answer
I guess that you're using the definition of supremum as the smallest upper bound.
Suppose $u$ is the supremum and take $\varepsilon>0$. Then $u-\varepsilon<u$, so $u-\varepsilon$ is not an upper bound of $A$, which implies that there exists $a\in A$ such that $a>u-\varepsilon$.
Suppose $u$ is not the supremum; then there exists an upper bound $v$ of $A$ such that $v<u$. Set $\varepsilon=u-v$. If $a\in A$, then $a\le v=u-\varepsilon$, so $u$ does not satisfy the condition.
About your proof: it's too long and full of repetitions, but essentially correct: the main ideas are there. Contradiction is not necessary: in the above, I proved the contrapositive.