Help with real analysis proof involving supremum Let $S\subseteq\Bbb R$ be nonempty. Prove that if a number $u \in \Bbb R$ has the properties:
(i) for every $n\in \Bbb N$ the number $u-1/n$ is not an upper bound of $S$, and
(ii) for every number $n\in \Bbb N$, the number $u+1/n$ is an upper bound of $S$,
then $u=\sup(S)$.
My attempt:
If $u-1/n$ is not an upper bound of $S$ for all $n\in N$, then there exists some value $s\in S$ such that $u-1/n<s$.
But $u-1/n<u$ and as $u-1/n$ is arbitrary, $u=\sup (S)$.
I know that this proof may be flawed. I need help on how to prove this. Any help would be appreciated.

Hint: The supremum of our set exists. Call it $b$. We want to show that $b=u$. 
Your analysis can be used to show that $b$ is between $u-\frac{1}{n}$ and $u+\frac{1}{n}$ for all $n$. To show that this forces equality of $u$ and $b$, suppose they are different. Let $\epsilon=|u-b|$. Use the fact that there is an $n$ such that $\frac{1}{n}\lt \epsilon/2$.  
