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.
 A: You need to use both pieces of information.
The first condition tells you that $u$ cannot be decreased and still be an upper bound. But something that was not an upper bound to begin with, if you decrease it by $1/n$ would not be an upper bound. 
The second condition tells you that increasing $u$ by arbitrarily small amounts is an upper bound. Any crude upper bound will have this property
The two combined should tell that $u$ is the smallest upper bound, i.e. the supremum.
First we prove that $u$ is an upper bound. If $x\in S$ and $x>u$ then take $n>1/(x-u)$. This gives us $u+1/n<x$, which contradicts the second assumption.
Assume that $r<u$ is an upper bound for $S$. Take $n>1/(u-r)$. Then, $r<u-1/n$. But we are told there are $x\in S$ with $u-1/n<x$. Then $r<u-1/n<x$. Therefore $r$ could not be an upper bound of $S$. 
Therefore $u$ is the smallest upper bound of $S$. It is its supremum.
A: 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 $\frac{1}{n}\lt \epsilon/2$.  
