Proving the supremum of a set Let A be a nonempty subset of $\mathbb R$ and let m be an upper bound of A. Prove that $m = supA$ iff $ \forall n \in \mathbb N, \exists a \in A, m < a + \frac 1n$. 
I don't really have any idea where to start with this one. I've tried the following:
Let $x<m$. Then, $m-x>0$. Since $m-x$ is positive, there exists $n \in \mathbb N$ such that $m-x > \frac 1n$. So, $ m > x + \frac 1n$.
But, this is the opposite of what I need to prove.
Any hints on where to start or the right path to take would be greatly appreciated, thanks!
 A: a)  Let $m= \sup A$ and $n \in \mathbb N$. Then $m - \frac{1}{n}<m$, hence $m - \frac{1}{n}$ is not an upper bound of $A$, thus there is $a \in A$ such that $a > m - \frac{1}{n}$ .
b) Suppose that we have  $\forall n \in \mathbb N, \exists a \in A: m < a + \frac 1n$.
Let $s= \sup A$ and assume, to the contrary,  that $s \ne m.$ Then $s<m$, hence $m-s>0$. There is $n \in \mathbb N$ such that $\frac1n<m-s$.
There is $a \in A $ with $m<a+\frac1n.$ This gives $a>m- \frac1n>s,$ a contradiction.
A: If $m=\sup A$ then $m-\frac1n$ is not an upper bound of $A$.
So some $a\in A$ exists with $m-\frac1n<a$ or equivalently $m\leq a+\frac1n$.
Conversely assume that $b$ is an upper bound of $A$ with $b<m$.
Then $0<\frac1n<m-b$ for $n$ large enough and for this $n$ we can find an $a\in A$ with $$m<a+\frac1n<a+m-b\leq m$$
The last inequality on base of $a\leq b$, since $b$ is an upper bound of $A$.
So a contradiction is found and we conclude that no upper bounds exist that are "smaller" than $m$.
This means that $m$ is the least upper bound of $A$.
