$s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$. Show $s = \sup A.$ 
Let $A \subset \mathbb R$ be non empty and bounded above, and let $s \in \mathbb R$ have the property that for all $n \in \mathbb N$, $s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$. Show $s = \sup A.$

I am having difficulty proving this statement. It is intuitively clear to me that it holds true but have no idea where to begin proving this. Thanks in advance for any help.

Since $s+1/n$ is an upper bound for $A$ we have $\sup A\leq s+1/n$ and further $s-1/n$ is not an upper bound so that there is an $a\in A$ which exceeds it and thus $s-1/n<a\leq \sup A$. It follows that we have $$s-\frac{1}{n}<\sup A\leq s+\frac{1}{n}$$ for all $n\in \mathbb {N} $. It is now almost obvious by Squeeze Theorem that $s=\sup A$. 
