$s + \frac{1}{n}$ 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.
 A: For all $a\in A$, $a\leq s+\frac1n$ for all $n$. Then $a\leq s$ (if it weren't, if you had $s<a$, you would be able to squeeze, for $n$ big enough, $s<s+\frac1n<a$, contradicting that $s+\frac1n$ is an upper bound for $A$). 
So $s$ is an upper bound for $A$. Given any $s'<s$, you can choose $n$ big enough so that $s'<s-\frac1n<s$. As $s-\frac1n$ is not an upper bound for $A$, there exists $a\in A$ such that $s-\frac1n<a$. Thus
$$
s'<s-\frac1n<a,
$$
showing that $s'$ is not an upper bound for $A$. In consequence, $s$ is the least upper bound of $A$: $s=\sup A$. 
A: Pure definitions.
The sup must exist as the reals have the least upper bound property.
$s - \frac 1n$ is not an upper bound for all $n$ so $\sup A > s + \frac 1n$ or all natural $n$.
All $s + \frac 1n$ is an upper bound so $\sup A \le s+ \frac 1n$ for all natural $n$.
So $s-\frac 1n < \sup A \le s+ \frac 1n$ for all natural $n$.
Now either $\sup A < s$ or $\sup A = s$ or $\sup A > s$.
If $\sup A < s$ then $s - \sup A > 0$ and there is an $n$ in $\mathbb N$ so that $\frac 1n < s - \sup A$ an $s < s - \frac 1n$ which is a contradiction.
If $\sup A > s$ then $\sup A - s > 0$ and there is an $n$ so that $\frac 1n < \sup A -s$ and $s > \sup a+\frac 1n$ which is a contradiction.
So $\sup A = s$.
Of course you may need to review why it is true that for all $x > 0$ theren is a natural $n$ so that $0 < \frac 1n < x$....
A: 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 $s=\sup A$. 
