Is the following argument correct? All cited results have been reproduced below.
Proposition. Let $A\subseteq R$ be nonempty and bounded above, and let $s\in \mathbf{R}$ have the property that for all $n\in N$, $s + \frac{1}{n}$ is an upper bound for $A$ and $s − \frac{1}{n}$ is not an upper bound for $A$, then $s = \sup A$.
Proof. Let $\epsilon>0$, then by $1.4.2$ there exists an $n\in\mathbf{N}$ such that $\frac{1}{n}<\epsilon$, by definition $\sup A$ is the least upper bound of $A$ and by hypothesis $s+\frac{1}{n}$ is an upper bound of $A$ implying $\sup A\leq s+\frac{1}{n}$.
In addition since $s-\frac{1}{n}$ is not an upper bound of $A$ then $s-\frac{1}{n}<a$ for some $a\in A$ but $a\leq\sup A$ thus $s-\frac{1}{n}<\sup A$ and by extension $$s-\frac{1}{n}<\sup A\leq s+\frac{1}{n}$$ Now since $\frac{1}{n}<\epsilon$ and therefore $-\epsilon<-\frac{1}{n}$ it follows that $s+\frac{1}{n}<s+\epsilon$ and $s-\epsilon<s-\frac{1}{n}$ which in conjunction with the above proposition implies
$$s-\epsilon<\sup A<s+\epsilon$$ equivalently $|\sup A-s|<\epsilon$.
Since our choice of $\epsilon\in\mathbf{R}^+$ was arbitrary it follows that $|\sup A-s|<\epsilon,\forall\epsilon>0$, appealing to $\textbf{1.2.6}$ then yields the required result.
$\blacksquare$
Theorem $1.2.6.$ Two real numbers $a$ and $b$ are equal if and only if for every real number $\epsilon > 0$ it follows that $|a − b| <\epsilon$.
Theorem $1.4.2$ (Archimedean Property).
- Given any number $x\in R$, there exists an $n\in N$ satisfying $n > x$.
- Given any real number $y > 0$, there exists an $n\in N$ satisfying $\frac{1}{n} < y$.