I am working on the following problem: If a set of integers $S \subset \mathbb{N}$ has a supremum, show that $\sup S \in S$.
My approach is as follows:
Let $s_0 = \sup S$ and suppose $s_0 \notin S$. It is a fact that for every $\epsilon > 0$ there exists $a \in S$ such that $ s_0 - \epsilon < a \leq s_0. $ However, since $s_0 \notin S$ and $a \in S$ we in fact have $ s_0 - \epsilon < a < s_0 $ But since $\epsilon > 0$ it follows that $ s_0 - \epsilon < a < s_0 + \epsilon \implies -\epsilon < a - s_0 < \epsilon \implies |a - s_0| < \epsilon $
Since this is true for every $\epsilon > 0$, $a - s_0 = 0 \implies a = s_0$, a contradiction. Therefore, $s_0 \in S$
My question then is whether this argument is valid. I believe it is correct but it is significantly different from the author's solution so I would like a second opinion.