In his book Calculus Vol. 1, Apostol writes the following proof:
Theorem I.32. Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > \sup S - h$$
Theorem I.33 Additive Property: Given nonempty subsets $A$ and $B$ of $\mathbb R$, let $C$ denote the set$$C = \left\{a + b \mid a \in A, b \in B\right\}.$$
(a) If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and $$\sup C = \sup A + \sup B.$$
Proof. Assume each of $A$ and $B$ has a supremum. If $c \in C$, then $c = a+b$, where $a \in A$ and $b \in B$. Therefore, $c \le \sup A +\sup B$; so $\sup A + \sup B$ is an upper bound for $C$. This shows that $C$ has a supremum and that $$\sup C \le \sup A + \sup B.$$
Now let $n$ be any positive integer. By Theorem I.32 (with $h = \frac{1}{n}$) there is an $a$ in $A$ and a $b$ in $B$ such that $$a > \sup A - \frac1n, \quad b > \sup B - \frac1n.$$
Adding those inequalities, we obtain $$a + b > \sup A + \sup B - \frac2n, \quad or \quad \sup A + \sup B < a + b + \frac2n \le \sup C + \frac2n,$$
since $a + b \le \sup C$. Therefore we have shown that $$\sup C \le \sup A + \sup B < \sup C + \frac2n$$
for every integer $n \ge 1$. By Theorem I.31, we must have $\sup C = \sup A + \sup B.$ This proves (a).
Theorem I.31 is Theorem 1.2 on the question I asked here:
Is my proof of "Theorem 1.2" correct?
My problem is, I don't understand the steps from adding the inequalities and onwards. Any insight you could provide would be most helpful.
