Just wanted to check my proof for Lemma 1.3.7. from Abbott's Understanding Analysis, which reads as follows:
Lemma 1.3.7. Assume $s \in R$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s=sup A$ if and only if for every choice of $\epsilon > 0$, there exists an element $a \in A$ such that $s - \epsilon < a$.
Proof:
($\rightarrow$): Assume $s=supA$ .Then take $s- \epsilon$ for some $\epsilon > 0$. Since $s - \epsilon < s, s - \epsilon$ must not be an upper bound, from definition of supremum. So then there must be some element $a \in A$ such that $a = s - \epsilon$ for the chosen $\epsilon$.
($\leftarrow$): Assume $s$ is an upper bound such that for all $\epsilon > 0$, there is some $a \in A$ such that $s - \epsilon < a$. Let $b$ be an upper bound for $A$. Then either $b < s$, or $b \geq s$. If $b < s$, then it follows that $s - \epsilon = b$ for some $\epsilon > 0$. So $b$ must not be an upper bound for $A$, but we assumed that it is, so this is a contradiction. So $b \geq s$. So $s$ satisfies the definition of supremum, so $s = supA$.
So the result is true.