Let $u$ be an upper bound of non-empty set $A$ in $\mathbb{R}$. Prove that $u$ is the supremum of $A$ if and only if for all $\epsilon > 0$ there is an $a \in A$ such that $u-\epsilon < a$.
Lemma
If $s = \sup(A)$, then $$ \exists \, a \in A, \, \, \forall \epsilon > 0 \, \, \text{such that} \, \, |s - a| < \epsilon$$
Proof:
Suppose $\nexists \, a \in A $ such that $\forall \, \epsilon > 0$, $|\sup(A) - a| < \epsilon$. Then $\forall \, a \in A, \, \exists \, \epsilon_0> 0$ such that $|\sup(A) - a| > \epsilon_0$. This implies that $\sup(A) > a + \epsilon_0$. Thus, $\exists\, s' \not\in A$ such that $\sup(A) > s' > a + \epsilon_1$ if we take $\epsilon_1 = \frac{\epsilon_0}{2}$. But we cannot have $s' < \sup(A) : s' \not\in A$ so we have arrived at a contradiction and thus the lemma is proved.
Proof:
$(i \implies ii)$ if $u = \sup(A)$, then, by the lemma, $\forall \, \epsilon > 0$, \, $\exists \, a \in A$ such that $|u-a|< \epsilon$. This implies $u-\epsilon < a$.
$(ii \implies i)$ If $\forall \, \epsilon > 0, \, \exists \, a \in A$ such that $u - \epsilon < a$, then $u-a < \epsilon$ and by the lemma, $u =\sup(A)$.
Is this a complete and correct proof?