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?
Best Answer
You can't base the proof of the statement on an obviously false lemma (at least with an incorrect formulation).
The proof is much simpler than that.
Assume that $u$ is an upper bound for $A$ with the property that, for every $\varepsilon>0$, there exists $a\in A$ with $u-\varepsilon<a$.
By definition, the supremum is the least upper bound. So if $u$ is not the supremum, there is an upper bound $v$ of $A$ with $v<u$.
Now take $\varepsilon=(u-v)/2$. By assumption there exists $a\in A$ with $u-\varepsilon<a$; this becomes $$ u-\frac{u-v}{2}<a $$ that is, $$ \frac{u+v}{2}<a $$ which is a contradiction, because from $a\le u$ and $a\le v$ we obtain $$ a\le\frac{u+v}{2} $$
For the converse, suppose $u=\sup(A)$ and let $\varepsilon>0$. Then $u-\varepsilon<u$, so $u-\varepsilon$ is not an upper bound for $A$, which means there exists $a\in A$ with $u-\varepsilon<a$.