Concerning the proof of an elementary theorem regarding $s = \sup A $.

Question

Is my argument to the proposition below correct?

Proposition. Assume $s\in\mathbf{R}$ is an upper bound for a set $A\subseteq R$. Then, $s = \sup A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a\in A$ satisfying $s − \epsilon < a$.

Proof. $(\Rightarrow)$. Assume $s = \sup A$ and let $\epsilon>0$, by definition $s$ is the smallest upper bound for $A$, so $s-\epsilon$ cannot be an upper bound for $A$ consequently for some $a\in A$ we have $s-\epsilon<a$.

$(\Rightarrow).$ For the converse assume on the contrary that $s-\epsilon<a,\forall\epsilon>0$ but $s\neq\sup A$, then $s>b$ where $b$ is some upper bound of $A$ but this implies that $s-b>0$ and thus for $\epsilon =s-b$ we have $s-\epsilon = b<a$ for some $a\in A$ contradicting the fact that $b$ is an upperbound for $A$.

$\blacksquare$

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Answer 1

It is correct, but I have two remarks:

1. The second $\Rightarrow$ should be a $\Leftarrow$ (this is a minor remark, of course).
2. There's a quantifier missing in the second proof. It should be $(\forall\varepsilon>0)\color{red}{(\exists a\in A)}:s-\varepsilon<a$.