Could someone verify my proof?
Definition: Suppose $s \in \mathbb{R}$ and upper bounded $A \subset \mathbb{R}$. For any $x \in A$, we have $x \leq s$. For any $v$ such that $x \leq v$ for any $x$, we have $s \leq v$. We say $s$ is $\sup A$.
Proposition: Let $A \subset \mathbb{R}$, not-empty and upper bounded by $s \in \mathbb{R}$. Prove that $s = supA$ iff , if $r < s$, then there exists $x \in A$ such that $r < x \leq s$.
Proof:
(necessary) Suppose $r \in \mathbb{R}$ such that $r < s$.
Assume, by contradiction, there doesn't exist $x$ such that$r < x \leq s$. So, for any $y \in A$, $y \leq r$. In this case, $r$ would be an upper bound of A. That's not possible because $r < s$ and $s = \sup A$. Contradiction!
(sufficient) Let $r, s \in \mathbb{R}$ such that $s$ is an upper bound of A. Suppose that, if $r < x$ then there exists $x \in A$ such that $r < x \leq s$. Once $x$ can be equal to $s$, $s$ just can be $\sup A$.
Best Answer
This is okay. This is direction $\implies$.
This is not good. You don't take $r$ along with $s$ intially. Seems to me that you are starting with the hypothesis, and not going exactly at the definition. Also, since we are in the beginning of the theory here, I wouldn't feel comfortable accepting a "$s$ just can be $\sup A$" as the end of the argument.
You have to prove that $s = \sup A$. The first thing to check is that $s$ is an upper bound for $A$. There is nothing to do, since this is a global hypothesis. Now, we have to check that for every upper bound $v$ of $A$, $s \leq v$. Suppose not. Suppose that $v_0$ is an upper bound for $A$ s.t. $v_0 < s$. Then, by hypothesis, exists $x \in A$ such that $v_0 < x \leq s$, contradicting that $v_0$ is an upper bound for $A$ (here I took $v_0$ as the $r$ in the proposition.)