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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 $\sup A$.

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(necessary) Suppose $r\in \Bbb R$ such that $r<s$. Assume, by contradiction, there doesn't exist $x$ such that $r<x≤s$. So, for any $y\in A$, $y≤r$. In this case, $r$ would be an upper bound of $A$. That's not possible because $r<s$ and $s=\sup A$. Contradiction!

This is okay. This is direction $\implies$.

(sufficient) Let $r,s\in \Bbb R$ such that $s$ is an upper bound of $A$. Suppose that, if $r<x$ then there exists $x\in A$ such that $r<x≤s$. Once $x$ can be equal to $s$, $s$ just can be $\sup A$.

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.)

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There's a piece missing from your assumptions; according to the way you stated the problem, there could be points in your set $A$ above $s$. You need another assumption, e.g.$s$ is an upper bound for $A$.

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  • $\begingroup$ Sorry, I don't understand. I defined $s$ as the least upper bound of A. $\endgroup$ – Guilherme Duarte Mar 16 '15 at 22:53
  • $\begingroup$ You stated your question as if and only if. The first part (s is the sup) doesn't necessarily hold if just the second condition in terms of $r$ holds. You need another assumption, like $s$ is an upper bound. $\endgroup$ – user2566092 Mar 16 '15 at 22:55
  • $\begingroup$ Let me rephrase: the property that $[\forall r\lt s,\,\exists x\in A,\,r\lt x\leqslant s]$ does not characterize $s=\sup A$. For example, if $A=[0,1]$, every $s$ in $A$ satisfies this property. $\endgroup$ – Did Mar 16 '15 at 22:56
  • $\begingroup$ So is there a problem with the second part of the proof? Or the first? $\endgroup$ – Guilherme Duarte Mar 16 '15 at 23:01
  • $\begingroup$ @Did, So you're saying there's a problem with the second part of the proposition. Therefore, the proposition is a "if" and not "if and only if". Do i understand it well? $\endgroup$ – Guilherme Duarte Mar 16 '15 at 23:17

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