Prove $s$ is a supremum iff for all $\epsilon>0$ there exists $a\in A$ such that $a>s-\epsilon$. Let $A\subseteq\Bbb{R}$ is a nonempty set and $s\in \Bbb{R}$ is an upper bound. Prove $s$ is the supremum iff for all $\epsilon>0$ there exists $a\in A$ such that $a>s-\epsilon$.
This is important to me to know I am correct in my attempt, because I always grant more when I do try.
Attempt: Suppose for all $\epsilon>0$ there exists $a\in A$ such that $a>s-\epsilon$. Assume $s$ is not the supremum. Then, since the supremum exists, let $s_1$ be the supremum. By the density of the real number,there exists $c$ such that $s_1<c<s$. Let us pick $\epsilon_1$ such that $s=c+\epsilon_1$ $\color{red}\Rightarrow c=s-\epsilon_1$ $\color{red}{\Rightarrow} s_1<s-\epsilon_1<s$ $\color{red}{\Rightarrow}$ $s>s_1+\epsilon_1$. Since for $\epsilon_1$ there exists $a\in A$ such that $a>s-\epsilon_1$. In particular $a>s_1+\epsilon_1-\epsilon_1$ meaning $a>s_1$. A contradiction. Therefore, $s$ is the supremum.
Now, suppose, $s$ is the supremum. Assume there exists $\epsilon$ such that for all $a\in A$, $a<s-\epsilon$. That would mean that $s-\epsilon$ is an upper bound that is smaller than $s$. A contradiction. Therefore, for all $\epsilon>0$ there exists $a\in A$ such that $a>s-\epsilon$.
 A: Looks great! There's a couple things that I might improve on to make some of the steps clearer. As a matter of personal taste, I would be be a bit more explicit about what exactly is being contradicted and I would try avoiding implications (preferring instead to chain them together when possible). Here's my version.

Given any nonempty $A \subseteq \mathbb R$, suppose that $s \in \mathbb R$ is an upper bound for $A$ so that $s \geq a'$ for all $a' \in A$ and (by the completeness property of $\mathbb R$) $A$ has some supremum, say $s' \in \mathbb R$. We will show that $s = s'$ iff for every $\epsilon > 0$, there exists some $a \in A$ such that $a > s - \epsilon$.


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*$(\Longrightarrow)$: Suppose that $s = s'$. Now suppose, towards a contradiction, that there exists some $\epsilon > 0$ such that for any $a \in A$, we have that $a \leq s - \epsilon$. Then $s - \epsilon$ is an upper bound for $A$. But since $s - \epsilon < s'$, this contradicts the minimality of $s'$, as $s'$ is supposed to be the least such upper bound. So for every $\epsilon > 0$, there exists some $a \in A$ such that $a > s - \epsilon$, as desired.

*$(\Longleftarrow)$: Suppose that for every $\epsilon > 0$, there exists some $a \in A$ such that $a > s - \epsilon$. Now suppose, towards a contradiction, that $s \neq s'$. Then since $s'$ is the least upper bound, we know that $s' < s$. Thus, by the density of $\mathbb R$, we know that there exists some $c \in \mathbb R$ such that $s' < c < s$. Now for $\epsilon = s - c > 0$, we know by hypothesis that there exists some $a \in A$ such that $a > s - \epsilon = s - (s - c) = c > s'$. Hence, $s'$ was not actually an upper bound for $A$, contradicting the definition of a supremum. So $s = s'$, as desired. $~~\blacksquare$
