Let $A\subset \mathbb{K}$, $s\in \mathbb{K}$. ($\mathbb{K}$ is an ordered field!). Show the equivalence of the following statements:
(i) $s=\sup A$
(ii) $s$ is an upper bound of $A$ and for all $\epsilon > 0$ exists a $x\in A$ with $x>s-\epsilon$
(iii) $s$ is an upper bound of $A$ and for all $\epsilon > 0$ exists a $x\in A$ with $x\geq s-\epsilon$
I wanted to show the the equivalence like that: $(i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (i)$ (don't know what it's called in English).
Definitions:
$s \in \mathbb{K}$ is an upper bound of $A$ if $\tilde{s}\geq x$ for all $x\in A$.
An upper bound is called Supremum of $A$ if $s\leq \tilde{s}$ for all upper bounds $\tilde{s}$ of $A$.
My attempt:
(i) $\Rightarrow$ (ii)
Since, by definition, $\tilde{s}\geq x \land s\leq \tilde{s}$, one can deduce that $s=x$. It should now be trivial to show that $x>s-\epsilon$ if $\epsilon>0$
(ii) $\Rightarrow$ (iii)
Since we have already proven that there exists a $x\in A$ with $x> s-\epsilon$, we solely have to prove that there are $x\in A$ with $x=s-\epsilon$. But I don't know how to.
(iii) $\Rightarrow$ (i)
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Best Answer
(i) $\implies$ (ii): It is clear that, if $s=\sup A$, then $s$ is an upper bound of $A$. Take $\varepsilon>0$. Since $s$ is the least upper bound of $A$, $s-\varepsilon$ is not an upper bound of $A$, which means that there is a $x\in A$ sych that $x\geqslant s-\varepsilon$.
(ii) $\implies$ (iii): There is some $x\in A$ such that $x>s-\varepsilon$. But then $x\geqslant s-\varepsilon$.
(iii) $\implies$ (i): Suppose that $s\neq\sup S$. Then here is some upper bound $s'$ of $A$ such that $s'<s$. Take $\varepsilon=\frac{s-s'}2$. There is some $x\in A$ such that $x\geqslant s-\varepsilon=s-\frac{s-s'}2=\frac{s+s'}2>s'$. This is absurd, since $s'$ is an upper bound of $A$.