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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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  • $\begingroup$ I dont understand: How does $a \ge b$ and $c \le a$ imply $b = c$? You don't say where $\tilde{s}, s,x$ come from so this is not true in general, i.e. consider $a = 4$, $b = 2$ and $c = 3$. $\endgroup$ – Viktor Glombik May 5 at 12:16
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(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 $A$.

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  • $\begingroup$ "(ii) $\Rightarrow$ (iii)": How is that a proof? Does $x>s-\varepsilon$ imply $x\geqslant s-\varepsilon$, and why? $\endgroup$ – Analysis May 5 at 12:23
  • $\begingroup$ Since $>$ means “larger than” and $\geqslant$ means “larger than or equalt to”, it is obvious that $a>b\implies a\geqslant b$. $\endgroup$ – José Carlos Santos May 5 at 12:26
  • $\begingroup$ Ahh, you're right...! $\endgroup$ – Analysis May 5 at 12:32
  • $\begingroup$ But why do you use $\geqslant$ instead of $\geq$? $\endgroup$ – Analysis May 5 at 12:39
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    $\begingroup$ Since $s'<s$, $\frac{s-s'}2>0$. So, I can take $\varepsilon=\frac{s-s'}2$. And I did take that $\varepsilon$ because, with it, I was able to reach a contradiction. $\endgroup$ – José Carlos Santos May 5 at 13:09

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