I am reading a book on Real Analysis which states as the axiom of completeness the following statement:
"We define two non empty subsets $A$ and $B$ of $\mathbb{R}$ to be separated if $a\leq b\ \forall a\in A,\ \forall b\in B$.
Axiom of Completeness: for every pair $A,B$ of non emtpy and separated subsets of $\mathbb{R}$ there exists at least one separating element, i.e. a real number $\xi$ such that $a\leq\xi\leq b\ \forall a\in A, \forall b\in B$."
Now, define a section of $\mathbb{R}$ to be a pair $(A,B)$ of separated subsets of $\mathbb{R}$ such that $A\cup B=\mathbb{R}$.
I would like to show the axiom of completeness as stated above is equivalent to the following statement:
(S2) "for every section $(A,B)$ of $\mathbb{R}$ there exists a unique separating element between $A$ and $B$"
I have provided a proof below and I would appreciate if someone would check it out and give me some feedback and/or point out other (perhaps better) ways to prove this. Thanks.
Proof. $\fbox{$\text{S2}\Rightarrow\text{AoC}$}$ Let $A,B$ be two separate non-empty subsets of $\mathbb{R}$; if $A\cup B=\mathbb{R}$ (i.e. $(A,B)$ is a section of $\mathbb{R}$) we immediately have that there is only one separating element between $A$ and $B$ and therefore it is true that there is at least one and we therefore have the statement of the completeness axiom. If $A\cup B\neq\mathbb{R}$ we consider $$B^+=\{x\in\mathbb{R}\setminus (A\cup B):x\geq a\ \forall a\in A \},$$ $\bar{B}=B\cup B^+$, $$A^-=\{x\in\mathbb{R}\setminus (A\cup \bar{B}):x\leq b\ \forall b\in \bar{B}\}$$ and $\bar{A}=A^-\cup A$. Then $(\bar{A},\bar{B})$ is a section of $\mathbb{R}$ ($^*$), so there is only one separator element $c$ between $\bar{A} $ and $\bar{B}$, therefore $a\leq c\leq b\ \forall a\in\bar{A},\ \forall b\in\bar{B}$ and this implies that, in particular , $a\leq c\leq b\ \forall a\in A,\ \forall b\in B$ therefore there is (at least) a separator element also for $A$ and $B$ ($c$ in this case) , which is exactly what the completeness axiom of $\mathbb{R}$ states.
($^*$ obviously $\bar{A}\cup\bar{B}\subset\mathbb{R}$ being $\bar{A}$ and $\bar{B}$ both subsets of $\mathbb{ R}$. Now let $x\in\mathbb{R}$. If $x\in A\cup B$ then $x\in \bar{A}\cup\bar{B}$; if $x\notin (A\cup B)$ then there are two possibilities: $x\geq a\ \forall a\in A$ or $\exists a\in A: x<a$ In the first case $x\in B^ +\subset\bar{A}\cup\bar{B}$ and in the second $x\notin B^+$ (if $x\in B^+$ we would have a contradiction since it must be $x\geq a\ \forall a\in A$) therefore $x\notin A\cup\bar{B}$ and being $a\leq b\ \forall b\in B$ since $A$ and $B$ are separated by hypothesis and $a \leq b\ \forall b\in B^+$ by definition of $B^+$ we have $x\leq b\ \forall b\in\bar{B}$ therefore $x\in A^-\subset \bar{A}\cup\bar{B}.$ In both cases we have that $x\in \bar{A}\cup\bar{B}$ therefore we conclude that $\mathbb{R}\subset (\bar{A}\cup\bar{B})$ and then $\bar{A}\cup\bar{B}=\mathbb{R}.$
Let $y\in\bar{A}$: then $y\in A$ or $y\in A^-$; if $y\in A$ then $y\leq b\ \forall b\in B$ $A$ and $B$ being separated by hypothesis and $y\leq b\ \forall b\in B^+$ by definition of $B^+$ therefore $y\leq b\ \forall b\in\bar{B}$. If $y\in A^-$ then $y\leq b\ \forall b\in\bar{B}$ by definition of $A^-$. From the arbitrariness of $y\in\bar{A}$ it follows that $a\leq b\ \forall a\in\bar{A},\ \forall b\in\bar{B}$, therefore $\bar {A}$ and $\bar{B}$ are also separate and thus a section.)
$\fbox{$\text{S2}\Leftarrow\text{AoC}$}$ Let $A$ and $B$ be two subsets of $\mathbb{R}$ such that $(A,B)$ is a section of $\mathbb{R}$. By the completeness axiom there exists $\xi\in\mathbb{R}$ such that $a\leq\xi\leq b\ \forall a\in A,\ \forall b\in B$. Suppose there is another separator element $c$ between $A$ and $B$, other than $\xi$. By trichotomy there are only two possibilities: either $\xi<c$ or $\xi>c$. Suppose $\xi<c$ and consider the number $d=\frac{c+\xi}{2}$; $a\leq \xi<d<c\ \forall a\in A$ so $d\notin A$ so it must be $d\in B$ (since $A\cup B=\mathbb{R}$) , which is a contradiction because then it must be $c\leq d$, $c$ being by hypothesis a separator element of $(A,B)$. Suppose then that $\xi>c$ and consider the number $e=\frac{c+\xi}{2}$; $c<e<\xi\leq b\ \forall b\in B$ then $e\in A$, which is a contradiction because then it must be $e\leq c$, $c$ being by hypothesis a separator element of $(A,B)$. Therefore there is no other separator element and this proves that if we assume the completeness axiom, then for each section $(A,B)$ of $\mathbb{R}$ there is a unique separator element. $\square$