I am required to prove that the Axiom of Completeness is equivalent to the Cut Property of Real Numbers.
The Following in my attempt so far
Proof. $(\Rightarrow)$. By hypothesis $A,B$ are non-empty sets bounded above and below respectively. Now define the set $H = \{h\in\mathbf{R}\ |\ h\ \text{ is a lower bound for }B\}$ and note that $A\subseteq H$. The axiom of completeness assures us that $\sup B$ exists and by exercise $\textbf{1.3.3}$ it is apparent that $\sup H = \inf B$. Now Let $c = \sup H = \inf B$ and assume that $a\in A$ and $b\in B$ respectively, then $b\geq c$ and since $a\in H$ it follows that $a\leq c$.
$(\Leftarrow).$ Now let $E\subseteq\mathbf{R}$ such that $E\neq\varnothing$ and is bounded above $\dots\dots$
I dont know where should i start for the converse in particular i am not sure how to partition $\mathbf{R}$ given the set $E$. Any hints and only hints would be appreciated.
The full statement of the cut property and exercise $\textbf{1.3.3}$ is as follows.
If $A$ and $B$ are nonempty, disjoint sets with $A\cup B = \mathbf{R}$ and $a < b$ for all $a\in A$ and $b\in B$, then there exists $c\in\mathbf{R}$ such that $x\leq c$ whenever $x\in A$ and $x\ge c$ whenever $x\in B$.
$\textbf{1.3.3}$ Let $A$ be non-empty and bounded below, and define $B = \{b\in\mathbf{R}|b\text{ is a lower bound for A}\}$, then $\sup B = \inf A$.