# Use Cut Property to prove Axiom of Completeness

I've already proved the Cut Property (CP) using the Axiom of Completeness (AoC), but want to now show the converse. Here are my book's definitions:

Cut Property: If $A, B$ are nonempty, disjoint sets with $A \cup B = \mathbb{R}$ and $a < b$ for all $a \in A$ and $b \in B$, then there exists $c \in \mathbb{R}$ such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$.

Axiom of Completeness: If $A$ is nonempty and bounded above, then $\sup A$ exists.

My attempt: Let $E$ be nonempty and bounded above. Then there exists some $b$ such that $e < b$ for all $e \in E$... From this point I tried to construct $A$ and $B$ as in the assumption in the Cut Property so that I can invoke it to pinpoint $c = \sup E$. So I tried "expanding" $E$ to go to left to $-\infty$ and construct $B = [b,\infty)$. However, there are possibly gaps between $E$ and $B$ that I don't know how to precisely allocate to $E$ and $B$ without invoking $\sup E$ (which I'm trying to show exists).

Hint: consider the set of all upper bounds of $E$.