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).
 A: Hint: consider the set of all upper bounds of $E$.
A: I know the OP already solved the problem, however it may be possible that other people will be redirected to this question and so I wanted to post an attempt to show the result using the property of the density of rational numbers on the set of real ones. Here goes my attempt (It's noteworthy to point out that I used the tip given by @Athar Abdul-Quader of considering the set of all upper bounds for the set $E$ that is bounded above):
Let $E$ be a nonempty set of real numbers that is bounded above. Let $B = \{x\in \mathbb{R}: \forall e\in E, x \geq e\}$ and $A=B^{c} = \{x\in \mathbb{R}:\exists e\in E, x < e \}$. Then $A$ and $B$ are disjoint sets such that $A\cup B = \mathbb{R}$ and $a<b$ whenever $a\in A$ and $b\in B$. By the Cut Property, there exists some $c\in \mathbb{R}$ such that $c\geq x$ for every $x\in A$ and $x\geq c$ for every $x\in B$.
Note that either $c\in A$ or $c\in B$. If $c\in A$, then there is some $y\in E$ such that $c<y$. However, by the density of rational numbers, there is some rational number $q$ such that $c<q<y$. Hence, $q\in A$, but this contradicts the fact that $c$ is greater or equal to every element in $A$. Therefore, $c\in B$ and so $c\geq e$  whenever $e\in E$ which implies that $c$ is an upper bound. Also, note that every element of $B$ is an upper bound of $E$ and $c\leq b$ whenever $b\in B$. Therefore, $c$ is the lowest upper bound of $E$.
Q.E.D
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