Abbott's book states that $\mathbb{R}$ possess the following property.
Let $A,B\subset\mathbb{R}$ be nonempty sets such that $A\cap B=\emptyset$, $A\cup B=\mathbb{R}$ and $a<b$ for all $a\in A$ and $b\in B$. Then, there exists $c\in\mathbb{R}$ such that $x\leq c$ if $x\in A$ and $x\geq c$ if $x\in B$.
An exercise asks to prove that the axiom of completeness (that is to say that all sets bounded above have a least upper bound) implies the above property (the Cut Property of $\mathbb{R}$) and then the other way around. While I understand the forward direction, there seems to be several holes in my reverse argument.
Here is my attempt:
Let $A\subset\mathbb{R}$ be a nonempty set that is bounded above, and assume that $\mathbb{R}$ has the cut property. Consider the set $B=\lbrace b\in\mathbb{R} | b\geq a \forall a\in A \rbrace$, of all upper bounds to $A$. We would like $A\cup B$ to fill $\mathbb{R}$. To accomplish this, define a set $A'=B^C=\bigcup\limits_{a\in A}(-\infty, a)$. We have that $A'\cup B=\mathbb{R}$, thus the cut property applies to these two sets. There exists a number $c\in\mathbb{R}$ such that $a\leq c$ for all $a\in A'$, and $b\geq c$ for all $b\in B$. As $B$ is the set of all upper bounds to $A$, it follows that $c=\sup(A)$.
Several points of the proof that I am unsure of:
- Is the construction of $A'$ appropriate here?
- I feel that I draw a conclusion too quickly, without enough rigor. Where am I skipping steps?