[Math] Use the Axiom of Completeness to prove the Cut Property

real numbersreal-analysis

I know that the Axiom of Completeness states

Every nonempty set of real numbers that is bounded above has a least upper bound.

Where a least upper bound, $s$, is defined as follow:
A real number $s$ is the least upper bound for a set $A\subseteq \mathbb{R}$ if it meets the following two criteria:

  1. $s$ is an upper bound for $A$;
  2. if $b$ is any upper bound for $A$, then $s\leq b$.

And I'm supplied with the following property (Cut Property):

If $A$ and $B$ 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 $x\in B$.

I am asked to:

  1. Use the Axiom of Completeness to prove the Cut property; and
  2. Use Cut Property to prove the Axiom of Completeness.

This seems to be right under my nose, but I'm not exactly sure what to assume for $(1)$.

Could anyone give me a push?

Best Answer

Suppose you have the axiom of completeness and assume you have $A$ and $B$ as in the statement of the cut property. Then, as $B$ is nonempty, $A$ has an upper bound. Let $c$ be the least upper bound for $A$.

  1. For $a\in A$, $a\le c$, because $c$ is an upper bound for $A$;
  2. For $b\in B$, $c\le b$, because $b$ is an upper bound for $A$ and $c$ is the least upper bound

Suppose you know the cut property. Consider a nonempty set $C$ with an upper bound. Then let $$ A=\{x\in\mathbb{R}: x<c\text{, for some $c\in C$}\} $$ and let $B$ be the complement of $A$. Since $C$ has an upper bound $b$, we have $b\notin A$, so $B$ is nonempty as well as $A$. The union of $A$ and $B$ is $\mathbb{R}$ by construction. Suppose $a\in A$ and $b\in B$. If $b\le a$, we have $b<c$ for some $c\in C$, so $b\in A$: a contradiction.

Now, the cut property provides $d$ so $a\le d$, for every $a\in A$, and $d\le b$, for every $b\in B$. Can you prove that $d$ is the least upper bound of $C$?

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