# Dedekind Property

Theorem: If $A$ and $B$ are two non-empty subsets of $\mathbb{R}$ such that (i) $A \cup B = \mathbb{R}$, (ii) $x \in A \wedge y \in B \Rightarrow x < y$, then either $A$ has the greatest member or $B$ has the least member.

Proof: By (ii), the non-empty set $A$ is bounded above. If $A$ has the greatest member, it establishes the theorem. If $A$ has no greatest member, then by the completeness axiom, in $\mathbb{R}$, $B$ being the set of upper bounds of $A$, it has a least member. Hence the theorem is proved.

Theorem: Dedekind’s Property is equivalent to the completeness axiom in $\mathbb{R}$.

Proof: In the previous theorem it is shown that the completeness axiom in $\mathbb{R}$ implies Dedekind’s property. It remains to show that Dedekind’s Property implies the completeness axiom.

Let $S$ be a non-empty set bounded above and sets $A$ and $B$ be defined by:

It evidently follows that $A$ and $B$ are non-empty, disjointed and (i) $A \cup B = \mathbb{R}$, (ii) $x \in A \wedge y \in B \Rightarrow x < y$. Therefore, by Dedekind’s Property either $A$ has the greatest member or $B$ has the least member. If possible, suppose $A$ has the greatest member, say $\alpha$. Then, $\alpha \in A \Rightarrow \alpha \notin B{\text{ }}\exists {\text{ }}a \in S$ such that $\alpha < a$, since $\alpha < \frac{{a + \alpha }}{2} \in B,{\text{ }}\frac{{a + \alpha }}{2}$ is an upper bound of $S$. This contradiction leads to the fact that $A$ has no greatest member. And so, $B$ has the least member. Hence, the set of upper bounds of a non-empty set $S$ bounded above has a least member, which is the completeness axiom in $\mathbb{R}$. Hence the theorem is proved.

Theorem: Between any two distinct real numbers there exist infinitely many rational numbers.

Theorem: Between any two distinct real numbers there exist infinitely many irrational numbers.