# 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:

\[A = \left\{ {x:x{\text{ is not an upper bound of }}S} \right\}\]

\[B = \left\{ {y:y{\text{ is an upper bound of }}S} \right\}\]

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.