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 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 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 the 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, disjoint 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 then 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.