Theorem: If  and  are two non-empty subsets of  such that (i) , (ii) , then either has the greatest member or has the least member.

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

Theorem: Dedekind’s Property is equivalent to completeness axiom in.

Proof: In the pervious theorem it is shown that the completeness axiom in implies Dedekind’s property. It remains to show that the Dedekind’s property implies the completeness axiom.
            Let  be a non-empty set bounded above and sets and be defined by
                       
                       
            It evidently follows that and are non-empty, disjoint and (i) , (ii) . Therefore, by Dedekind’s property either has the greatest member or has the least member. If possible, suppose has then greatest member, say. Then,  such that , since  is an upper bound of. This contradiction leads to the fact that has no greatest member. And so, has the least member. Hence, the set of upper bounds of a non-empty set bounded above has a least member, which is the completeness axiom in. Hence the theorem.

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.