Proof: Let .We need to show that . Now suppose that , then we have two cases:
Case1: . We know that and . If then since the set is written in nondescending order, we know that . If we add the inequalities together, we get . Now subtract 1 from both sides preserving the inequality to obtain which is a contradiction.
Case2: . Similarly, we have , so we get , and subtracting 1 from both sides we get ,which is again a contradiction.
Therefore, which proves that is injective. Since is both injective and onto, it is a bijection. QED
Please critique this proof. Thanks