Well-ordering principle

In mathematics, the well-ordering principle states that every non-empty subset of nonnegative integers contains a least element. In other words, the set of nonnegative integers is well-ordered by its "natural" or "magnitude" order in which \(x\) precedes \(y\) if and only if \(y\) is either \(x\) or the sum of \(x\) and some nonnegative integer (other orderings include the ordering \(2, 4, 6, ...\); and \(1, 3, 5, ...\)).

The phrase "well-ordering principle" is sometimes taken to be synonymous with the "well-ordering theorem", according to which every set can be well-ordered. On other occasions it is understood to be the proposition that the set of integers \(\{\ldots, -2, -1, 0, 1, 2, 3, \ldots \}\) contains a well-ordered subset, called the natural numbers, in which every nonempty subset contains a least element.

Depending on the framework in which the natural numbers are introduced, this (second-order) property of the set of natural numbers is either an axiom or a provable theorem. For example:

• In Peano arithmetic, second-order arithmetic and related systems, and indeed in most (not necessarily formal) mathematical treatments of the well-ordering principle, the principle is derived from the principle of mathematical induction, which is itself taken as basic.
• Considering the natural numbers as a subset of the real numbers, and assuming that we know already that the real numbers are complete (again, either as an axiom or a theorem about the real number system), i.e., every bounded (from below) set has an infimum, then also every set \(A\) of natural numbers has an infimum, say \(a^*\). We can now find an integer \(n^*\) such that \(a^*\) lies in the half-open interval \((n^*-1,n^*]\), and can then show that we must have \(a^* = n^*\), and \(n^*\) in \(A\).
• In axiomatic set theory, the natural numbers are defined as the smallest inductive set (i.e., set containing 0 and closed under the successor operation). One can (even without invoking the regularity axiom) show that the set of all natural numbers \(n\) such that "\(\{0,\ldots,n\}\) is well-ordered" is inductive, and must therefore contain all natural numbers; from this property one can conclude that the set of all natural numbers is also well-ordered.

In the second sense, this phrase is used when that proposition is relied on for the purpose of justifying proofs that take the following form: to prove that every natural number belongs to a specified set \(S\), assume the contrary, which implies that the set of counterexamples is non-empty and thus contains a smallest counterexample. Then show that for any counterexample there is a still smaller counterexample, producing a contradiction. This mode of argument is the contrapositive of proof by complete induction. It is known light-heartedly as the "minimal criminal" method and is similar in its nature to Fermat's method of "infinite descent".

Garrett Birkhoff and Saunders Mac Lane wrote in A Survey of Modern Algebra that this property, like the least upper bound axiom for real numbers, is non-algebraic; i.e., it cannot be deduced from the algebraic properties of the integers (which form an ordered integral domain).

Theorem: Every integer greater than one can be factored as a product of primes. This theorem constitutes part of the Prime Factorization Theorem.

Proof (by well-ordering principle). Let \(C\) be the set of all integers greater than one that cannot be factored as a product of primes. We show that \(C\) is empty.

Assume for the sake of contradiction that \(C\) is not empty. Then, by the well-ordering principle, there is a least element \(n \in C\); \(n\) cannot be prime since a prime number itself is considered a length-one product of primes. By the definition of non-prime numbers, \(n\) has factors \(a, b\), where \(a, b\) are integers greater than one and less than \(n\). Since a, b < n\), they are not in \(C\) as \(n\) is the smallest element of \(C\). So, \(a, b\) can be factored as products of primes, where \(a = p_1p_2...p_k\) and \(b = q_1q_2...q_l\), meaning that \(n = p_1p_2...p_k \cdot q_1q_2...q_l\), a product of primes. This contradicts the assumption that \(n \in C\), so the assumption that \(C\) is nonempty must be false.

Theorem: \(1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}\) for all positive integers \(n\).

Proof. Suppose for the sake of contradiction that the above theorem is false. Then, there exists a non-empty set of positive integers \(C = \{n \in \mathbb N \mid 1 + 2 + 3 + ... + n \neq \frac{n(n + 1)}{2}\}\). By the well-ordering principle, \(C\) has a minimum element \(c\) such that when \(n = c\), the equation is false, but true for all positive integers less than \(c\). The equation is true for \(n = 1\), so \(c > 1\); \(c - 1\) is a positive integer less than \(c\), so the equation holds for \(c - 1\) as it is not in \(C\). Therefore,

\(\begin{aligned} 1 + 2 + 3 + ... + (c - 1) &= \frac{(c - 1)c}{2} \\ 1 + 2 + 3 + ... + (c - 1) + c &= \frac{(c - 1)c}{2} + c\\ &= \frac{c^2 - c}{2} + \frac{2c}{2}\\ &= \frac{c^2 + c}{2}\\ &= \frac{c(c + 1)}{2} \end{aligned}\),

which shows that the equation holds for \(c\), a contradiction. So, the equation must hold for all positive integers.