Infinite Sequence Property of Well-Founded Relation
Theorem
Let $\struct {S, \RR}$ be a relational structure.
Then $\RR$ is a well-founded relation if and only if there exists no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
- $\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Proof
Reverse Implication
Suppose $\RR$ is not a well-founded relation.
So by definition there exists a non-empty subset $T$ of $S$ which has no minimal element.
Let $a \in T$.
Since $a$ is not minimal in $T$, we can find $b \in T: \paren {b \mathrel \RR a} \text { and } \paren {b \ne a}$.
This holds for all $a \in T$.
Hence the restriction $\RR \restriction_{T \times T}$ of $\RR$ to $T \times T$ is a right-total endorelation on $T$.
So, by the Axiom of Dependent Choice, it follows that there is an infinite sequence $\sequence {a_n}$ in $T$ such that:
- $\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
It follows by the Rule of Transposition that if there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
- $\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
then $\RR$ is a well-founded relation.
$\Box$
Forward Implication
Let $\RR$ be a well-founded relation.
Aiming for a contradiction, suppose there exists an infinite sequence $\sequence {a_n}$ in $S$ such that:
- $\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Let $T = \set {a_0, a_1, a_2, \ldots}$.
Let $a_k \in T$ be a minimal element of $T$.
That is:
- $\forall y \in T \setminus \set {a_k}: \neg \paren {y \mathrel \RR a_k}$
But we have that:
- $a_{k + 1} \mathrel \RR a_k$ and $a_{k + 1} \ne a_k$.
So $a_k$ is not a minimal element.
It follows by Proof by Contradiction that such an infinite sequence cannot exist.
$\blacksquare$
Axiom of Dependent Choice
This theorem depends on the Axiom of Dependent Choice, by way of Infinite Sequence Property of Well-Founded Relation/Reverse Implication.
Although not as strong as the Axiom of Choice, the Axiom of Dependent Choice is similarly independent of the Zermelo-Fraenkel axioms.
The consensus in conventional mathematics is that it is true and that it should be accepted.
Sources
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.5$: Relations: Lemma $1.5.1$
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $9$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(f)}$