Infinite Sequence Property of Well-Founded Relation/Reverse Implication/Proof 1

Theorem

Let $\struct {S, \RR}$ be a relational structure.

Let $\RR$ be such that there exists no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:

$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$

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:

$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$

$\blacksquare$

This theorem depends on the Axiom of Dependent Choice.

The consensus in conventional mathematics is that it is true and that it should be accepted.