# Infinite Sequence Property of Well-Founded Relation/Reverse Implication

## 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:

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

Then $\RR$ is a well-founded relation.

## Proof 1

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.

## Proof 2

Suppose $\RR$ is not a well-founded relation.

Hence there exists $T \subseteq S$ such that $T$ has no minimal element under $\RR$.

Let $a_0 \in T$.

We have that $a_0$ is not minimal in $T$.

So:

- $\exists a_1 \in T: \paren {a_1 \mathrel \RR a_0} \text { and } a_1 \ne a_0$

Similarly, $a_1$ is not minimal in $T$.

So:

- $\exists a_2 \in T: \paren {a_2 \mathrel \RR a_1} \text { and } a_2 \ne a_1$

Let $a_{k + 1}$ be an arbitrary element for which $a_{k + 1} \mathrel \RR a_k$ and $a_{k + 1} \ne a_k$.

In order to allow this to be possible in the infinite case, it is necessary to invoke the Axiom of Dependent Choice as follows:

Let $a_k \in T$.

Then as $a_k$ is not minimal in $T$:

- $\exists a_{k + 1} \in T: \paren {a_{k + 1} \mathrel \RR a_k} \text { and } \paren {a_{k + 1} \ne a_k}$

Hence by definition $\RR$ is a right-total relation.

So, by the Axiom of Dependent Choice, it follows that:

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

Thus we have been able to construct an infinite sequence $\sequence {a_n}$ in $T$ such that:

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

It follows by the Rule of Transposition that if there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:

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

then $\RR$ is a well-founded relation.

## Axiom of Dependent Choice

This theorem depends on the Axiom of Dependent Choice.

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.