Infinite Sequence Property of Well-Founded Relation

Theorem
Let $\left({S, \preceq}\right)$ be a poset.

Then $\left({S, \preceq}\right)$ is well-founded iff there is no infinite sequence $\left \langle {a_n}\right \rangle$ of elements of $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

That is, no infinite sequence $\left \langle {a_n}\right \rangle$ such that $a_0 \succ a_1 \succ a_2 \succ \cdots$.

Proof

 * Suppose $\left({S, \preceq}\right)$ is not well-founded.

Let $T \subseteq S$ have no minimal element.

Let $a_0 \in T$.

Since $a_0$ is not minimal in $T$, we can find $a_1 \in T: a_1 \prec a_0$.

Now suppose then, that $a_k \in T$ such that $a_k \prec a_{k-1}$.

Then as $a_k$ is not minimal in $T$, we can find $a_{k+1} \in T: a_{k+1} \prec a_k$.

So, by induction, it follows that $\forall n \in \N: \exists a_n \in T: a_{n+1} \prec a_n$.

Thus we have been able to construct an infinite sequence $\left \langle {a_n}\right \rangle$ in $T$ such that $\forall n \in \N: a_{n+1} \prec a_n$.


 * Now suppose there exists an infinite sequence $\left \langle {a_n}\right \rangle$ in $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

We let $T = \left\{{a_0, a_1, a_2, \ldots}\right\}$.

Clearly $T$ has no minimal element.