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

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

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 $b \in T: b \prec a$.

Since this holds for all $a \in T$, $\prec \restriction_T$, the restriction of $\prec$ to $T$, is a right-total relation on $T$.

So, by the Axiom of Dependent Choice, it follows that we can construct an infinite sequence $\left \langle {a_n}\right \rangle$ in $T$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

Sufficient Condition
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.

Thus by definition $S$ is not well-founded.