Infinite Sequence Property of Well-Founded Relation

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

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 $a_0 \succ a_1 \succ a_2 \succ \cdots$.

Forward Implication
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.