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.