Exercise A0.V.3.02

Proposition

A totally ordered set $Z$ is well-ordered if and only if every descending chain in $Z$ stabilizes.

Proof

Suppose $Z$ is well-ordered. Let $a_1 \geq a_2 \geq a_3 \geq \ldots$ be a descending chain. By well-ordering, there exists the least element $\hat{a}$ in this chain. Thus, the chain stabilizes at $\hat{a}.$

Suppose instead that every descending chain in $Z$ stabilizes and let $S \subseteq Z$ be non-empty. Let $a \in S$ be any element. We construct a sequence with $a_0 = a,$ $a_{n + 1} = a_n$ if $S$ has no elements smaller than $a_n,$ otherwise $a_{n + 1} \in \{ b \in S \mid b < a_n\}.$

This gives us a descending chain $a_0 \geq a_1 \geq \ldots.$ By hypothesis, this chain stabilizes, say at $\hat{a}.$ By construction, no element of $S$ is smaller than $\hat{a}.$ Thus, $\hat{a}$ is the least element of $S.$

$\blacksquare$