## A poset has the descending chain condition if and only if every nonempty subset has a minimal element

Let $P$ be a nonempty set partially ordered by $\leq$. Then the following are equivalent.

1. Every nonempty subset of $P$ contains a minimal element.
2. If $a : \mathbb{N} \rightarrow P$ is a sequence in $P$ such that $a_i \geq a_{i+1}$ for all $i$, then there exists $n \in \mathbb{N}$ such that for all $i \geq n$, $a_i = a_n$. (The descending chain condition.

$(1) \Rightarrow (2)$ Suppose every nonempty subset of $P$ contains a minimal element, and let $a$ be a descending chain in $P$. In particular, $\{p_i\}_\mathbb{N}$ is a nonempty subset of $P$, which contains a minimal element $a_n$. We have $a_n \leq a_i$ for all $i$ as desired.

$(2) \Rightarrow (1)$ Suppose $P$ satisfies the descending chain condition, and let $S \subseteq P$ be a nonempty subset. Suppose $S$ does not have a minimal element. Define a descending sequence $a : \mathbb{N} \rightarrow S$ inductively as follows. Let $a_0$ be any element of $S$. (We can do this since $S$ is not empty.) Given that $a_i$ is defined, since $a_i \in S$ is not minimal, there exists $a_{i+1} < a_i$. Now $a$ is a descending chain, and for all $i$, $a_i > a_{i+1}$; a violation of the descending chain condition.