Increasing Sequence in Ordered Set Terminates iff Maximal Element

Theorem
Let $(P,\leq)$ be an ordered set.

Then the following are equivalents:


 * 1. Every increasing sequence $x_1 \leq x_2 \leq x_3 \leq \cdots$ with $x_i \in P$ eventually terminates: there is $n \in \N$ such that $x_n = x_{n+1} = \cdots$.


 * 2. Evert non-empty subset of $P$ has a maximal element.

Proof
1. $\implies$ 2.

Suppose 1. and pick $\emptyset \neq S \subseteq P$.

Let $x_1 \in S$ be arbitrary.

Given $x_k \in S$, pick $x_{k+1} \in S$ strictly bigger that $x_k$.

By hypothesis the process must eventually terminate, say $x_n$ is the last element.

Then by construction there are no larger elements than $x_n$, i.e. $x_n$ is a maximal element of $S$.

2. $\implies$ 1.

Suppose 2. and let $(x_k)_{k \in \N}$ be an increasing sequence of elements of $P$.

By hypothesis, the set $\{x_k\}$ has a maximal element, say $x_n$.

Then since $x_m \geq x_n$ for all $m \geq n$, we must have $x_n = x_{n+1} = \cdots$.