# Increasing Sequence in Ordered Set Terminates iff Maximal Element

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## Theorem

Let $\left({P, \le}\right)$ be an ordered set.

The following are equivalent:

- $(1): \quad$ 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 $x_n = x_{n+1} = \cdots$.

- $(2): \quad$ Every non-empty subset of $P$ has a maximal element.

## Proof

### $(1)$ implies $(2)$

Suppose $(1)$ holds.

Pick $\varnothing \ne S \subseteq P$.

Let $x_1 \in S$ be arbitrary.

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

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 $S$.

$\Box$

### $(2)$ implies $(1)$

Suppose $(2)$ holds.

Let $\left({x_k}\right)_{k \in \N}$ be an increasing sequence of elements of $P$.

By hypothesis, the set $\left\{{x_k: k \in \N}\right\}$ has a maximal element, say $x_n$.

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

$\blacksquare$