Let $\le$ be an ordering on a set $S$ and let $A \subseteq S$ Then, with respect to the ordering $\le$

• $a \in A$ is the least element of $A$ if $a \le x$ for all $x \in A$
• $a \in A$ is a minimal element of $A$ if there exists no $x \in A$ such that $x \le a$ and $x \ne a$
• $a \in A$ is the greatest element of $A$ if $x \le a$ for all $x \in A$
• $a \in A$ is a maximal element of $A$ if there exists no $x \in A$ such that $a \le x$ and $x \ne a$

Examples.

• The natural numbers $\mathbb{N}$ ordered by divisibility ($\mid$ have a least element, $1$ The natural numbers greater than 1 ($\mathbb{N} \setminus \{1\}$ have no least element, but infinitely many minimal elements (the primes.) In neither case is there a greatest or maximal element.
• The negative integers ordered by the standard definition of $\le$ have a maximal element which is also the greatest element, $-1$ They have no minimal or least element.
• The natural numbers $\mathbb{N}$ ordered by the standard $\le$ have a least element, $1$ which is also a minimal element. They have no greatest or maximal element.
• The rationals greater than zero with the standard ordering $\le$ have no least element or minimal element, and no maximal or greatest element.