Maximum and minimum of a set

Let $S \subseteq \mathbb{R}$ and $m \in \mathbb{R}$.

• $m$ is a maximum (or greatest element) of $S$ if $m \in S$ and $s \le m$ for every $s \in S$.
• $m$ is a minimum (or least element) of $S$ if $m \in S$ and $m \le s$ for every $s \in S$.

A set has at most one maximum: if $m_1$ and $m_2$ are both maxima then $m_1 \in S$ gives $m_1 \le m_2$ and $m_2 \in S$ gives $m_2 \le m_1$, so $m_1 = m_2$ by antisymmetry of the order, which is immediate from the trichotomy axiom of an ordered field (Ordered field, Complete ordered field (least-upper-bound property)). The same argument applies to minima, so we may write $\max S$ and $\min S$.

• A maximum is precisely an upper bound of $S$ (Complete ordered field (least-upper-bound property)) that happens to lie in $S$; a minimum is a lower bound of $S$ (Lower bound, bounded below, bounded set) that lies in $S$. In particular a set with a maximum is bounded above and a set with a minimum is bounded below.
• The empty set has neither a maximum nor a minimum, because the requirement $m \in S$ cannot be met.
• The membership requirement $m \in S$ is exactly what separates a maximum from a supremum, and it is the theme of this page. A supremum is a bound on the set and is not asked to belong to it; a maximum is an element of the set. The two agree exactly when the supremum happens to be attained (The supremum is attained exactly when a maximum exists), and they genuinely differ in general (FALSE: the supremum of a set belongs to the set).
• Every nonempty finite subset of $\mathbb{R}$ has both a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum), which is what licenses the notation $\max\{a_1, \dots, a_n\}$. Infinite sets need not: the failure of attainment is an infinitary phenomenon.