Upper bound, least upper bound, and strict upper bound

Let $(P,\le )$ be a poset (Partial order and partially ordered set) and $S\subseteq P$.

An element $u\in P$ is an upper bound of $S$ if $s\le u$ for every $s\in S$.

An element $u\in P$ is a least upper bound (or supremum) of $S$ if $u$ is an upper bound of $S$ and $u\le v$ for every upper bound $v$ of $S$. When it exists we write $u=\sup S$.

An element $u\in P$ is a strict upper bound of $S$ if $s<u$ for every $s\in S$.

• A least upper bound is unique when it exists. If $u$ and $u^{\mathrm{\prime }}$ are both least upper bounds of $S$ then each is an upper bound and each is below the other, so $u\le u^{\mathrm{\prime }}$ and $u^{\mathrm{\prime }}\le u$, whence $u=u^{\mathrm{\prime }}$ by antisymmetry (Partial order and partially ordered set). This is what makes the notation $\sup S$ legitimate. Antisymmetry is not peculiar to this argument: the same two-inequality step gives uniqueness of a greatest element (Maximal element and greatest element), and it is used essentially in Bourbaki–Witt fixed point theorem, whose fixed point is obtained by passing from $f(c)\le c$ and $c\le f(c)$ to $f(c)=c$. Drop antisymmetry and it is the conclusion, not merely the notation, that goes: on two distinct elements each below the other, every subset still has a least upper bound, yet the map exchanging the two satisfies $x\le f(x)$ and has no fixed point.
• Every element of $P$ is an upper bound of the empty set, vacuously. Consequently $\sup \mathrm{\varnothing }$, when it exists, is the least element of $P$.
• An upper bound of $S$ need not belong to $S$, and $S$ may have many upper bounds and no least one. In $\mathbb{Q}$ with its usual order, the set $\{q\in \mathbb{Q}:q^2<2\}$ has upper bounds but no least upper bound.
• In a poset, a strict upper bound is exactly an upper bound outside $S$. If $u$ is strict then $u\mathrm{\notin }S$, since $u<u$ is impossible. Conversely, if $u$ is an upper bound and $u\mathrm{\notin }S$, then every $s\in S$ satisfies $s\le u$ and $s\mathrm{\ne }u$, hence $s<u$. This distinction from an arbitrary upper bound matters in Zorn's lemma, where the argument must produce one outside the chain.