Supremum and Infimum are Unique

Theorem
Let $\left({S, \preceq}\right)$ be a poset.

Any non-empty subset of $S$ admits at most one supremum and one infimum in $S$.

Proof
Let $c$ and $c\,'$ both be suprema of $T$ in $S$.

From the definition of supremum, $c$ and $c\,'$ are upper bounds of $T$ in $S$.

By that definition:


 * $c$ is an upper bound of $T$ in $S$ and $c\,'$ is a supremum of $T$ in $S$ implies that $c\,' \preceq c$
 * $c\,'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S$ implies that $c \preceq c\,'$.

So $c\,' \preceq c \land c \preceq c\,'$ thus $c = c\,'$ by the antisymmetry of a partial ordering.

A similar argument applies to establish the same of the infimum.