Supremum is Unique

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $T$ be a non-empty subset of $S$.

Then $T$ has at most one supremum 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:
 * $c' \preceq c \land c \preceq c'$

and thus by the antisymmetry of the ordering $\preceq$:
 * $c = c'$