Supremum and Infimum Unique
From ProofWiki
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 antisymmety of a partial ordering.
A similar argument applies to establish the same of the infimum.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 14$: Theorem $14.3$