Supremum and Infimum Unique

From ProofWiki
Jump to: navigation, search

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

Personal tools
Namespaces
Variants
Actions
Navigation
ProofWiki.org
ToDo
Toolbox
Google AdSense