Supremum is Unique

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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 $\preceq$:

$c = c'$

$\blacksquare$


Sources