Upper Closure is Smallest Containing Upper Section

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $T \subseteq S$.

Let $U = T^\succeq$ be the upper closure of $T$.

Then $U$ is the smallest upper section containing $T$ as a subset.

Proof

Follows from Upper Closure is Closure Operator and Set Closure is Smallest Closed Set/Closure Operator.

$\blacksquare$

Also see

• Lower Closure is Smallest Containing Lower Section