Upper Closure is Upper Section
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a subset of $S$.
Let $U$ be the upper closure of $T$.
Then $U$ is an upper section.
Proof
Let $a \in U$.
Let $b \in S$ with $a \preceq b$.
By the definition of upper closure, there is a $t \in T$ such that $t \preceq a$.
By transitivity, $t \preceq b$.
Thus, again by the definition of upper closure:
- $b \in U$
Since this holds for all such $a$ and $b$, $U$ is an upper section.
$\blacksquare$