Strict Upper Closure is Upper Section
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Let $p^\succ$ denote the strict upper closure of $p$.
Then $p^\succ$ is an upper section.
Proof
Let $u \in p^\succ$.
Let $s \in S$ with $u \preceq s$.
Then by the definition of strict upper closure:
- $p \prec u$
Thus by Extended Transitivity:
- $p \prec s$
So by the definition of strict upper closure:
- $s \in p^\succ$
Since this holds for all such $u$ and $s$, $p^\succ$ is an upper section.
$\blacksquare$