Upper Closure is Closure Operator
Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $T^\succeq$ be the upper closure of $T$ for each $T \subseteq S$.
Then $\cdot^\succeq$ is a closure operator.
Proof
Inflationary
Let $T \subseteq S$.
Let $t \in T$.
Then since $T \subseteq S$, $t \in S$ by the definition of subset.
Since $\preceq$ is reflexive, $t \preceq t$.
Thus by the definition of upper closure, $t \in T^\succeq$.
Since this holds for all $t \in T$, $T \subseteq T^\succeq$.
Since this holds for all $T \subseteq S$:
- $\cdot^\succeq$ is inflationary.
$\Box$
Order-Preserving
Let $T \subseteq U \subseteq S$.
Let $x \in T^\succeq$.
Then by the definition of upper closure: for some $t \in T$, $t \preceq x$.
By the definition of subset:
- $t \in U$
Thus by the definition of upper closure:
- $x \in U^\succeq$
Since this holds for all $x \in T^\succeq$:
- $T^\succeq \subseteq U^\succeq$
Since this holds for all $T$ and $U$:
- $\cdot^\succeq$ is order-preserving.
$\Box$
Idempotent
Let $T \subseteq S$.
By Upper Closure is Upper Section, $T^\succeq$ is an upper section.
Thus by Upper Section: Definition 3:
- $\paren {T^\succeq}^\succeq = T^\succeq$
Since this holds for all $T$:
- $\cdot^\succeq$ is idempotent.
$\blacksquare$