Weak Lower Closure in Restricted Ordering
Theorem
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
- $t^{\preccurlyeq T} = T \cap t^{\preccurlyeq S}$
where:
- $t^{\preccurlyeq T}$ is the weak lower closure of $t$ in $\left({T, \preccurlyeq \restriction_T}\right)$
- $t^{\preccurlyeq S}$ is the weak lower closure of $t$ in $\left({S, \preccurlyeq}\right)$.
Proof
Let $t \in T$, and suppose that $t' \in t^{\preccurlyeq T}$.
By definition of weak lower closure $t^{\preccurlyeq T}$, this is equivalent to:
- $t' \preccurlyeq \restriction_T t$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
- $t' \preccurlyeq t \land t' \in T$
as it is assumed that $t \in T$.
The first conjunct precisely expresses that $t' \in t^{\preccurlyeq S}$.
By definition of set intersection, it also holds that:
- $t' \in T \cap t^{\preccurlyeq S}$
if and only if $t' \in T$ and $t' \in t^{\preccurlyeq S}$.
Thus, it follows that the following are equivalent:
- $t' \in t^{\preccurlyeq T}$
- $t' \in T \cap t^{\preccurlyeq S}$
and hence the result follows, by definition of set equality.
$\blacksquare$