Weak Upper Closure in Restricted Ordering
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Theorem
Let $\struct {S, \preccurlyeq}$ 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 $t \in T$:
- $t^{\succcurlyeq T} = T \cap t^{\succcurlyeq S}$
where:
- $t^{\succcurlyeq T}$ is the weak upper closure of $t$ in $\struct {T, \preccurlyeq \restriction_T}$
- $t^{\succcurlyeq S}$ is the weak upper closure of $t$ in $\struct {S, \preccurlyeq}$.
Proof
Let $t \in T$.
Suppose that:
- $t' \in t^{\succcurlyeq T}$
By definition of weak upper closure $t^{\succcurlyeq T}$, 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 $t' \in t^{\succcurlyeq S}$.
By definition of set intersection, it also holds that:
- $t' \in T \cap t^{\succcurlyeq S}$
if and only if $t' \in T$ and $t' \in t^{\succcurlyeq S}$.
Thus it follows that the following are equivalent:
- $t' \in t^{\succcurlyeq T}$
- $t' \in T \cap t^{\succcurlyeq S}$
and hence the result follows, by definition of set equality.
$\blacksquare$