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:


 * $t' \in T \cap t^{\preccurlyeq S}$

precisely when $t' \in T$ 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.