Strict Lower Closure in Restricted Ordering

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.

Then for all $t \in T$:


 * $\mathop{\downarrow_T} \left({t}\right) = T \cap \mathop{\downarrow_S} \left({t}\right)$

where $\downarrow_T$ is the strict lower closure in $\left({T, \preceq \restriction_T}\right)$, and $\downarrow_S$ is the strict lower closure in $\left({S, \preceq}\right)$.

Proof
Let $t \in T$, and suppose that $t' \in \mathop{\downarrow_T} \left({t}\right)$.

By definition of $\downarrow_T$, this is equivalent to:


 * $t' \preceq \restriction_T t \land t \ne t'$

By definition of $\preceq \restriction_T$, the first condition comes down to:


 * $t' \preceq t \land t' \in T$

as it is assumed that $t \in T$.

In conclusion, $t' \in \mathop{\downarrow_T} \left({t}\right)$ is equivalent to:


 * $t' \in T \land t' \preceq t \land t \ne t'$

These last two conjuncts precisely express that $t' \in \mathop{\downarrow_S} \left({t}\right)$.

By definition of set intersection, it also holds that:


 * $t' \in T \cap \mathop{\downarrow_S} \left({t}\right)$

precisely when $t' \in T$ and $t' \in \mathop{\downarrow_S} \left({t}\right)$.

Thus, it follows that the following are equivalent:


 * $t' \in \mathop{\downarrow_T} \left({t}\right)$
 * $t' \in T \cap \mathop{\downarrow_S} \left({t}\right)$

and hence the result follows, by definition of set equality.