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 $t \in T$:

$t^{\prec T} = T \cap t^{\prec S}$

where:

$t^{\prec T}$ is the strict lower closure of $t$ in $\left({T, \preceq \restriction_T}\right)$

$t^{\prec S}$ is the strict lower closure of $t$ in $\left({S, \preceq}\right)$.

Proof

Let $t \in T$, and suppose that $t' \in t^{\prec T}$.

By definition of strict lower closure, 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, $t' \in t^{\prec T}$ is equivalent to:

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

These last two conjuncts precisely express that $t' \in t^{\prec S}$.

By definition of set intersection, it also holds that:

$t' \in T \cap t^{\prec S}$

if and only if $t' \in T$ and $t' \in t^{\prec S}$.

Thus, it follows that the following are equivalent:

$t' \in t^{\prec T}$

$t' \in T \cap t^{\prec S}$

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

$\blacksquare$