# Strict Upper Closure in Restricted Ordering

## Theorem

Let $\struct {S, \preceq}$ 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 $t \in T$:

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

where:

$t^{\succ T}$ is the strict upper closure of $t$ in $\struct {T, \preceq \restriction_T}$
$t^{\succ S}$ is the strict upper closure of $t$ in $\struct {S, \preceq}$.

## Proof

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

By definition of strict upper 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, $t' \in t^{\succ T}$ is equivalent to:

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

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

By definition of set intersection, it also holds that:

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

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

Thus, it follows that the following are equivalent:

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

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

$\blacksquare$