Intersection of Strict Upper Closures in Toset

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

Let $a, b \in S$.

Then:


 * $a^\succ \cap b^\succ = \left({\max \left({a, b}\right)}\right)^\succ$

where:
 * $a^\succ$ denotes strict upper closure of $a$
 * $\max$ denotes the max operation.

Proof
As $\left({S, \preceq}\right)$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$.

Since both sides are seen to be invariant upon interchanging $a$ and $b$, WLOG let $a \preceq b$.

Then it follows by definition of $\max$ that:
 * $\max \left({a, b}\right) = b$

Thus, from Intersection with Subset is Subset, it suffices to show that $b^\succ \subseteq a^\succ$.

By definition of strict upper closure, this comes down to showing that:


 * $\forall c \in S: b \prec c \implies a \prec c$

So let $c \in S$ with $b \prec c$, and recall that $a \preceq b$.

By Strictly Precedes is Strict Ordering, $a \prec c$.

Also see

 * Intersection of Strict Lower Closures in Toset