Intersection of Strict Upper Closures in Toset

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

Let $a, b \in S$.

Then:


 * ${\dot\uparrow} a \cap {\dot\uparrow} b = {\uparrow} \left({\max \left({a, b}\right)}\right)$

where $\dot\uparrow$ denotes strict up-set and $\max$ 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 $b$, let WLOG $a \preceq b$.

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 ${\dot\uparrow} b \subseteq {\dot\uparrow} a$.

By definition of $\dot\uparrow$, 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 Down-Sets in Toset