Intersection of Weak Upper Closures in Toset

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

Let $a,b \in S$.

Then:


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

where $\bar \uparrow$ denotes weak upper closure, 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 $\mathop{\bar \uparrow} \left({b}\right) \subseteq \mathop{\bar \uparrow} \left({a}\right)$.

By definition of $\bar \uparrow$, this comes down to showing that:


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

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

Now as $\preceq$ is a total ordering, it is in particular transitive.

Hence $a \prec c$.