Intersection of Weak Upper Closures in Toset

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

Let $a, b \in S$.

Then:


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

where:
 * $a^\succcurlyeq$ denotes weak upper closure of $a$
 * $\max$ denotes the max operation.

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

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

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:
 * $b^\succcurlyeq \subseteq a^\succcurlyeq$

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


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

So let $c \in S$ with $b \preceq c$.

Recall that $a \preceq b$.

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

Hence $a \preceq c$.

Also see

 * Intersection of Weak Lower Closures in Toset