Intersection of Weak Lower Closures in Toset

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

Let $a, b \in S$.

Then:


 * $a^\preccurlyeq \cap b^\preccurlyeq = \left({\min \left({a, b}\right)}\right)^\preccurlyeq$

where:
 * $a^\preccurlyeq$ denotes the weak lower closure of $a$
 * $\min$ denotes the min 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 $b \preccurlyeq a$.

Then it follows by definition of $\min$ that $\min \left({a, b}\right) = b$.

Thus, from Intersection with Subset is Subset, it suffices to show that:
 * $b^\preccurlyeq \subseteq a^\preccurlyeq$

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


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

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

Recall that $b \preccurlyeq a$.

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

Hence $c \preccurlyeq a$.

Also see

 * Intersection of Weak Upper Closures in Toset