Intersection of Weak Lower Closures in Toset

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

Let $a,b \in S$.

Then:


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

where $\bar \downarrow$ denotes weak lower closure, and $\min$ denotes the min 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 $b \preceq a$.

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 $\mathop{\bar \downarrow} \left({b}\right) \subseteq \mathop{\bar \downarrow} \left({a}\right)$.

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


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

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

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

Hence $c \preceq a$.