Intersection of Strict Lower Closures in Toset

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

Let $a,b \in S$.

Then:


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

where $\dot\downarrow$ denotes strict down-set, 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 ${\dot\downarrow} b \subseteq {\dot\downarrow} a$.

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


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

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

By Strictly Precedes is Strict Ordering, $c \prec a$.

Also see

 * Intersection of Strict Up-Sets in Toset