Intersection of Strict Lower Closures in Toset
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Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $a,b \in S$.
Then:
- $a^\prec \cap b^\prec = \left({\min \left({a, b}\right)}\right)^\prec$
where:
- $a^\prec$ denotes strict lower closure of $a$
- $\min$ 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 $b^\prec \subseteq a^\prec$.
By the definition of strict lower closure, 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, $c \prec a$.
$\blacksquare$