Intersection of Weak Lower Closures in Toset

Theorem

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

Let $a, b \in S$.

Then:

$a^\preccurlyeq \cap b^\preccurlyeq = \paren {\min \set {a, b} }^\preccurlyeq$

where:

$a^\preccurlyeq$ denotes the weak lower closure of $a$

$\min$ denotes the min operation.

Proof

As $\struct {S, \preccurlyeq}$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.

Both sides are seen to be invariant upon interchanging $a$ and $b$.

Without loss of generality, let $b \preccurlyeq a$.

Then it follows by definition of $\min$ that $\min \set {a, b} = b$.

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 $c \preccurlyeq a$.

$\blacksquare$

Also see

• Intersection of Weak Upper Closures in Toset