Intersection of Weak Upper Closures in Toset

Theorem

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

Let $a, b \in S$.

Then:

$a^\succcurlyeq \cap b^\succcurlyeq = \paren {\map \max {a, b} }^\succcurlyeq$

where:

$a^\succcurlyeq$ denotes weak upper closure of $a$

$\max$ denotes the max 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 $a \preccurlyeq b$.

Then it follows by definition of $\max$ that:

$\map \max {a, b} = b$

Thus, from Intersection with Subset is Subset, it suffices to show that:

$b^\succcurlyeq \subseteq a^\succcurlyeq$

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

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

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

Recall that $a \preccurlyeq b$.

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

Hence $a \preccurlyeq c$.

$\blacksquare$

Also see

• Intersection of Weak Lower Closures in Toset