Convex Set Characterization (Order Theory)

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Theorem

Let $\left({S, \preceq, \tau}\right)$ be an ordered set.

Let $C \subseteq S$.


The following are equivalent:

\((1)\)   $:$   $C$ is convex.             
\((2)\)   $:$   $C$ is the intersection of an upper set with a lower set.             
\((3)\)   $:$   $C$ is the intersection of its upper closure with its lower closure.             


Proof

$(2)$ implies $(1)$

Follows from Upper Set is Convex, Lower Set is Convex, and Intersection of Convex Sets is Convex Set (Order Theory).

$\Box$


$(1)$ implies $(3)$

Let $C$ be a convex set in $S$.

Let $U$ and $L$ be the upper and lower closures of $C$, respectively.

Since $C \subseteq U$ and $C \subseteq L$:

$C \subseteq U \cap L$.

Let $p \in U \cap L$.

Then $a \preceq p \preceq b$ for some $a, b \in C$.

Since $C$ is convex, $p \in C$.

Since this holds for all $p \in U \cap L$:

$U \cap L \subseteq C$.

Since we know that $C \subseteq U \cap L$, $C = U \cap L$.

$\Box$


$(3)$ implies $(2)$

Follows from Upper Closure is Upper Set and Lower Closure is Lower Set.

$\blacksquare$