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.
$\blacksquare$