Intersection of Convex Sets is Convex Set (Order Theory)

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $\CC$ be a set of convex sets in $S$.

Then $\ds \bigcap \CC$ is convex.

Proof

Let $a, b, c \in S$.

Let $a, c \in \ds \bigcap \CC$.

Let $a \prec b \prec c$.

By the definition of intersection:

$\forall T \in \CC$: $a, c \in T$

Since each $T \in \CC$ is convex:

$\forall T \in \CC$: $b \in T$

Thus by the definition of intersection:

$b \in \ds \bigcap \CC$

Thus $\ds \bigcap \CC$ is convex.

$\blacksquare$