Intersection of Convex Sets is Convex Set (Order Theory)

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Theorem

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

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


Then $\displaystyle \bigcap \mathcal C$ is convex.


Proof

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

Let $a, c \in \displaystyle \bigcap \mathcal C$.


Let $a \prec b \prec c$.

By the definition of intersection:

$\forall C \in \mathcal C$: $a, c \in C$

Since each $C \in C$ is convex:

$\forall C \in \mathcal C$: $b \in C$.

Thus by the definition of intersection:

$b \in \displaystyle \bigcap \mathcal C$

Thus $\displaystyle \bigcap \mathcal C$ is convex.

$\blacksquare$