Union of Non-Disjoint Convex Sets is Convex Set

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

Let $\mathcal C$ be a set of convex sets of $S$ such that their intersection is non-empty:
 * $\displaystyle \bigcap \mathcal C \ne \varnothing$

Then the union $\displaystyle \bigcup \mathcal C$ is also convex.

Proof
Let $x, y, z \in S$ be arbitrary elements of $S$ such that $x \prec y \prec z$.

Let $x, z \in \displaystyle \bigcup \mathcal C$.

First let $x, z \in C$ where $C \in \mathcal C$.

Then as $C$ is convex, $y \in C$.

Hence, by definition of union, $y \in \displaystyle \bigcup \mathcal C$.

Now let $x \in C_1, z \in C_2$ where $C_1, C_2 \in \mathcal C$.

We have that $\displaystyle \bigcap \mathcal C \ne \varnothing$.

Thus $C_1 \cap C_2 \ne \varnothing$.

Then $\exists a \in C_1 \cap C_2: x < a < z$.

Hence one of the following cases holds:


 * $(1): \quad x < y < a < z$, whence $y \in C_1$, by convexity of $C_1$


 * $(2): \quad x < a < y < z$, whence $y \in C_2$, by convexity of $C_2$


 * $(3): \quad y = a$, whence $y \in C_1$ and $y \in C_2$, by definition of $a$.

Thus in all cases $y \in \displaystyle \bigcup \mathcal C$.

Thus $\displaystyle \bigcup \mathcal C$ is convex by definition.