Union of Overlapping Convex Sets in Toset is Convex/Infinite Union

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

Let $\mathcal A$ be a set of convex subsets of $S$.

For any $P, Q \in \mathcal A$, let there be elements $C_0, \dotsc, C_n \in \mathcal A$ such that:
 * $C_0 = P$
 * $C_n = Q$
 * For $k = 0, \dots, n-1: C_k \cap C_{k+1} \ne \varnothing$

Then $\bigcup \mathcal A$ is convex in $S$.

Proof
Let $a, c \in \bigcup \mathcal A$.

Let $b \in S$.

Let $a \prec b \prec c$.

Since $a, c \in \bigcup \mathcal A$, there are $P, Q \in \mathcal A$ such that $a \in P$ and $c \in Q$.

By the premise, there are elements $C_0, \dots, C_n \in \mathcal A$ such that:
 * $C_0 = P$
 * $C_n = Q$
 * For $k = 0, \dots, n-1: C_k \cap C_{k+1} \ne \varnothing$

Applying Union of Overlapping Intervals is Interval inductively:


 * $\displaystyle \bigcup_{k \mathop = 0}^n C_k$ is convex.

Since $\displaystyle a, c \in \bigcup_{k \mathop = 0}^n C_k$, by the definition of convexity:
 * $\displaystyle b \in \bigcup_{k \mathop = 0}^n C_k$

Thus:
 * $\displaystyle b \in \bigcup \mathcal A$

Since this holds for all such triples $a, b, c$, it follows that $\mathcal A$ is convex.