Union of Overlapping Convex Sets in Toset is Convex/Infinite Union
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Theorem
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\AA$ be a set of convex subsets of $S$.
For any $P, Q \in \AA$, let there be elements $C_0, \dotsc, C_n \in \AA$ such that:
- $C_0 = P$
- $C_n = Q$
- For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
Then $\bigcup \AA$ is convex in $S$.
Proof
Let $a, c \in \bigcup \AA$.
Let $b \in S$.
Let $a \prec b \prec c$.
Since $a, c \in \bigcup \AA$, there are $P, Q \in \AA$ such that $a \in P$ and $c \in Q$.
By the premise, there are elements $C_0, \dots, C_n \in \AA$ such that:
- $C_0 = P$
- $C_n = Q$
- For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
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Applying Union of Overlapping Intervals is Interval inductively:
- $\ds \bigcup_{k \mathop = 0}^n C_k$ is convex.
Since $\ds a, c \in \bigcup_{k \mathop = 0}^n C_k$, by the definition of convexity:
- $\ds b \in \bigcup_{k \mathop = 0}^n C_k$
Thus:
- $\ds b \in \bigcup \AA$
Since this holds for all such triples $a, b, c$, it follows that $\AA$ is convex.
$\blacksquare$