Union of Non-Disjoint Convex Sets is Convex Set
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Theorem
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\CC$ be a set of convex sets of $S$ such that their intersection is non-empty:
- $\ds \bigcap \CC \ne \O$
Then the union $\ds \bigcup \CC$ is also convex.
Proof
Let $x, y, z \in S$ be arbitrary elements of $S$ such that $x \prec y \prec z$.
Let $x, z \in \ds \bigcup \CC$.
First let $x, z \in C$ where $C \in \CC$.
Then as $C$ is convex, $y \in C$.
Hence, by definition of union, $y \in \ds \bigcup \CC$.
Now let $x \in C_1, z \in C_2$ where $C_1, C_2 \in \CC$.
We have that $\ds \bigcap \CC \ne \O$.
Thus $C_1 \cap C_2 \ne \O$.
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 $y \in \ds \bigcup \CC$.
Thus $\ds \bigcup \CC$ is convex by definition.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $2$