Union of Chain of Convex Sets in Vector Space is Convex

Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $\Gamma$ be a chain of convex sets.

Let:

$\ds C = \bigcup \Gamma$

Then $C$ is convex.

Proof

Let $t \in \closedint 0 1$ and $x, y \in C$.

Then there exists $C_1, C_2 \in \Gamma$ such that $x \in C_1$ and $y \in C_2$.

Since $\Gamma$ is a chain, we have $C_1 \subseteq C_2$ or $C_2 \subseteq C_1$.

Without loss of generality suppose that $C_1 \subseteq C_2$.

Then $x, y \in C_2$.

Since $C_2$ is convex, we have $t x + \paren {1 - t} y \in C_2$.

So $t x + \paren {1 - t} y \in \bigcup \Gamma = C$.

So $C$ is convex.

$\blacksquare$