Closure of Convex Set in Topological Vector Space is Convex

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\Bbb F$.

Let $C \subseteq X$ be convex.

Then the closure $C^-$ of $C$ is convex.

Proof
Let $t \in \closedint 0 1$.

Since $C$ is convex, we have:


 * $t C + \paren {1 - t} C \subseteq C$

We show that:


 * $t C^- + \paren {1 - t} C^- \subseteq C^-$

We have:

Also see

 * Closure of Convex Subset in Normed Vector Space is Convex