Convex hull of an open set is an open set

Question

I have to prove this. Actually, I have the proof, but I don't understand one part.

It says:
"Since $\operatorname{co}A$ is intersect of all convex sets that contain set A, it follows that $\operatorname{co}A\subset (\operatorname{co}A)^\circ$".

Answer 1

I assume that an earlier part of the proof showed that the interior of a convex set is convex, so that $(\operatorname{co}A)^\circ$ is convex. Also, $A$ is open so it is contained in its own interior, and therefore in the interior of the larger set $\operatorname{co}A$.

So now $(\operatorname{co}A)^\circ$ is a convex set containing $A$, and therefore it also contains $\operatorname{co}A$.