Closed convex hull = closure of convex hull? If the "closed convex hull" of A is the intersection of all closed convex sets containing A, is this the same as the closure of the convex hull of A? Many have asked whether the closure of the convex hull is the same as the convex hull of the closure (answer: no), but I think this is a bit different. I feel like this should have a simple answer, either based on set logic (if it's true) or a simple counterexample (if it's false).

They are equal.

Let $C$ be the closed convex hull of $A$, defined as the intersection of all closed convex sets containing $A$. (Closed half-spaces are enough here.)

Let $B$ be the convex hull of $A$, defined as the intersection of all convex sets containing $A$. (Half-spaces, either open or closed, are not enough here.)

Clearly $B\subset C$. Since $C$ is closed, it follows that $\overline{B}\subset C$.

Conversely, $\overline{B}$ is both closed and convex, hence $C\subset \overline{B}$.