A sufficient condition for being the boundary of one’s convex hull?

Let $Asubsetmathbb R^n$ be such that:

  1. every non-zero linear functional is maximized by a unique point of $A$

  2. every point of $A$ is a point where some linear functional achieves its maximum over $A$ (i.e., every point of $A$ is an exposed point, though normally exposed points are only defined for convex sets)

  3. the boundary of $A$‘s convex hull is contained in the closure of $A$.

Does it follow that $A$ is closed, and hence equal to the boundary of its convex hull?

It doesn’t follow in the absence of (3). Let $A = { e^{itheta}: 0letheta<pi } cup { e^{itheta} – i : piletheta <2pi } subseteq mathbb C = mathbb R^2$.