TITLE: Cube/Circle/etc without one point on the boundary convex?
QUESTION [7 upvotes]: Reading this paper here, about convex optimization (photo page 2-3).

What about if the object misses the corner point? I think it is still convex but I am not sure whether my intuition fails me here. Is it convex or not? Is a ball without the boundary convex? I doubt it because if I select a point, I can always go nearer-and-nearer (never really reaching the point without extended Number system ofc). I am not sure about the general case with limit points like if I have non-dense space like missing points $\mathbb R^{N} \text{ (exclude sign here please) } \mathbb Q^{N}$ from the space $\mathbb C^{N}$.

REPLY [6 votes]: The boundary of a convex set has special points, called extreme points, which generalize corners. You can remove any of those and the remaining set will be convex. If you remove an ordinary boundary point that is not an extreme point then you may end up with a set that is not convex.
All points in the boundary of a ball are extreme and so you can remove any or all of them. In the wikipedia illustration, the extreme points appear in red; you can remove any or all of them. But if you leave the endpoints of the red segments then you cannot remove any of the ordinary boundary points without breaking convexity.
A related result is the Krein–Milman theorem, which says that a compact convex set is the closure of the convex hull of its extreme points. Informally, you can recover the shape of a compact convex set from its extreme points.