A compact convex subset in with nonempty interior is homeomorphic to .
Without loss of generality we may suppose the origin is an interior point of . We claim that the map maps the boundary homeomorphically onto . By convexity, is homeomorphic to the cone on , and therefore to the cone on which is .
Shouldn’t the claim be obvious?
The restricted map is continuous.
It’s surjective: contains a ball in its interior, and for each nonzero , the positive ray through intersects in a bounded half-open line segment. For the extreme point on this line segment, . Thus every unit vector is of the form for some extreme point , and such extreme points lie in .
It’s injective: for this we need to show that if are distinct points, then neither is a positive multiple of the other. Supposing otherwise, we have for , say. Let be a ball inside containing ; then the convex hull of is contained in and contains as an interior point, contradiction.
So the unit vector map, being a continuous bijection between compact Hausdorff spaces, is a homeomorphism.
Any compact convex set of is homeomorphic to a disk.
has nonempty interior relative to its affine span which is some -plane, and then is homeomorphic to by the theorem.
I guess I’d now like to add a little more to this little article, since it started out being about compact convex sets.
Let be a compact convex set of . The essential interior of is the set of points that are interior points of relative to its affine span. The essential boundary of is the set of points that do not belong to the essential interior of .
If belongs to the essential boundary of , then there is some functional such that and for every essential interior point .
In this case, is compact, convex, and has dimension strictly less than the dimension of .
(Minkowski-Carathéodory) A compact convex subset of is the convex hull of its extreme points. Indeed, every point is a convex combination of of its extreme points if the affine span of is -dimensional.
This of course is a baby version of the Krein-Milman theorem.
By induction on . The case for is trivial.
(to be continued)