I'm trying to understand the concept of faces in convex set. But, I'm stuck with it.

Namely, if I have a convex set such that $C = \{(x,y) \mid x^2 + y^2 \leqslant 1 \}$, then it is clearly convex set. But, I don't know how to find the faces in this set.

Any ideas or explanation would be helpful to me.

  • $\begingroup$ What definition are you using for "face of a convex set"? Is a point on the boundary of the disc a face according to this definition? $\endgroup$ – Brian Borchers Sep 18 '17 at 2:30
  • $\begingroup$ In good MathJax usage, the entire expression $$ C = \{(x,y) \mid x^2 + y^2 \leqslant 1 \} $$ should be within just one pair of dollar signs or double dollar signs. I edited the question accordingly. $\endgroup$ – Michael Hardy Sep 18 '17 at 2:41

If you are willing to understand what are the faces of a general convex set $C$, you might start to look at what are called the exposed faces of a convex set, which is a notion easy to handle. Exposed faces are a subclass of faces which are particularly easy to define: $F \subset C$ is an exposed face if it is the intersection of the boundary of $C$ together with some hyperplane being tangent to $C$.

In the case of your disc, you see that any hyperplane will be tangent to the disc at a unique point on the boundary, meaning that each point of the unit circle is a face.

Maybe are you familiar with the notion of face for polyhedra? It happens that for polyhedra the notion of face and exposed face coincide. For example, you can find all the faces of a square by intersecting its boundary with appropriate hyperplane.


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