I am used to the following definition of a (proper) face of a polytope:

A nonempty convex subset $F$ of a polytope $C$ is called a face of $C$ if $\alpha x + (1-\alpha) y \in F$ with $x, y \in C$ and $0 < \alpha < 1$ imply $x, y \in F$. $F$ is a proper face of $C$ if $F$ is a face of $C$ and $F \neq C$.

I often read the following definition of a proper face:

A nonempty subset $F$ of a polytope C is a proper face of $C$ if there is closed half-space $H$ containing $C$ such that $F = C \cap \partial H$, where $\partial H$ is the affine hyperplane defined by the boundary of $H$.

How can I show that these two definitions are equivalent?

  • One direction should be clear: If $F = C \cap \partial H$ for some half-space $H = \{f \leq t\} \supset C$, where $f$ is a linear form, then $F$ is a face of $C$. Suppose that $x \in F$ is of the form $(1-\alpha)p + \alpha q$ with $0 \lt \alpha \lt 1$ and $p,q \in C$ then $H \supset C$ gives $f(p) \leq t$ and $f(q) \leq t$ while we know $t = f(x) = (1-\alpha) f(p) + \alpha f(q)$, so we must have $f(p)=t=f(q)$, hence $p,q \in F$. – t.b. Nov 18 '11 at 7:48

In an infinite-dimensional real topological vector space $V$ with a discontinuous linear functional $f$, you could take $C$ to be the half-space $\{x \in V: f(x) \ge 0\}$, and $F = \{x \in V: f(x) = 0\}$ is a proper face according to the first definition but not according to the second.

  • Indeed, thanks. I've edited my question, restricting things to the finite-dimensional case. – Tom Jonathan Nov 18 '11 at 7:37

In ${\mathbb R}^2$, let $C$ be the union of the open lower half plane and the nonnegative $x$ axis, and let $F$ consist of the origin. Then $F$ is a proper face of $C$ according to your first definition, but is not of the form $C \cap \partial H$.

  • You're right again, thanks. I'm making a fool of myself. I actually read the second definition in books about polytopes, so I hope the two definitions are equivalent in the case of polytopes. I've edited my question again, restricting things to polytopes now. – Tom Jonathan Nov 20 '11 at 3:20

The general answer for convex sets is: no.

Your first definition corresponds to what is indeed called the face of a convex set.

The second definition (the one with hyperplanes) defines what is called the exposed faces of a convex set. As its name suggests, any exposed face is a face.

The fact is that: 1) There is convex sets whose faces can not be exposed. This typically holds when the boundary is flat and then becomes slowly curved. If you want a visual example, look at Figure 1.a) in https://arxiv.org/abs/1107.2319 where each point indicated with a star is a face (a singleton face), but is not exposed. You can see this because if you intersect the set with an horizontal hyperplane you will get a bigger face, the one relying both stars.

2) For polytopes, it happens that all the faces are exposed. This is why in this case both definitions coincide. Note that this is also the case for any stricly convex set.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.