# Two definitions of a face of a convex set: are they equivalent?

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?

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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

## 2 Answers

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.

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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 $C \cap \partial H$.

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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