Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S \subseteq \mathbb{R}^d$ be a $d$-dimensional convex set (i.e. $\exists d+1$ affinely independent points in $S$). Let the origin of the coordinate system lie in the interior of $S$ and let: $$S^* = \underset{a\in S}\bigcap K(a) $$ where $K(a) = a \cdot x \le 1$

Then, $S^*$ is the polar of $S$. Also, if $S$ is a $d$-polytope in $\mathbb{R}^d$, then $S^*$ is a dual of $S$.

Since a $d$-polytope can be represented by a countably infinite set of points, this implies that the dual (which is also a polytope) is formed from the intersection of a countably infinite number of halfspaces --- which clashes with the definition of a polytope as the intersection of finitely many closed halfspaces.

What am I doing wrong here?

share|cite|improve this question
up vote 1 down vote accepted

If the polytope can be represented by a countably infinite set of points, the dual can be represented by the intersection of countably infinite number of halfspaces.

Which is true and not in contradiction with the existence of a finite representation.

share|cite|improve this answer
OK, I see. So that means there exists a representation of the dual with finitely many halfspaces as well? – Jacob Oct 30 '11 at 18:29
@P23 Yes, it is enough to take the halfspaces corresponding to the vertices of the original polytope. This is analogous to the fact that the original polytope is the convex hull of all its points, but also the convex hull of the finitely many vertices. – Phira Oct 30 '11 at 18:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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