Prove that a convex $d$-polytope has at least $d+1$ facets

Question

This seems trivial but I can't come up with a formal proof.

I think there should be a way to do this inductively but I can't figure out how$\ldots$

Any help much appreciated

Answer 1

The polar of the given $d$-polytope is a $d$-polytope, which by dimension considerations has at least $d+1$ vertices.

Answer 2

We will prove by an induction

i) If $P$ is bounded convex $d$-dimensional polytope in $\mathbb{E}^d$, then the case of $d=2$ is clear

ii) Assume that $d=k$ holds Now consider $d=k+1$ Fix a vertex $v$ Then there is a hyperplane $H$ passing through $v$ s.t. $P\cap H=\{v\}$ Consider a parallel hyperplane $H_\epsilon$ wrt $H$ so that $H_\epsilon$ divides $P$ into two $d$-dimensional convex polytopes $P_i$ If $P_1$ contains $v$, then $${\rm number\ of\ facets\ in}\ P_1 \leq {\rm number\ of\ facets\ in}\ P$$

Note that $P_1\cap H_\epsilon$ is $k$-dimensional polytope Hence $P_1$ has at least $k+2$ facets

Answer 3

If you want to prove this more elementary(i.e. without polarity) I would suggest the following approach: Let $P$ be a $d$-polytope with at most $d$ facets. Thus we can write $P = \{x: Ax \leq b\}$ such that $A$ has at most $d$ rows (why?). Then show that $P$ is not a polytope to yield a contradiction. To do so find an unbounded ray contained in $P$. It might be easier to apply an (invertible) linear transformation to $P$ first.