Polyhedra has more corners than facets

Question

Let $P$ be a polyhedron. Is it true that $P$ has always more/as many corners than facets? I haven't found a counterexample in $\mathbb R^2$ and $\mathbb R^3$ and intuitively I think the statement is true. Is there a proof?

Answer 1

In $\Bbb R^2$ the claim is trivial, since every polygon always has the same number of sides and vertices.

In $\Bbb R^3$, it's definitely false. For every polyhedron that has $F$ facets and $C$ corners, there is a so-called “dual” polyhedron that has $C$ facets and $F$ corners.

For example, the cube has 6 facets and 8 corners. Its dual is the regular octahedron, which has 8 facets but only 6 corners.

Did I misunderstand what you meant by "facets" and "corners"?