Proving the easy direction of Steinitz's theorem.

Question

Long time lurker here, first time poster. I am trying to prove that the graph of a convex polyhedron is 3-connected, as a homework problem.

So far I know that every convex polyhedron has a tetrahedron as a minor, and I think I also know that every vertex of a convex polyhedron has at least degree 3. I am unable to put together a proof using this information though, and I feel as if there must be some fact about convex polyhedra that I simply am not aware of.

Can anyone provide a hint or some advice?

Answer 1

Here's a sketch of a proof:

Given a vertex $v$, all edges not containing $v$ of all faces containing $v$ form a simple cycle in the graph and any edge containing $v$ has the other vertex on that circle.

Now we remove 2 arbitrary vertices and consider any simple path between some other 2 vertices. There are 4 cases:

• No removed vertices were lying on the path
• Only one of the removed vertices was lying on the path
• Both vertices are lying on the path and are not next to each other
• Both vertices are lying on the path and are next to each other.

Show that you can still connect the vertices by some path. Use the fact that vertices connected to the removed ones lie on the correspondent circles. For the last one show that the circles intersect.