Convex polyhedron with 5 vertices/10 edges, convex polytope with 6 vertices/12 edges. Does there exist a convex polyhedron in $\mathbb{R}^3$ with 5 vertices and 10 edges? How about a convex polytope in $\mathbb{R}^4$ with 6 vertices and 12 edges?
 A: The answers are no and no. For the first, consider the graph of the polyhedron. It has $5$ edges and $10$ vertices – but the only graph on $5$ vertices with that many edges is $K_5$. This is a contradiction since the graph of a polyhedron is necessarily planar (which $K_5$ is not): projection from the inside of the polyhedron to a sphere outside it embeds the graph without crossings in $S^2$, stereographic projection then embeds it without crossings in $\mathbb{R}^2$.
For the second the proof gets a little bit more involved. First note that there are at least $4$ edges that connect to each vertex of a $4$-polytope. (And similarly in higher dimensions.) On the other hand, each edge connects $2$ vertices. By counting vertex-edge incidences in two ways, we see that there are at least twice as many edges as vertices. But in this case, we have equality, so each vertex lies on exactly $4$ edges.
Now choose any facet of the $4$-polytope. This will be a $3$-polytope, and so have at least $4$ vertices. Clearly, it can not have $6$ vertices, for then the polytope would not be full-dimensional. If it had $5$ vertices, then the polytope would be a cone over this facet (since there is only a single vertex left). But this vertex would be connected to all 5 other vertices, which is a contradiction. So the facet is a $3$-simplex. But consider the graph of this polytope. Each vertex of the $3$-simplex is connected by an edge. This takes $6$ edges. Each of the remaining $2$ vertices is connected to $4$ others, which takes at least $7$ edges. But $6 + 7 > 12$, contradiction.
