Polyhedron in $E^n$ with infinitely many sides must contain a side that is a polyhedral wedge So this is from Ratcliffe's text on hyperbolic manifolds. In it there is a part of a proof where he just states that for a polyhedron in $E^n$ with infinitely many sides it must contain a side that is a polyhedral wedge. I don't understand why this is true. **DEFINITION** : The definition of a polyhedral wedge is a polyhedron such that the intersection of all its sides is nonempty(i.e. think of a triangular wedge in $E^3$ with two planes meeting at a line). **EDIT** : I've included a copy of the proof in question. ![enter image description here]( **EDIT** : Sorry guys. I totally forgot to read all the assumptions in the lemma.

You don't seem to have understood the structure of the argument, which is by contradiction. Suppose $P$ is an _infinite sided_ polyhedron so that _all but finitely many sides_ of $P$ are wedges. That hypothesis already implies that there is at least **ONE** side which is a polyhedral wedge.