How does a graph with one vertex have one face? I understand Euler's theorem polyhedron $V-E+F=2$. However, I don't understand what a polytope and hyperplane is so I can't understand the topologically complicated definition of face. If I lookup a dumbed-down version of what a "face" is, it's too hand-wavy.
Let's say I have the following graph with $6$ vertices, $6$ edges, and therefore $2$ faces. I see how the triangular-like region formed by $4$ vertices makes up a face. I however don't understand how the extra 1 face was counted. See image drawn below:
Next, let's remove an edge below. Now the transformed graph below has $1$ face total. How the heck did this happen? How does it have $1$ face and not $2$?
Generally, a face is a component polygon, polyhedron, or polytope. A two-dimensional face thus has vertices and edges, and can be used to make cells. More formally, a face is the intersection of an $n$-dimensional polytope with a tangent hyperplane. Zero-dimensional faces are known as polyhedron vertices (nodes), one-dimensional faces as polyhedron edges, $(n-2)-D$ faces as ridges, and $(n-1)$-dimensional faces as facets.
and the "Face" in dumbed-down speak
is regions bounded by edges, including the outer, infinitely large region
What infinitely large region does wikipedia talk about on https://en.wikipedia.org/wiki/Planar_graph? Is every graph bounded by an infinitely larger graph? Does that mean that an infinitely large graph is a non-existent because it is an element and subset of itself and therefore not an element of itself? Russell's paradox