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 $2$?
See below:
Recall wolfram's:
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
Best Answer
In your second image, the floating disconnected vertex is throwing off your total. Euler's Theorem only applies to connected graphs -- otherwise you could arbitrarily add as many isolated vertices as you want and make $F-E+V$ come out to any arbitrary whole number greater than or equal to $2$.