Euler's formula doesn't work for null graph?

Question

Given the null graph with no edges or vertices, we have a connected planar graph as no edges cross when this graph is drawn in the plane, and the fact that any two distinct vertices have a path between them is vacuously true. However, Euler's formula doesn't work: plugging into $v+f= e+2$, we have $1=2$. Why is this the case? Can we not apply Euler's formula here?

Answer 1

We can slightly generalize Euler's formula as: $V-E+F-C = 1$, where $C$ is the number of components. Most of the time $C=1$, which gives us the familiar formula. But it works great if $C>1$. And the question is about the "trivial" case where $C=0$. Now of course $V=E=0$ and $F=1$, giving us the right answer.

There is a standard inductive proof of Euler's formula which involves either removing an edge and a face, or removing an edge and a vertex, and observing that the invariant remains the same. See here. But these proofs are quite complicated, because they are trying to preserve connectedness.

With the generalized formula, the proof can be much cleaner (still skating over elementary topology). Remove any edge. There are two disjoint possibilities:

(1) We merge two distinct faces, so: $E-1; F-1$

(2) The edge always had the same face on both sides, so removing the edge instead splits a component into two: $E-1; C+1$

In either case the invariant $V-E+F-C$ remains the same. So we remove all the edges, and then have a bunch of isolated vertex/components in a single face, so $V=C$ and hence $V-0+1-C = 1$.

Always nice when a formula can apply all the way down to $0$.