CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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For a plane graph with $c$ components, $|V|-|E|+|F|=1+c$ , including the null graph

Statement

If a plane graph has $c$ connected components, then

$|V|-|E|+|F|=1+c.$

This includes the null graph, for which $c=0$ and the complement is its single face. Component sets and their finite cardinalities are those of The connected components of a graph partition its vertex set and are its maximal connected subgraphs and The cardinality $\lvert A\rvert$ of a finite set, and face finiteness is A plane graph has finitely many faces and exactly one unbounded face.

Facts & Assumptions

Given: A plane graph $G$ with components $G_1,\ldots,G_c$ when $c>0$ .

[L1]

For every connected plane graph, $|V|-|E|+|F|=2$ (Euler's formula $|V|-|E|+|F|=2$ for every connected plane graph).

[L2]

The vertex sets of the connected components of a graph are nonempty, cover the vertex set, and any two are equal or disjoint (The connected components of a graph partition its vertex set and are its maximal connected subgraphs).

[L3]

A finite graph is connected if and only if it has a spanning tree (A finite graph is connected if and only if it has a spanning tree).

[L4]

A tree on $n$ vertices has $n-1$ edges (For every forest, $|V|=|E|+c$ , where $c$ is the number of connected components).

[L5]

Every plane forest has exactly one face (Every plane forest has exactly one face).

[L6]

A cycle edge in a plane graph is incident with two distinct faces (Face frontiers are unions of whole edges; a cycle edge borders two faces and a bridge borders one).

Proof

technique · direct

1.1

Insert the connected component drawings one at a time in their inherited positions. Before a component is inserted, its connected drawing lies in one face of the components already inserted. By [L3] choose a spanning tree; [L4] gives it $|V(G_i)|-1$ edges, and [L5] shows that inserting this tree does not split the containing face. Add the remaining edges in their inherited drawing order. Each joins vertices already connected by the tree, hence lies on a cycle, so [L6] shows that its insertion splits one face into two. There are $|E(G_i)|-|V(G_i)|+1=|F(G_i)|-1$ such edges, where the last equality is [L1]. Starting from the one face of the empty drawing therefore gives $|F|=1+\sum_i(|F(G_i)|-1)$ . Vertices and edges add disjointly by [L2].

L1L2L3L4L5L6

2.1

Therefore $|V|-|E|+|F|=\sum_i(|V(G_i)|-|E(G_i)|+|F(G_i)|)-c+1=2c-c+1=1+c$ .

step 1.1algebra

3.1

For the null graph, $|V|=|E|=c=0$ and $|F|=1$ , so the same formula reads $1=1$ .

step 2.1∎

Depends on

Used by

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Sources

R. Diestel, Graph Theory, 6th ed., Chapter 4, Section 4.2 (standard reference, not scraped)

For a plane graph with ccc components, ∣V∣−∣E∣+∣F∣=1+c|V|-|E|+|F|=1+c∣V∣−∣E∣+∣F∣=1+c, including the null graph