Five Color Theorem

Theorem
Any planar graph $$G \ $$ can be assigned a $k$-coloring such that $$k \le 5$$.

Proof
It is obvious the theorem is true for a graph with only one vertex. We will induct on the number of vertices.

Since $$G \ $$ is planar, $$\chi(G) = 2 \ $$ by the Euler Polyhedron Formula.

Suppose every vertex of $$G \ $$ is incident on $$6$$ edges or more.

Then $$2 = \chi(G) = v - e + f \le v - (3v) + f \ $$, so $$f \ge 2+2v \ $$.

But each face is bounded by at least $$3$$ edges, and each edge bounds at most $$2$$ faces, so so $$f \le \tfrac{3e}{2} \ $$ and hence $$3 e \ge 4 + 4v \ $$

Therefore, $$G \ $$ has at least one vertex with at most $$5$$ edges.

Remove that vertex $$x \ $$ from $$G \ $$ to create another graph, $$G ' \ $$.

By the induction hypothesis, $$G' \ $$ is five-colorable.

If all five colors were not connected to $$x \ $$, then we can give $$x$$ the missing color and thus five-color $$G \ $$.

If all five colors were connected to $$x \ $$, we examine the five vertices $$x \ $$ was adjacent to, and call them $$y_1, y_2, y_3, y_4, y_5 \ $$.