Kuratowski's Theorem

Theorem
The following conditions on a graph $$\Gamma$$ are equivalent:

(a) $$\Gamma \ $$ is planar;

(b) $$\Gamma \ $$ contains no subdivision of either the complete graph $K_5 \ $ or the complete bipartite graph $K_{3,3} \ $.

Proof
Let $$\Gamma$$ be a graph with $$V$$ vertices, $$E$$ edges and $$F$$ faces.

The proof proceeds in two parts: (a)$$\Rightarrow$$(b) and (b)$$\Rightarrow$$(a).

(a)$$\Rightarrow$$(b)
First we consider $$K_5 \ $$.

$$K_5 \ $$ has 5 vertices and 10 edges, so by the Euler Polyhedron Formula, a planar embedding would have 7 faces.

But each face has at least 3 edges, while each edge bounds at most two faces.

If we count the incident edge-face pairs, the number of faces is at most (Edges)* 2/3 = 6 + 2/3, a contradiction. Hence $$K_5 \ $$ is non-planar.

Now consider $$K_{3,3} \ $$.

This graph has 6 vertices and 9 edges, and hence 5 faces if it is planar.

Since this graph is bi-partite, each closed Euler trail has an even number of edges.

Hence any face must have at least 4 edges, and so the number of faces is at most (Edges)*2/4 = 4.5, a contradiction.

Hence $$K_{3,3} \ $$ is non-planar.

Now consider a subdivision of either graph.

The arguments as above continue to work as before, since both $$V$$ and $$E$$ have increased by one in the Euler formula, and so the expected value of $$F$$ is unchanged.

Hence if a graph $$\Gamma \ $$ contains a subdivision of $$K_5 \ $$ or $$K_{3,3} \ $$, then $$\Gamma \ $$ is not planar, and taking the contrapositive, we have:

$$\Gamma \ $$ is a planar graph $$\Rightarrow \Gamma \ $$ does not contain a subdivision of $$K_5 \ $$ or $$K_{3,3} \ $$.