# Generalisation of Kuratowski's theorem

So I've recently read the infinite graph version of Kuratowski's theorem. It says that a graph $G$ is planar if and only if the following three conditions holds:

1. $|V(G)| \le |\mathbb{R}|$
2. $G$ has at most countably many vertex with degree at least 3
3. $G$ has neither $K_{3,3}$ nor $K_5$ subdivision

It is clear that if a graph $G$ is planar, then all three conditions must hold. To show its converse, I assume that there is a graph that satisfies all three condition and yet it is not planar. I have eliminated condition 1 and 3 as a source of non-planarity of $G$, but I cannot derive a contradiction with 2. Any help would be greatly appreciated.

• I don't see how you can reduce to a graph with countably many vertices and edges. For example, let us start with the graph consisting of vertices $a$, $b$, and continuum many degree-2 vertices connected to $a$ and $b$. How does the reduction proceed? – Emil Jeřábek supports Monica Apr 14 '18 at 6:54