If $G =(V,E)$ is a simple, undirected graph (finite or infinite), and $\kappa \neq \emptyset$ is a cardinal, we say that the complete graph $K\_\kappa$ is a **minor** of $G$ if there is a collection ${\frak S}$ of connected and pairwise disjoint subsets of $G$ such that


1. $|{\frak S}| = \kappa$, and
2. whenever $S\neq T\in {\frak S}$, then there is $e\in E$ such that $(S\cap e) \neq \emptyset \neq (T\cap e)$.


The [Hadwiger-Nelson graph](https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem) has $\mathbb{R}^2$ as its ground set, and two elements $x,y\in\mathbb{R}^2$ form an edge if and only if their Euclidean distance equals $1$.


**Question.** Is the set of cardinals $\kappa \neq \emptyset$ such that $K\_\kappa$ is a minor of the Hadwiger-Nelson graph finite? If yes, what is its greatest member?


**Note.** Since the [Hadwiger conjecture](https://en.wikipedia.org/wiki/Hadwiger_conjecture_(graph_theory)) is correct for $k\leq 6$, the set asked for in the question is a super-set of $\{1,2,3,4,5\}$, because it is known that the chromatic number of the Hadwiger-Nelson graph is a member of $\{5,6,7\}$.



Hadwiger number of the Hadwiger-Nelson graph on $\mathbb{R}^2$
No, the set of such cardinals is infinite.


Take $\kappa$ points on the interval $(0, 1) \times \{0\}$. Every two points $(u, 0)$ and $(v, 0)$ have a common neighbour $$\left(\frac{u+v}{2}, \sqrt{1 - \frac{(u-v)^2}{4}}\right)$$ which is not adjacent to any other vertex on the interval. Thus the Hadwiger-Nelson graph has a $K\_\kappa$ minor.


Note that we can pick $\kappa$ to be any natural number (as well as $\aleph\_0$ or $2^\aleph\_0$), thus there are infinitely many cardinals $\kappa$ such that $K\_\kappa$ is a minor of the Hadwiger-Nelson graph.


