Given a positive integer $n\in \mathbb{N}$ we define the Hadwiger-Nelson graph $\text{HN}\_n$ by


1. $V(\text{HN}\_n) = \mathbb{R}^n$;
2. $E(\text{HN}\_n) = \{\{v\_1, v\_2\}: v\_1, v\_2 \in \mathbb{R}^n \text{ and } |v\_1-v\_2| = 1\}$.


Is it true that the clique number $\omega(\text{HN}\_n)$ equals $n+1$ for all $n\in \mathbb{N}$?



Hadwiger-Nelson problem in higher dimensions
Yes. The moral reason is that cliques are 1-skeletons of regular simplices. If you want convincing quickly then it's straightforward to show inductively that there is exactly one clique in $\mathbb R^n$ of order $n+1$ up to the obvious symmetries, because when you go up a dimension you have exactly two choices for where to place the extra point, and can't take both of them because they're not the correct distance apart.


As a crude upper bound we have $\chi(\mathbb R^n) \leq (\sqrt n + 1)^n$. Tile $\mathbb R^n$ by cubes of side length $1/\sqrt n$, give a $(\sqrt n + 1) \times \cdots \times (\sqrt n + 1)$ block of cubes $(\sqrt n + 1)^n$ different colours and extend periodically.


