Let $G=(V, E)$ be the following graph:


1. $V=\ell^\infty = $ set of bounded real sequences, with the norm $$\|x\|\_\infty = \sup\_{n\in\mathbb{N}}|x\_n|,$$
2. $E = \big\{\{x,y\}: x,y\in \ell^\infty \text{ and }\|x-y\|\_\infty = 1\big\}$.


The unit vectors $e\_i$ (defined by $e\_i(n) = 1$ for $i=n$ and $e\_i(n) = 0$ otherwise) form a clique, so $\chi(G) \geq \aleph\_0$. But do we have $\chi(G)=\aleph\_0$?



Hadwiger-Nelson problem for $\ell^\infty$
No. The set of all $\{0,1\}$-sequences is also a clique in $G$. Thus, $\chi(G) \geq 2^{\aleph\_0}$. On the other hand, the set of all bounded real sequences has size $2^{\aleph\_0}$, so $\chi(G)=2^{\aleph\_0}$. 


