The Hadwiger-Nelson problem: the Euclidean plane is not five-colorable

Every coloring of $\mathbb{R}^2$ with five colors, with arbitrary (possibly non-measurable) color classes, has two points at distance $1$ of the same color. With the classical seven-coloring this gives $6\le\chi(\mathbb{R}^2)\le 7$. The route is a transfer theorem showing that a proper $k$-coloring exists if and only if a 'weak measurable' $k$-coloring (monochromatic unit pairs of measure zero) exists, for every finite $k$, followed by a geometric obstruction to weak measurable five-colorings. It does not decide between six and seven, and it does not give a finite non-five-colorable unit-distance graph (one exists by de Bruijn-Erdos compactness, but none is exhibited).

Every coloring of $\mathbb{R}^2$ with five colors, with arbitrary (possibly non-measurable) color classes, has two points at distance $1$ of the same color. With the classical seven-coloring this gives $6\le\chi(\mathbb{R}^2)\le 7$. The route is a transfer theorem showing that a proper $k$-coloring exists if and only if a 'weak measurable' $k$-coloring (monochromatic unit pairs of measure zero) exists, for every finite $k$, followed by a geometric obstruction to weak measurable five-colorings. It does not decide between six and seven, and it does not give a finite non-five-colorable unit-distance graph (one exists by de Bruijn-Erdos compactness, but none is exhibited).