I am studying some theorems on combinatorial set theory, especially Ramsey theorem and Hindman's theorem. I think I am going to ask a silly question, but I am too much involved in the subject to think clear about it.
As in the title, I am searching for a finite colouring $c: \mathbb{N} \rightarrow \{1, \ldots, k\}$ such that there is no monochromatic set which is closed under addition.
I am convinced one can do this using only two colours. I have thought about this for too long now, but I found some necessary conditions.
The set of all even numbers can't be colored with the same colour, otherwise, it would be an infinite monochromatic set closed under addition. The same thing with every natural number k, the set $k\mathbb{N} = \{kn \mid n \in \mathbb{N} \}$ is closed under addition so you have to colour it with both colours. (every 'tail' has to contain two elements with different colours)
I have thought about defining a function $h: \mathbb{N} \rightarrow \mathbb{N}$, where $h(n)$ is the highest power of two in $n$, and by this function defining a colouring $c: \mathbb{N} \rightarrow \{1,2\}$ where $c(n) = 1$ iff $h(n)$ is even. But when I have to give a rigorous proof, I get stuck. Hopefully someone can help me out.
Color a number according to whether it is an odd number times an even power or an odd power of two. Note that $x$ and $2x=x+x$ have different colors. (Compare with this MO question.)
On the other hand, as you know, it follows from a result of Schur that for any finite coloring of $\mathbb N$ we can find monochromatic $x,y,z$ with $x+y=z$ and, in fact, Hindman's theorem gives us that there is a fixed color $c$ and an infinite set $H$ such that for any finite non-empty $S\subset H$, the sum of the elements of $S$ has color $c$.