This is inspired by an older (as yet unanswered) [question](https://mathoverflow.net/questions/383891/graph-of-number-pairs-summing-to-a-square-number).


Let us call a set $S\subseteq\omega$ *thin in the 1st sense* if $$\lim\sup\_{n\to\infty}\frac{|S\cap n|}{n+1}=0$$ where $\omega$ is the first infinite cardinal, and $n=\{0,\ldots,n-1\}$ for all $n\in\omega$ with $n>0$.


Moreover, to $S\subseteq \omega$ we associate a simple, undirected graph $G\_S=(\omega, E\_S)$ where $$E\_S = \big\{\{a,b\}:a\neq b\in \omega \textrm{ and }a+b\in S\big\}.$$


Let us call $S\subseteq \omega$ *thin in the 2nd sense* if the chromatic number $\chi(G\_S)$ is finite.


**Question.** Are there any implications between these two notions of thinness?



A notion of thinness for subsets of $\omega$, using chromatic number
The two notions are incomparable.


To see that the first notion does not imply the second, let's construct a set $S$ with asymptotic density $0$, but with infinite chromatic number. We place infinitely many increasingly large intervals into $S$, but spaced very far apart, so that the density is zero. If $S$ has an interval centered at $n$ of size $k$, then all numbers within $k/2$ of $n/2$ will be connected to the others in your graph. This will cause a complete subgraph in $G\_S$ of size $k$. Since $k$ becomes as large as desired, the chromatic number of $G\_S$ will be infinite.


Conversely, to see that the second notion does not imply the first, consider the set $S$ of odd numbers, which has density $1/2$. If $a+b$ is odd, then the parities of $a$ and $b$ must differ. So every edge in $G\_S$ connects an odd number with an even number, and never two even numbers or two odd numbers. So the graph is $2$-colorable.


