We say that $S\subseteq \mathbb{N}$ is *meager* if $$\text{lim sup}\frac{S\cap\{1,\ldots, n\}}{n} = 0.$$


Given $S\subseteq \mathbb{N}$, we associate to $S$ the *sum-graph* $G\_S = (\mathbb{N}, E)$ where $$E = \big\{\{m,n\}: m,n \in \mathbb{N} \text{ and } m+n\in S\big\}.$$


Is there a set $S\subseteq \mathbb{N}$ such that


* $S$ is meager;
* $G\_S$ is connected;
* $\text{diam}(G\_S)$ is infinite?



Diameter of sum-graph over a meager set
Let $S=\{2^k-1: k \in \mathbf{N}\}=\{1\_2,(11)\_2,(111)\_2,...\}$. 


Clearly $S$ is sparse. 


Writing $m,n \in \mathbf{N}$ in their binary expansion, $m$ is connected to $n$ in $G\_S$ if $m>n$ and $n$ is obtained by bitwise inverting $m$. For example, $(11001)\_2$ is connected to $(00110)\_2=(110)\_2$. 
This shows that $G\_S$ is connected by repeatedly applying this procedure to an arbitrary element to find a path to 1.


To show that there is no bound on the diameter of $G\_S$, consider the sequence $(a\_k)$ given by $1,(101)\_2, (10101)\_2, (1010101)\_2,...=\{(2^{2k}-1)/3: k \in \mathbf{N}\}.$
Consider a path from $a\_n$ to 1. Whenever a digit of a number on this path is flipped, all digits to the right are flipped too. This means that if the left-most digit of $a\_n$, which is initially 1, is flipped $r$ times, then the second digit, which is initially 0, must be flipped at least $r+1$ times. By induction, $a\_n$ has distance precisely $2n-2$ from 1. 


