If $V$ is a subset of $\mathbb{N}$ such that for every pair of numbers $2n-1$ and $2n$ exactly one number is in $V$ , then $V\in\mbox{CES}$ with $\gamma(V)=\tfrac{1}{2}$ . Let $V_{1}$ be the set of odd numbers. Let $V_{2}$ be the set that includes $1$ , and all even numbers $2^{2k-1}<n\le2^{2k}$ , and all odd numbers $2^{2k}<n\le2^{2k+1}$ . So, $$V_{1}=\{1,4,5,7,10,12,14,16,17,19,\ldots,31,\ldots\}.$$ Then, the intersection $V_{1}\cap V_{2}$ includes $1$ and all odd numbers $2^{2k}<n\le2^{2k+1}$ . Let $$r_{n}=\tfrac{\#(V_{1}\cap V_{2}\cap\{1,2,3,\ldots,n\})}{n}.$$ Then, $$\begin{eqnarray*} r_{2^{1}} & = & \tfrac{1}{2},\\ r_{2^{2}} & = & \tfrac{1}{4},\\ r_{2^{3}} & = & \tfrac{3}{8},\\ r_{2^{4}} & = & \tfrac{3}{16},\\ r_{2^{5}} & = & \tfrac{11}{32},\\ r_{2^{6}} & = & \tfrac{11}{64},\\ & \cdots \end{eqnarray*}$$ In particular, $$\begin{eqnarray*} r_{2^{2k+1}} & = & \tfrac{1+2+\ldots+2^{2k-3}+2^{2k-1}}{2^{2k+1}}\\ & = & \tfrac{1}{3}\tfrac{2^{2k+1}+2}{2^{2k+1}}\\ & \to & \tfrac{1}{3},\\ r_{2^{2k+2}} & = & \tfrac{1}{2}r_{2^{2k+1}}\\ & \to & \tfrac{1}{6}. \end{eqnarray*}$$ Therefore, the limit $\gamma(V_{1}\cap V_{2})$ does not exist.