Does my set have empty interior?

Question

I am trying to figure out if my set has empty interior or not. It is defined as follows. Let $(q_n)_{n \in \mathbb{N}} = \mathbb{Q}^2$, we define $U = \bigcup_{n \in \mathbb{N}} B_{\frac{1}{2^n}}(q_n) \subset \mathbb{R}^2$ and $A = U^C$. Now we fix $x \in \mathbb{R}\setminus \mathbb{Q}$ and define $C = A \cap (\{x\} \times \mathbb{R})$. Then we can view $C$ as subset of $\mathbb{R}$ and I wonder if $C$ has any interior points in $\mathbb{R}$. I mean I can always find a point $q_n \in U$ which is arbitrary close to $C$ in $\mathbb{R}^2$. But the radius of its ball $\frac{1}{2^n}$ could be even smaller. I can't say if in every neighbourhood of a $y \in C$ is an intersection with a ball $B_{\frac{1}{2^n}}(q_n)$. Does anyone have ideas how to tackle that?

Answer 1

In general, there may be no interior points. Let $(r_n)=\mathbb Q$ be an enumeration of the rationals. Let me now construct balls that will contain the point $(x,r_n)$. For each $n$ there is a rational number $s_n$ such that $|s_n-x| \le \frac1{2^{2n+1}}$. Now set $q_{2n}:=(s_n,r_n)$, and fill all other rational points of $\mathbb R^2$ into $(q_{2n+1})$. Then $(x,r_n) \in B_{\frac1{2^{2n}}}(q_{2n})= B_{\frac1{2^{2n}}}(s_n,r_n) $. The points $(x,r_n)$ are dense in $\{x\}\times \mathbb R$ but not in $C$. So $C$ has no interior points.

[This is a similar construction to proof that subsets of separable sets are separable.]