Construct Discrete Sequence in Complex space Consider an arbitrary open subset $U \subset \mathbb{C}$. I intend to construct a sequence $(a_i)_{i \in \mathbb{N}}$ contained in $U$ with following two properties:


*

*for every rational point $q=q_R+q_I i \in U$ (therefore $q_R,q_I \in \mathbb{Q}$) with positive distance to the boundary $dist(q, \partial U):= r_q >0$ the ball $B_{r_q}(q)$ contains infinitely many $a_k$

*$(a_i)_{i \in \mathbb{N}}$ is discrete in U (so the sequence doesn't contain a subsequence which konverges to a point in $U$)
My ideas: 
since we consider only rational points in $U$ their set countable so of same cardinality as $\mathbb{N}$.
Can we construct a "zigzag"-sequence (as in proof that the $\mathbb{Q}$ have the same cardinality as $\mathbb{N}$ but with one "$\mathbb{N}$"-axis and and one axis containing all rationals from $U$?
Then on the "$\mathbb{N}$"-axis the $a_i$ "running" outwards from $q$ to a point $\partial B_r(q) \cap \partial U$. 
Does this work?
I is there exist a more elegant/ canonical way to abtain a sequence with desired properties?
 A: For $n \in \mathbb N$, let $C_n=\{ z \in U : d(z,\partial U)=\frac{1}{n}\}$.
We will define the set $A=(a_i)_{i \in I}$ as a disjoint union of countable discrete sets $A_n \subset C_n$. It is easy to see that any set $A$ defined in this way has no accumulation point in $U$.
Let $E_n$ denote the set of balls $B(q, c_q)$ where $c_q:=d(q,\partial U)$ and $q \in U$ has rational real and imaginary parts, such that $B(q,c_q) \cap C_n \neq \emptyset$ (or equivalently such that $2c_q > \frac{1}{n}$). 
We choose $A_n$ so that for every ball $B \in E_n$, $A_n \cap B$ contains at least one point, but in a way that $A_n$ is discrete in $C_n$. This can be done in the following way: choose an enumeration $B_1, \ldots, B_k \ldots $ of $E_n$. Define $C_k$ inductively so that $C_0=\{q_0\}$, and $C_{k+1}=C_k \cup \{q_{k+1}\}$ if $C_k \cap B_{k+1} = \emptyset$ (here $q_k$ is the center of $B_k$). Then take $A_n = \bigcup_{k \geq 0} C_k$. This set is indeed discrete in $C_n$ since all points are at distance at least $\frac{1}{2n}$ from each other by construction.
Finally, $A=\bigcup_{n \geq 1} A_n$ works. Indeed, it is countable, discrete, and for any $B(q,c_q)$ with rational $q \in U$, $B_q$ belongs to an infinity of $A_n$ and therefore contains infinitely many points of $A$.
