Show that a discrete set is at most countable How to show that a discrete subset $A$ of $\mathbb C$ is at most countable? Ok I read the relevant questions here but I still don't understand how I should create an injection from $A$ to $\mathbb Q$. 
If $a \in A$ then I can find a rational number $q(a)$ such that $D(a,q(a)) \cap A = $ {$a$}. But the map $a \mapsto q(a)$ is not necessarily injective because for example take {$1,2$}: it is discrete and I can for example take the rational number associated to each number to be $1/4$. So we don't necessarily have an injection unless something is done. Please explain in detail without hints how to construct this injection 
thanks
 A: Since $\Bbb Q$ is countable, so is $\Bbb Q^4$, which means the set of all open rectangles with rational endpoints, $\{(a,b)\times(c,d)\mid a,b,c,d \in \Bbb Q\}$ is also countable. Such sets form a topological basis for $\Bbb R^2 \equiv \Bbb C$. Since $A$ is discrete, for each element $a$, there is a basis set containing $a$ that does not contain any other element of $A$. Therefore the cardinality of $A$ is less than or equal to the cardinality of $\{(a,b)\times(c,d)\mid a,b,c,d \in \Bbb Q\}$.
A: Assume that $\phi:\>{\mathbb N}\to{\mathbb Q}^2$ is a counting of the points with rational coordinates in ${\mathbb R}^2$.
By assumption the set $A\subset{\mathbb R}^2$ is discrete. This means that for any point $p\in A$ there is a $$\delta(p):=\inf\bigl\{{\textstyle{1\over2}}|z-p|\>\bigm|\>z\in A, \ z\ne p\bigr\}>0\ .$$
It follows that $U_{\delta(p)}(p)$ contains  points $z$ with rational coordinates and among them one with smallest number $\phi^{-1}(z)$. In other words, we have a function
$$\psi:\quad A\to{\mathbb N},\qquad p\mapsto\min\bigl\{\phi^{-1}(z)\>|\>z\in {\mathbb Q}^2\cap U_{\delta(p)}(p)\bigr\}\ .$$
Since $U_{\delta(p)}(p)\cap U_{\delta(q)}(q)=\emptyset$ when $p\ne q$ it follows that $\psi$ is injective, hence bijective onto $\psi(A)\subset{\mathbb N}$. This proves that $A$ is countable.
