Infinite sets of nonintersecting discs is countable on a plane Prove that any infinite set of non-intersecting discs on the plane is countable. I know that every disk contains rational points and hence there is an objective function from the set of disk to the set of points with rational coordinates. But how would I show that each disk can be mapped to one point with rational coordinates 
 A: If you're concerned about how to pick one point with rational coordinates out of the infinitely many such points in the disk (and you don't want to rely on the Axiom of Choice), take a particular enumeration of the pairs of rationals and choose the first one that is in your disk.
A: Yes, you can solve it using that fact.  If we define a surjection from the rationals onto this set of disks, we will have shown it to be countable.  For each rational $q$, if $q$ is contained in a disk then map $q$ to that disk.  Otherwise, map it to any disk you like.  Since the disks are disjoint, this map is well-defined.  The map hits every disk because every disk contains a rational (as you pointed out).
A: for any disc $D \in \mathbb{D}$ choose an arbitrary origin and let $\chi_r(D)=1$ if $D$ is entirely contained in the ball of radius $r$ centered at the origin, and $\chi_r(D)=0$ otherwise. define $\rho(D)$ as the radius of a disc.
now define
$$
C_n = \{D: \chi_n(D)=1 \land \rho(D) \ge \frac1{n}\} 
$$
then, since the discs do not overlap, a simple constraint on areas implies that
$$
|C_n| \lt n^4
$$
thus each disc in $C_n$ may be assigned a unique number $M_n(D)$ less than $n^4$ 
for each disc define $L(D) = \min \{n|D \in C_n\}$ and set $L'(D) = M_{L(D)}(D)$
in this way every disc is assigned a unique pair of integer 'co-ordinates' $(L(D),L'(D))$
