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In the Euclidean plane $\mathbb{R^2}$, the set of points inside a circle is a disk. Can we claim that every set of non-overlapping disks in the plane is at most countable?

My intuition says it must be countable by choosing some rational points from the disk but I am not sure about my claim.

Thanks for your help .

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That's absolutely the way to go! Enumerate the rational points in some sequence, and for each disk $D$ let $n_D$, be the least positive integer $n$ such that the $n$th rational point in the sequence lies in $D$. The map $D\mapsto n_D$ gives an injection into the positive integers from the set of disks.

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$ – monalisa Apr 19 '13 at 10:46
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We need an injective function from a given set of disks into the countable set $Q \times Q$. The idea is to pick in each disk a point both of whose coordinates are rational numbers. Since the disks are disjoint, no pair of rational numbers is repeated for di fferent disks, so the association disk to (element of $Q \times Q$) is one-to-one.

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