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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 .

monalisa
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  • For a generalization to separable spaces, and for a counterexample which shows that a space need not be separable if it has this property, see http://math.stackexchange.com/questions/356334/separable-space-and-countable – Stefan Hamcke Apr 19 '13 at 10:02

2 Answers2

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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 different disks, so the association disk to (element of $Q \times Q$) is one-to-one.

Srijan
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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.

Cameron Buie
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