# Is the number of circles in the Apollonian gasket countable?

Is it correct to say that the number of circles in an Apollonian gasket is countable becuase we can form a correspondence with a Cantor set, as their methods of construction are similar?

What about if we apply the Apollonian gasket construction inside of a fractal like the Koch snowflake? (I think that will still be countable.)

What if you did the Apollonian gasket construction between f(x) = sin(1/x) and g(x) = 2 - sin(1/x) between -1 and 1?(I think that will still be countable too, but it's not matching my intuition... which says "no way is that countable!")

Is there any closed curve that would result in the number of circles being uncountable?

What if we consider the Apollonian gasket made of spheres in $\mathbb{R}^3$?

(Please keep in mind I have only had two courses in Analysis. My apologies if any of this is too naive.)

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The interior of the circles/spheres are disjoint, and a set of disjoint open sets in a Euclidean space is at most countable. – Mariano Suárez-Alvarez Dec 9 '10 at 23:17
So, this is more of a topology question, then? Thanks. – futurebird Dec 9 '10 at 23:23
Your first sentence is puzzling, because the number of points in the Cantor set is not countable. – Rahul Dec 9 '10 at 23:32
@Rahul: I think the OP is referring to the steps in the construction of the Cantor set, not its points. – Qiaochu Yuan Dec 9 '10 at 23:35
Also, see if you can get a copy of Indra's Pearls klein.math.okstate.edu/IndrasPearls (or take one out from the CUNY math library) – deoxygerbe Dec 10 '10 at 2:03

The reason a collection of disjoint open sets in $\mathbb{R}^n$ is at most countable is that $\mathbb{R}^n$ is separable: it has a countable dense subset given by the points with rational coordinates, and any open set must contain such a point that the others don't contain, so there can be at most countably many of them.
There is also a measure-theoretic argument relying on the fact that $\mathbb{R}^n$ with the Lebesgue measure is $\sigma$-finite, which might come closer to agreeing with geometric intuition: if you could fit uncountably many open balls into $\mathbb{R}^n$, some closed $n$-cube (WLOG $[0, 1]^n$) would contain uncountably many of them, each of which would have positive measure. But a standard argument shows that this is impossible: if $S_k$ is the set of balls in $[0, 1]^n$ of radius at least $\frac{1}{k}$, then since $\bigcup S_k$ is uncountable it follows that some $S_k$ is uncountable, hence countable unions of its elements have unbounded measure; contradiction.
You can do it using measure, instead of topology, if you like. The circles (actually, open disks) in this collection are disjoint, but have positive measure. Space $\mathbb R^2$ is sigma-finite for $2$-dimensional Lebesgue measure. Therefore we have at most countably many disks in this collection.