# Reference: Euclidean Plane is not union of disjoint circles

Recently I came across a post over at MO (https://mathoverflow.net/questions/162324/covering-the-space-by-disjoint-unit-circles) that claimed Sierpinski proved that $$\mathbb{R}^2$$ cannot be expressed as the disjoint union of circles. I have been unable to find a reference, does anyone have an idea where one could find the details to this result?

EDIT: A general argument is fine, too, but I'm also interested in Sierpinski's original proof.

• Are you trying to find the proof by Sierpinski or just a general proof of this fact? Mar 20, 2019 at 4:51
• I've also tried, with no luck, to find Sierpinski's original proof, so I'm going to try adding a bounty. I've seen some general arguments before and I'll try to post a link to some sometime. Mar 22, 2019 at 18:04
• A more challenging problem is to show that $R^2$ cannot be partitioned as a disjoint union of topological circles. (Still true: math.stackexchange.com/questions/1657094/…) Mar 28, 2019 at 3:18
• You can look for Sierpiński’s papers (mainly in French) here. Mar 29, 2019 at 7:12
• I asked our local historians of mathematics (among them one topologist) about this paper and they suggested that this result may belong to Janiszewski. Apr 4, 2019 at 1:36

I think it should go like this. Suppose by contradiction that we can write $$\mathbb{R}^2$$ as a disjoint union of (nondegenerate) circles. Pick one such circle. It bounds a disk in $$\mathbb{R}^2$$, so call the closure of that disk $$K_1$$. Any point in the (nonempty) interior of the disk is on some other circle. By disjointness, that circle is entirely in the interior of $$K_1$$, so we get another compact set $$K_2 \subsetneq K_1$$. Actually we can construct an infinite chain of such sets, $$K_1 \supsetneq K_2 \supsetneq \dotsb$$. Notice that $$\operatorname{diam}(K_n) := \sup\{d(x,y) \mid x, y \in K_n\}$$ is positive, finite, and for $$n \ge 2$$ we can choose it to be strictly less than $$\frac12\operatorname{diam}(K_{n-1})$$. If all circles in the interior of $$K_{n-1}$$ are concentric this is obvious, and if not, by disjointness the sum of the diameters of two nonconcentric circles has to be smaller than $$\operatorname{diam}(K_{n-1})$$, so one of them has diameter smaller than $$\frac 12\operatorname{diam}(K_{n-1})$$.
By Cantor's intersection theorem, the intersection $$\bigcap_{n=1}^\infty K_n$$ is nonempty. I claim it has exactly one point $$p$$, since $$\lim_{n\to\infty}\operatorname{diam}(K_n) = 0$$. By the above argument, the circle containing $$p$$ must be contained in each $$K_n$$, so it would have to be contained in the intersection. This is a contradiction.
• I think you need a bit more footwork to show that $\operatorname{diam}(K_n)\to 0$ -- it looks like your construction only makes the sequence strictly decreasing. But if you replace "any point in the interior of the disk" with "the center of the disk", then you can squeeze it by a geometric sequence. Mar 22, 2019 at 18:58
• Note : to choose $K_n$ with the appropriate diameter, you don't actually have to make a case distinction, you can just pick a circle that contains the center of $K_{n-1}$: it has to have a small enough diameter if it doesn't meet the outer circle of $K_{n-1}$ Mar 22, 2019 at 19:09
• I was able to find scans of both papers of Sierpinski referenced on the Wikipedia pages for Sierpinski Triangle and Sierpinski Carpet. I don't speak French, but Google Translate convinces me that neither of them discuss decomposing $\mathbb{R}^2$ into circles. This one had cute pictures, so I'll leave the link babel.hathitrust.org/cgi/… Mar 22, 2019 at 22:24