# One Puzzle with Two Totally Different Solutions

Peter Winkler‘s excellent book Mathematical Puzzles: A Connoisseur’s Collection has in it the problem of finding a partition of $\mathbb{R}^3$ into disjoint non-trivial circles. (Here “non-trivial” means “not a point.”) Winkler gives a very clever solution which is purely geometric.

Later, I read the same problem in Krzysztof Ciesielski‘s excellent book Set Theory for the Working Mathematician. In that book Ciesielski gives an almost purely set-theoretic solution.

I’ll discuss both solutions below. 　(Don’t read on yet if you want to think about the puzzle first.)

Geometric Solution: First observe that you can partition a 2-sphere (e.g., $\{(x,y,z) \mid x^2 + y^2 + z^2 = 1\}$) minus two points into disjoint circles. The easiest way to see this is to see that you can do it if you remove the north and south poles from the sphere by taking the circles to be the lines of latitude. Then observe that you can drag the two holes at the two poles to any other two locations on the sphere you want and allow the circles to follow. (For example, say that the two holes are still on different hemispheres. Then the circles will still radiate out from the holes, but their centers will gradually become closer and closer to the north or south pole, depending on the hemisphere that the hole is in.)

Given that, the partition of $\mathbb{R}^3$ is as follows: The first circles in the partition will be those in the $xy$-plane with center $(x,0,0)$ where $x\cong 1\text{ (mod }4)$ and with radius 1. Now notice that every 2-sphere centered at the origin meets these circles in exactly two points. Thus we may complete the partition by taking a partition of each of these 2-spheres into circles separately.

There is another (probably much better) explanation of this solution at cut-the-knot.

Set-theoretic solution: Let $\alpha$ be the first ordinal of cardinality $2^{\aleph_0} = |\mathbb{R}| = |\mathbb{R}^3|$. Pick a well-ordering $\{w_\beta\}_{\beta < \alpha}$ of $\mathbb{R}^3$ of length $\alpha$. We will build the partition of $\mathbb{R}^3$ by transfinite recursion along $\{w_\beta\}$. At each step $\beta$, we will define a circle $C_\beta$ such that $w_\beta\in C_\beta$ and $C_\beta\cap C_\gamma = \emptyset$ for $\gamma < \beta$ if $C_\beta\ne C_\gamma$. Hopefully, it’s clear that this suffices.

Here’s what to do at step $\beta$. First of all, if $w_\beta$ is in some $C_\gamma$ where $\gamma < \beta$ then let $C_\beta = C_\gamma$ and stop. Otherwise, pick a plane $P$ passing through $w_\beta$ which is not coplanar with any $C_\gamma$ for $\gamma < \beta$. This is possible since there are $2^{\aleph_0}$ planes through $w_\beta$ but only $|\beta|< 2^{\aleph_0}$ circles $C_\gamma$.

Now, each circle $C_\gamma$ intersects $P$ in at most 2 points. Since there are $2^{\aleph_0}$ circles in $P$ containing $w_\beta$ but only $|\beta|< 2^{\aleph_0}$ circles $C_\gamma$ there must be a circle in $P$ which contains $w_\beta$ and is disjoint from each $C_\gamma$ for $\gamma<\beta$. Let this circle be $C_\beta$. This completes the proof.

## 4 thoughts on “One Puzzle with Two Totally Different Solutions”

1. This is a fun observation! The set-theoretic solution gives a hint to the obvious follow-up question: Why 3? I don’t see how to get that information from the (very clever) geometric solution. It seems odd that the non-constructive solution gives more information than the constructive one. Perhaps the geometric solution is too clever?

2. mkoconnor says:

Great observation, I hadn’t thought of that. It did occur to me that the set-theoretic solution gives the extra information that any family of $\kappa$ disjoint circles can be extended to a partition when $\kappa < 2^{\aleph_0}$. The problem of getting a geometric solution even for $\kappa = \omega$ or $\kappa$ finite seems like it’s probably very hard.

3. Gauster says:

Indeed…and what is even more surprising, if I got it right, is that it’s even possible to have all circles to be the same size with the second construction. And here the geometric solution seems impossible.

4. Wojowu says:

Unfortunatelly, this proof uses axiom of choice. Without it, we can’t prove that space (or any other continuum) is well-orderable