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Sierpinski showed that, on the assumption of CH (in fact, equivalently to it), each point in the plane can be coloured (say) black or white so that every section of the plane parallel to the $x$ axis is "almost" white $-$ in the sense that all but countably many points of it are white $-$ while every section parallel to the $y$ axis is almost black. Equivalently, given CH, every line through the origin can be almost white, while every circle centred on the origin is almost black. Is there a corresponding result for three-dimensional space? For example, assuming CH, can we bicolour space so that every plane through the origin is almost white, while every sphere centred at the origin is almost black?

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up vote 7 down vote accepted

The answer to the last sentence in the question is no, for a silly reason, namely that a plane through the origin and a sphere about the origin have an uncountable intersection, in which only countably many points could be white and countably many black. The obvious way to evade that silliness is to replace planes and spheres with two families of sets, say $X$ and $Y$, such that the intersection of any set in $X$ and any set in $Y$ is countable. If you assume that, and if each of the families $X$ and $Y$ contains only $\aleph_1$ sets, then Sierpinski's construction will work.

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@Andreas Blass. Thanks! Yes, of course you are right. What I was wondering was whether there was a decomposition into two families, as you described, that were geometrically recognizable. Just mapping the plane to 3D space via some arbitrary bijection would carry over the planar decomposition intact: but this wouldn't be interesting. I think that your answer points at least to $X$ and $Y$ not being of the same dimensionality. – John Bentin Jan 2 '11 at 22:04

Here is a way to realize a positive version of the phenomenon:

Theorem. If the CH holds, then there is a coloring of $\mathbb{R}^3$ into six colors, such that every line parallel to one of the coordinate axes uses only three of the colors and moreover is almost monochromatic, meaning that it uses only one color for all but countably many points. (And each major axis direction will give rise to distinct two of the six colors.) Furthermore, the coloring on every plane parallel to the principal axis planes uses only two colors (except on countably many lines) and exhibits the two-dimensional Sierpinski property mentioned in the question.

The proof idea is simply to extend the two-dimensional proof idea to three dimensions. Namely, if CH holds, then well-order the reals in order type $\omega_1$. In the two dimensional case, we may color a pair $(x,y)$ black or white according to whether $x\lt y$ or not with respect to that well-ordering. Every vertical line is almost black, since all but countably many $y$ are above any given $x$ in the well-order, and every horizontal line is almost white, since only countably many $x$ are below any fixed $y$.

In three dimensions, we color a triple $(x,y,z)$ according to the order type of the triple with respect to the well-order, that is, according to whether $x\leq y\leq z$ or $y\leq x\leq z$ and so on, using the well-order relation here (not the usual order on $\mathbb{R}$), for each of the six possibilities. (We may use the first occurring case when the coordinates are not distinct, since these exceptional cases don't affect the final property.) Any line parallel to one of the principal axes corresponds to fixing two of the variables. The third variable will range over all the remaining reals. The point now is that there are only three ways for a given real to sit with respect to two fixed reals, so every principal line has only three colors. Furthermore, once we fix two reals, say $x$ and $y$, then all but countably many $z$ will exceed both $x$ and $y$, meaning that only one color is used for almost all $z$ on that line. Thus, all such lines are monochromatic outside a countable set.

Similarly, every plane parallel to a principal axis plane amounts to fixing one of the variables. If we fix $x$, say, then the coloring on that plane corresponds to the colorings of $(y,z)$, which outside the exceptional lines occuring when $y\leq x$ or $z\leq x$ in the well-order, will be determined solely by the relation of $y$ and $z$ in the well-order, putting us exactly in the two-dimensional situation.

A similar phenomenon arises in every higher dimension $\mathbb{R}^n$, simply by using more colors: one should simply color $(x_1,\ldots,x_n)$ by the order type of these elements with respect to the fixed well-ordering.

Finally, let me say that one can avoid the CH assumption in this entire discussion, if one simply re-interprets the phrase "almost all" as meaning "fewer than continuum many" rather than "countably many". One can achieve this by well-ordering the reals in order type continuum rather than order type $\omega_1$, and then applying the same argument. The crucial fact is that every initial segment of such a well-ordering has size less than the continuum.

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@Joel: Thank you very much for this interesting answer, which has given me much to study and think about. As a first thought, from Andreas's answer, it seems there is no geometrically coherent 2-colouring, while your answer shows a geometrically nice 6-colouring. Perhaps the latter is optimal, in the sense of using the fewest colours (while being geometrically simple). – John Bentin Jan 3 '11 at 0:11

Hamkins' theorem can be pushed down to 3 colors. Give $(x_1,x_2,x_3)$ color $i$ where $x_i$ is the least coordinate according to the $\omega_1$-type ordering. (Break ties by choosing the least possible $i$.) Every plane perpendicular to the $x_i$-axis will be colored $i$, except possibly on countably many lines. Every line parallel to the $x_i$-axis will colored $i$ on at most countably many points; the rest of the line will be monochromatic in another color. If we switch to spherical coordinates, then we get an analogous theorem involving spheres, cones, planes, rays, and circles.

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Hi David. Welcome to MO! – Andrés E. Caicedo Jan 9 '11 at 2:35
@David Milovich: Thank you! This is a very nice answer. – John Bentin Jan 9 '11 at 11:45
Yes, very nice! – Joel David Hamkins Jan 9 '11 at 18:46

Sierpinski himself proved the following striking version of his theorem for the plane:

CH is equivalent to the statement that $\mathbb R^3$ can be written as $A_x\cup A_y\cup A_z$ where each $A_u$, $u\in\{x,y,z\}$, has a finite intersection with every line parallel to the $u$-axis.

(See, respectively [Simms, John C.; Sierpinski’s theorem. Simon Stevin 65 (1991), no. 1-2, 69–163]. I believe this theorem is proved in Sierpisnki's 1951 paper in Fund. Math. 38, pp. 1--13, but I would have to actually walk to the library to verify this (and decipher enough of the french in the article).)

There is actually a planar version of this result of Sierpinski, due to Komjath. A set $A\subseteq\mathbb R^2$ is a cloud if there is $p\in\mathbb R^2$ such that every line through $p$ meets $A$ in only finitely many points.
CH is equivalent to $\mathbb R^2$ being the union of 3 clouds.

In both cases the striking fact is that we go down from countable to finite, which is somehow less arbitrary, in the light of Joel's comment that the previous constructions can be done in ZFC if countable is replaced by $\lneq 2^{\aleph_0}$.

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Stefan, you can access Sierpinski's paper at the following URL: The theorem of his you mention is Theorem 2 of his paper. – Ali Enayat Jul 22 '11 at 17:27

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