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parallel to one of the coordinate axes uses only three of the colors and moreover is almostvariable will range over all the remaining reals, but allbut countably many of these reals will have the samerelation . The point now is that there are only three ways for a given real to those sit with respect to two fixed reals, so every principal line has only three colors. For exampleFurthermore, if once we fix two reals, say $x$ and $y$, then for all but countably many $z$ above both of them in thewell-order, the order type of $(x,y,z)$ will be the same.And it is the same if we fix exceed both $x$ and $z$ or $y$ and y$, meaning that only one color is used for almost all$z$.z$ on that line. Thus, all such lines are monochromatic outside a countable

It follows that all the planes

Similarly, every plane parallel to the a principal axis planes will exhibit plane amounts to fixing one of the two-dimensional variables. If we fix $x$, say, then the coloring property mentioned 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 questionwell-order, will be determined solely by the relation of $y$ and $z$ in the well-order, putting us exactly in the two-dimensional situation.

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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 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 pointslines) 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, but all but countably many of these reals will have the same relation to those two fixed reals. For example, if we fix $x$ and $y$, then for all $z$ above both of them in the well-order, the order type of $(x,y,z)$ will be the same. And it is the same if we fix $x$ and $z$ or $y$ and $z$. Thus, all such lines are monochromatic outside a countable set.

It follows that all the planes parallel to the principal axis planes will exhibit the two-dimensional coloring property mentioned in the question.

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.

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 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 points) 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. (using We may use the first occurring case when the coordinates are not distinct, since this case doesn't actually matter). 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, but all but countably many of these reals will have the same relation to those two fixed reals. For example, if we fix $x$ and $y$, then for all $z$ above both of them in the well-order, the order type of $(x,y,z)$ will be the same. And it is the same if we fix $x$ and $z$ or $y$ and $z$. Thus, all such lines are monochromatic outside a countable set.

It follows that all the planes parallel to the principal axis planes will exhibit the two-dimensional coloring property mentioned in the question.

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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