Partition of plane into parabolas

The plane is partitioned into parabolas (each point belongs to exactly one parabola). Does it follow that their axes have the same direction?

Solution 1:

Lemma 1: If parabola A is inside parabola B, then A and B must have parallel axes.

("Inside B" means on the side that B curves towards, in other words, the convex hull of B is B and its "interior".)

Zooming out, a parabola looks more and more like a ray. For example, if we look at $y=x^2$, then zooming out by a factor of 100 (i.e. $y_{zoom}=y/100$ and $x_{zoom}=x/100$) turns it into $y_{zoom}=100x_{zoom}^2$.

If A's axis is not parallel to B's, then if we zoom out sufficiently, we will see A leaving the interior of B. But since A and B share no points, this cannot happen, so A's axis must in fact be parallel to B's.

(From the projective point of view, having non-parallel axes means touching different points on the line at infinity, which clearly can't happen if one is inside the other.)

Parabolic Regions

Given a parabola $A$ in the partitioning of the plane, let $R(A)$ be the union of all parabolae which contain $A$ or are contained by $A$. The notion of containment provides an ordering on these parabolae.

If the region $R(A)$ is the entire plane, then by lemma 1 all the axes are parallel.

If the region $R(A)$ is not the entire plane, then (considering the supremum under the ordering) it must be convex with a parabolic boundary (either open or closed).

[[EDIT: $R(A)$ could also be an open half-plane. If two regions are open half-planes, then no parabolae can fit in the gap between them, so at most one region can be an open half-plane. The following paragraph rules out even this possibility (since although the open half-plane region is not strictly convex, the other regions still are).]]

If it is open (not including its parabolic boundary), then every point on the boundary must be a member of another region. Since the regions are strictly convex, no two points on the boundary can be members of the same other region, and so an uncountable number of other regions are needed. However, since each region includes a rational point in its interior, there can only be a countable number of regions. Therefore, there cannot be an open region.

A Reduced Question

So the original question reduces to the question of whether it is possible to partition the plane into convex closed parabolic regions.

Such a partitioning is not possible. (We will follow Cantor here rather than Sierpiński). Since we can enumerate the rational points, this yields an enumeration of the parabolic regions (ordered by earliest interior rational point). This lets us define a sequence of shrinking open intervals on an arbitrary line: Starting with any interval I$_0$ that spans multiple parabolic regions, we define the next open interval I$_{n+1}$ as the interval between the first two parabolic regions (according to the enumeration) to intersect the interval I$_n$.

These intervals shrink towards either a limit point or a limit interval, and either way provides a point contained in all intervals. This point cannot be in any parabolic region -- the parabolic region would have appeared in the enumeration and so would have been used at some point to reduce the interval.

So the answer to the reduced question is no, it is not possible to partition the plane into convex closed parabolic regions.

Therefore the answer to the original question is yes, if the plane is partitioned into parabolas, then it follows that their axes must all have the same direction.

Solution 2:

Yes, they have a common axis direction.

Since two parabolas whose direction differs by an infinitesimal amount will always intersect, directions have to differ by some finite value. So you can consider families of parabolas with the same direction, and for these I will show that the axes have to indeed agree, which in turn leads to parabolas which intersect.

Note that I'm not overly concerned with the limit lines between the different families. According to your question, every such point on the boundary should belong to one of its adjacent families, but I make no such assumption. If you do enforce that assumption, some contradictions will become even easier to prove. My proof should still work if you allow for a set of points of measure zero which belongs to no parabola, which is a more general requirement.

I'm making two attempts to explain why parabolas within one family must intersect, one less and one more projective in nature.

Non-projective explanation

Suppose you had different directions for your axes. Take two in such a way that there is no other direction in between them. Consider the area between these two axes. A Parabola through a poin there will either curve left and belong to a parabola for the left axis, or it will curve right and belong to a parabola for the right axis. This is because curvature doesn't change sign along a parabola. Since parabolas don't intersect, the boundary between the two families can't belong to several different parabolas. Therefore the boundary has to be shaped like the limit of one of these families. And the limit between left-curving and right-curving parabolas is a straight line. So the boundary between the two families has to be a straight line.

When you concentrate on a single axis direction and the family that goes with it, the same argument will hold in the neighbour in the other direction as well. So you have a family of parabolas with the same axis direction and delimited by straight lines on both sides. These straight lines have to be tangents for every parabola in the family. If the axis direction is known, and two tangents are known, then there is only one parameter left to uniquely define every member of the family. For example, you could use the curvature at the vertex as that parameter. That curvature and the axis direction together define the parabola up to translation, and with the two known tangents the translation is uniquely determined. So you have a one parameter family of parabolas. Up to an affine transformation, this family has to look as follows:

As you can see, the parabolas all intersect, so this is a contradiction to your assumed partition. Therefore there can't be more than one direction for the axes of parabolas.

Now all that remains is showing that two different axes for the same direction are impossible as well. I'll leave that as an excercise for now, will add more later if I find the time. If someone else wants to edit this answer, feel free to do so.

Projective explanation

This is the first explanation I had. Here I'm thinking about this in terms of projective geometry, which might be the cause for my choice of vocabulary. A parabola in projective geometry is a conic which is tangent to the line at infinity. Its sole infinite point describes the direction of the axis of the parabola. Suppose there were two different parabolas in your family with different directions, and also suppose that between these two directions there are no others convered by your family. Then you'd have two different points at infinity covered by your family, and none between these two. In the plane, you'd have some points where the corresponding parabolas bend left to end up at the left inifinite point, and others where they bend right to the right infinite point. In between these two, there have to be points where there is no bend at all, i.e. a straight line. So the boundary between those parabolas with one direction and those with the other direction has to be a straight line.

For a given direction, the associated parabolas are limited by a line on both sides, since directions are cyclic. So you are talking about a family of conics which have three tangents in common, and even have the touching point for one tangent in common as well, since they have a common direction. This amounts to four conditions, or four real degrees of freedom. Choosing a conic you have five degrees of freedom, so your family is a one parameter family. After a projective transformation, it has to look like this:

The red point represents the point at infinity, the black lines are the common tangents. As you can see, these conics do intersect. There is no choice, hence the assumption of different directions leads to a contradiction to the assumed partition.

Showing that two different axes for the same direction are impossible was again left as an excercise.