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

user64494
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  • Let the families of such parabolas be y=$x^2$+t; t $\in \mathbb R$

    For any $x_o,y_o$ on the plane there exists one and only one parabola given by $t=y_o-x_o^2$

     – ARi Oct 05 '13 at 09:09
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    @ARi: The question is not to find a partition of the plane into parabolas. – user64494 Oct 05 '13 at 10:29
  • If their axis are not in the same direction then the curves will intersect and the point of intersection belong to two parabolas. – ARi Oct 05 '13 at 13:43
  • @ ARi: Consider $y=x^2$ and $x=y^2+1$. – user64494 Oct 05 '13 at 15:16
  • Using AC may give you an elegant proof here. – ARi Oct 07 '13 at 14:11
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    Degenerate parabolas (doubly covered rays) have to be excluded. Consider the following example: Parabolas of the form $y=x^2-1+p$, $p>0$, and rays through the origin, cut off inside the parabola $y=x^2-1$. – Christian Blatter Jul 05 '14 at 12:29
  • Let $A$ be the parabola family $y=tx^2+x+t,~t>0$. Note that no two parabolas in $A$ intersect, and every point above the line $y=x$ lies on exactly one parabola, given by $t=(y-x)/(x^2+1)$. Similarly, let $B$ be the family $x=ty^2+y+t,~t>0$. The parabolas in $A$ point up, while the parabolas in $B$ point to the right. Together, they cover the entire plane except for the line $y=x$. So we could say the answer is within a hair's breadth of "no". – Matt Jul 08 '14 at 22:48

2 Answers2

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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.

Matt
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  • The boundary of a parabolic region could be a straight line, e.g. the set of upwardly opening parabolas above the $x$ axis, chosen correctly so no two intersect (with shared axis of symmetry the $y$ axis). Another thing, even if the boundary is a parabola in the usual sense, the points on that parabola might be partitioned into two sets $A,B$ where points of $A$ are actually on one of the parabolas forming the region, but points of $B$ are not. (At least so it seems without further reasoning.) – coffeemath Jul 06 '14 at 02:07
  • @coffeemath: That is a good point, that the supremum could be a half plane. I don't see the second objection. Pick a point X of your set A, and let P be the parabola it is on. There can't be any parabola (in R(P)) outside P -- it would contain P (by definition of R(P)), so X would be in P's interior, not on the boundary of R(P). Therefore P is itself the boundary of R(A), and the set B is empty. (Which doesn't strictly contradict your claim!) – Matt Jul 06 '14 at 14:40
  • I pretty much agree the second "objection" I gave in the comment likely isn't relevant. Note also that, unless one restricts the parabola "insides" to be all on the same side, a partial family of parabolas might be the complement of another parabolic region. (That idea also can be treated by your lemma applied to the said complement.) But I now think your idea shows the claim (+1) – coffeemath Jul 06 '14 at 14:53
  • I see that you follow a stricter sense of partition, which is very much in agreement with the letter of the question. I guess that based on my background, I tend to allow for sporadic degenerate cases. Do you think your approach could also work if you allowed for a set of measure zero to be not contained within any parabola? The limit lines used in my own answer would be such a set. Don't take this comment as criticism, I'm merely curious. – MvG Jul 07 '14 at 17:41
  • @MvG: I think in that case the final part would need to be changed. The proof as given would work for closed square regions, which can cover all but a set of measure zero. But the parabolic shape makes it much harder to tile the plane -- the parabolae need to come in from every direction (every point on the line at infinity), since every region looks, when zoomed out, like a ray, so to fill the "pie slice" between two regions, it seems you would always need parabolae at intermediate angles, with serious problems under their cusps. But I haven't been able to wrestle that into an argument. – Matt Jul 07 '14 at 21:59
  • @MvG: It feels like trying to cover the upper half-plane with disks that touch the x-axis. Super-impossible. There should be an extremely simple proof of the impossibility of this, not even using much topology. The problem at hand is dealing with ellipses, not disks, and the line at infinity, not the x-axis, but is otherwise very similar. I like your projective plane view, but to me one of my first intuitions is that the parabola directions would need to be dense, so I have trouble with your approach. Or maybe your approach is justifying the idea that the directions need to be dense! – Matt Jul 07 '14 at 22:32
  • @MvG: If you allow a single line of points to be not contained within any parabola, then you can cover the rest of the plane with parabolas having axes in two different directions. I've put the details in a comment to the question. – Matt Jul 08 '14 at 22:53
  • @Matt : You wrote "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". Could you found that statement in detail? – user64494 Jul 12 '14 at 03:18
  • @user64494: The previous paragraph pointed out that zooming out will make a parabola look like a ray. Here there are only 2 parabolae, so we can zoom out so they both look like rays, indeed, their foci will both look to be in about the same place. So we see two "rays", A and B, pointing in different directions. Since they point in different directions, there is an angular gap between them. (For small angles, we have to zoom out a lot to make the "rays" thin enough to see this gap clearly.) This gap contradicts our assumption that A is inside B. ((The next paragraph gives an alternate proof.)) – Matt Jul 12 '14 at 07:38
  • @Matt: Sorry, I don't find it to be a math proof. – user64494 Jul 12 '14 at 11:09
  • @user64494: Two proofs of the lemma were supplied. You have given no indication of what problem you have with either proof. A third proof could be given using sin & cos to transform coordinates. A fourth proof could be given by considering that the axis of symmetry of the interior parabola must exit the "enclosing" parabola in both directions. All of these are math proofs. – Matt Jul 17 '14 at 12:47
  • @user64494: The amusing thing (to me) is, I answered this question just to see if you accept geometry answers at all, since for the linked "flowing curves" question, where I gave a detailed, nontrivial, to-the-point answer after over a year of no answers to your question, I found your response a bit bizarre: you insulted the answer and voted it down! – Matt Jul 17 '14 at 12:56
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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:

Family of parabolas

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:

Illustration of family of ellipses

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.

MvG
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  • I don't understand it at all. Could you express your attempt in usual terms of geometry? – user64494 Oct 05 '13 at 08:39
  • @user64494: Does my edit help? – MvG Oct 05 '13 at 12:32
  • Someone downvoted this answer. Why? If there is a problem with this proof, I'd like to see it expressed. – MvG Oct 05 '13 at 12:33
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    The first remark. You wrote:"Take two in such a way that there is no other direction in between them". The directions may be everywhere dense in $[0,2\pi).$ – user64494 Oct 05 '13 at 12:48
  • @user64494: Dense directions should be easy to rule out. If two directions are arbitrarily close together, then the curvature at the apex has to become arbitrarly great to avoid overlaps. In the limit, all parabolas would degenerate to rays. – MvG Oct 05 '13 at 13:22
  • @ MvG: Don't understand your arguments. You wrote "If two directions are arbitrarily close together, then the curvature at the apex has to become arbitrarly great to avoid overlaps". The distance between two fixed directions is a fixed number. Why do you think that all parabolas have the common apex? – user64494 Oct 05 '13 at 15:12
  • let us continue this discussion in chat – MvG Oct 05 '13 at 16:43
  • I agree that the axes have to have the same direction, but not that the axes have to be the same. Consider the family $y=max(t,1)x^2+x+t$. – Matt Jul 05 '14 at 00:16
  • @Matt: The “same axis” was a consequence of “same direction and two distinct common tangents” for one family, and these two tangents were a consequence of the assumption of different directions for different families. So while your family has different axes, it also has only a single limit line for $t=0$, if you consider that a limit at all, and it's not even a tangent of all parabolas. It doesn't satisfy one of my underlying assumptions. – MvG Jul 05 '14 at 08:39
  • @MvG: Then I'm confused, because you claim this in your introductory sentence, before you have assumed anything. – Matt Jul 05 '14 at 09:13
  • @Matt: You are correct, I made a mistake in my introduction. Hope this is more clear now. – MvG Jul 07 '14 at 17:34
  • @MvG: It is clearer now, but I still have trouble with the idea that the parabola directions are not dense. You can place disks in the upper half-plane tangent to each rational point on the x-axis so that the disks do not touch. Transform the x-axis into the line at infinity, and you have non-overlapping parabolas pointing in every rational direction. Also, the parabola family in my comment above tiles the plane but they do not share an axis. There is no limit line -- the max prevents the parabolas from getting "flat". The parameter $t$ is in $(-\infty,\infty)$. – Matt Jul 07 '14 at 22:47
  • @MvG: I also don't see why you say "These straight lines have to be tangents for every parabola in the family." For the only case I know of where a family has a straight line as a boundary (families like $y=tx^2+t,t>0$), it isn't true -- no parabola is tangent to the family boundary. – Matt Jul 08 '14 at 09:10