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Let E be the Euclidean plane. Does there exist a collection C of subsets of E whose union is E and which are all straight lines such that (1) No two distinct straight lines belonging to C are parallel (2) Every subset of E which is a straight line not belonging to C is parallel to exactly one straight line belonging to C and (3) No three pairwise distinct straight lines belonging to C are concurrent? The problem is trivial, of course, without condition (3). All the straight lines belonging to C can intersect in the same point. If such a collection as C exists, could anybody give an example of one. Or is some version of the Axiom of Choice needed to prove its existence? Finally, while still requiring conditions (1) and (2) to be fulfilled, one could ask whether it is possible for the union of all the straight lines belonging to C can be a closed proper subset of E-rather than E itself. But this, I suspect, is impossible-even with the Axiom of Choice and even though we have not imposed condition (3).

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Can one show that there is no continuous foliation of such lines in some appropriate sense? Or no Borel collection of such lines? See my related (unanswered) question at: mathoverflow.net/questions/93601/… –  Joel David Hamkins Nov 24 '12 at 4:31

4 Answers 4

up vote 9 down vote accepted

Such a set of lines can be constructed by transfinite induction of length $\mathfrak c = 2^{\aleph_0}$.

Enumerate all possible directions and all possible points in order type $\mathfrak c$. (This uses a well-order of the reals.)

In the i-th step, do the following two steps:

Step (1,i): If the i-th point is covered by the lines selected so far, do nothing. Otherwise, choose a line through this point which has a direction that was not used by previously selected lines, and that does not meet any intersection point of any two previous lines.

Step (2,i): If a line with the i-th direction has already been selected, do nothing. Otherwise, choose a line in the i-th direction which does not contain meet any intersection point of lines chosen so far.

Each of the choices above uses a well-order of the reals. Each such choice is possible because there are $\mathfrak c$ many possibilities a priori, and fewer than $\mathfrak c$ are disqualified by the requirement to avoid certain points.

Unless I have made a mistake, this construction by transfinite induction is quite standard. If I recall correctly, Ciesielski's book "Set theory for the working mathematician" contains more examples of such constructions. ADDED: Also, Joel's question mentions several similar (plus a few dissimilar) constructions, both on mathoverflow and elsewhere.

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That is a really neat construction! I thought this was an especially simple example of how the Axiom of Choice is able to "define" a (non-empty) set of "undefinable" elements. Another well known example is the set of all real-valued functions of a real variable f such that f(x+y)=f(x)+f(y) for all real x and y. Such functions are easy to define. But if we impose the further condition that they be non-linear, then-just as occurs with my condition (3)-we need the Axiom of Choice to prove that such functions exist and no single one of them is "definable". –  Garabed Gulbenkian Nov 25 '12 at 19:47
    
One further "paradoxical" set may perhaps be worth mentioning here. A theorem of Besicovitch-related to the Kakeya problem-suggests that there might possibly exist a subset S of E such that (1) Every subset of E that is a straight line is either a subset of S or is parallel to a straight line which is a subset of S and (2) S has zero two-dimensional Lebesgue measure. –  Garabed Gulbenkian Nov 26 '12 at 19:57

Let $C$ the curve defined by $y=x^x$ (for $x>0$) union the point $x=0,y=1$. The set of tangents to $C$ (including the $y$ axis) satisfies the hypothesis, as the slope of those tangents grow continuously from $-\infty$ to $+\infty$, and it is impossible to draw three distinct tangents from any point of the plane. So what is wrong with this example?

I just forgot the union of the lines must be $E$ . Time to go to bed ; please delete this answer if you can. On the other hand, in this case, the union of the lines is a closed subset of $E$, so it may still be worth something

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Very nice! I thought my second question could not have a "yes" answer. But it seems that you have proven me wrong-and without either using the Axiom of Choice or giving up condition (3). –  Garabed Gulbenkian Nov 28 '12 at 19:10

(Too long to be a comment, but:) It is impossible to, say, make the union of $C$ either the complement of an open concave region (trivial) or the complement of an open circle (there's only one possible choice of line for each point on the boundary of the circle, and then any other line has to be parallel to one of these tangent lines).

On the other hand, I think it should be possible - certainly with $AC+CH$ - to make the union of elements of $C$ be, say, the complement of an open triangle $T$. Work in $\omega_1$-many stages, where at each stage $\eta$ you have countably many lines already drawn, and you're trying to get a new line which goes through the point $(x_\eta, y_\eta)$ (assuming this point isn't in $T$ and doesn't lie on an existing line already) and is "legal" (i.e., doesn't intersect $T$, doesn't intersect an existing point of intersection between already-chosen lines, and isn't parallel to any given line). Initialize the construction by starting with the three lines containing the three edges of the triangle; then it seems that, by our assumption that $CH$ holds, we can continue this construction to hit every point not in $T$. [EDIT: As Emil points out below, CH can be dropped by considering instead the cardinality of the set of points to be avoided at each given stage. So only $AC$ is used.]

A key step in this proof is that the set of lines through a given point not in the closure of $T$ can be identified with a set of reals of positive measure, so the countably many lines we've already determined can't get in the way of continuing the construction; and the boundary of $T$ is covered by finitely many lines, so that won't cause unavoidable trouble either. This last condition fails if we try to avoid an open disc.

On the other hand, it does seem possible to avoid a closed disc, by basically the same argument as above. I think, in general, it is possible to have the union of the lines in $C$ be the complement of any convex, bounded, open region; and it is possible to have the union of lines in $C$ be the complement of any closed, convex region whose boundary is a union of countably many line segments; so long as we assume $AC+CH$. ($CH$ can be replaced by the appropriate cardinal characteristic requirement that any set of reals of cardinality $<2^{\aleph_0}$ has measure 0; see Do sets with positive Lebesgue measure have same cardinality as R? for why this isn't a consequence of ZFC alone.) I have no idea how much choice is required for any of this.

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Correction: clearly we cannot avoid any closed polygon such that for three of its sides, the lines containing those sides all intersect. –  Noah S Nov 22 '12 at 21:05
    
I don’t think you need CH, you can just use cardinalities instead of Lebesgue measure as in Goldstern’s answer. –  Emil Jeřábek Nov 23 '12 at 11:46
    
Quite right, good point. –  Noah S Nov 23 '12 at 19:19

How about tangents to a deltoid, http://mathworld.wolfram.com/Deltoid.html ?

It do look like the tangents really do cover the entire plane, and all of those are non-parallel.

http://imageshack.us/photo/my-images/42/deltoidtangents.png

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This fails condition (3)... –  Dan Petersen Nov 23 '12 at 9:35
    
Yes, I realized that, however, might it be possible to distort the figure a tiny bit? –  Per Alexandersson Nov 24 '12 at 11:39

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