Is there a set $X \subset \mathbb{R}^2$ such that every straight line in the plane is ordinary in relation to it? i.e. if $r$ is any straight line then $r \cap X=2$.
1 Answer
The answer is yes, by an argument using the axiom of choice.
There are exactly continuum many lines in the plane, and so by the axiom of choice, we may enumerate them in a wellordered sequence of length continuum.
Let's build the set $X$ in stages, so that by stage $\alpha$ we've included two points from all the lines in the enumeration up to $\alpha$ and furthermore, we've done so in such a way that no three points of the set $X$ are collinear.
If we've done this up to stage $\alpha$, then consider the next line $\ell$. If we've already got two points from $\ell$ in $X$, then we're done; we've already handled this line. Otherwise, consider the fewer than continuum many lines through any two points that we've already placed into $X$. These lines intersect $\ell$ in fewer than continuum many points, leaving plenty of other points on $\ell$ that we are free to add to $X$ without violating the nothreepointscollinear requirement. If we've already got one point from $\ell$, then we can add another in such a way that it is not on any of these other lines, thereby maintaining the nothreepointscollinear property. And if we have no points yet from $\ell$ in $X$, then we simply add two such points from $\ell$ not on any of those other lines. In this way, we add points to $X$ so as to make $X\cap\ell$ of size two, while maintaining the nothreepointscollinear property.
Thus, by this transfinite recursive process, we build the desired set $X$, which has exactly two points from every line.
Essentially the same argument works in $\mathbb{R}^3$ or in any finite dimension, or indeed, even in $\mathbb{R}^\omega$.
The construction also works, suitably modified, with circles or other kinds of shapes instead of lines. The important point being that distinct circles intersect in at most two points. So there is a set in the plane containing exactly three points from every circle (or exactly four, whatever is desired).

$\begingroup$ I'm not sure if there is a direct construction of such an $X$. Meanwhile, the AC construction seems quite flexible. For example, you could ensure exactly three points on every line, or exactly four or whatever, any specific cardinal less than the continuum. $\endgroup$ Dec 29, 2016 at 5:25

3$\begingroup$ (1) I think you mean exactly three points (or four etc.) from every circle, since any three noncollinear points determine a circle. (2) Do you know what is the reference for this classical result, that there is a set in the plane meeting each line in exactly two points? (3) Is it still an open problem whether such a set can be a Borel set? $\endgroup$– bofDec 29, 2016 at 6:10

2$\begingroup$ According to "Settheoretic constructions of twopoint sets" by Ben Chad, Robin Knight, and Rolf Suabedissen, the first" twopoint set" was constructed (with AC) by S. Mazurkiewicz; they refer to his 1914 paper (in Polish) and its translation into French. $\endgroup$– bofDec 29, 2016 at 6:28

$\begingroup$ Yes, @bof, that is right, and thanks for the reference, of which I was not aware. As for Borel, I'm not at all sure, and I once asked a MO question about this: mathoverflow.net/q/93601/1946 $\endgroup$ Dec 29, 2016 at 12:25

$\begingroup$ The paper I linked to in my comment—I don't see a date on it but the references go up to 2008 so I guess it's fairly recent—says that the question about Borel sets is still open, all that's known is that a $2$point set can't be an $F_\sigma$ set. $\endgroup$– bofDec 29, 2016 at 12:31