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 $|r \cap X|=2$.



Can all lines in the euclidian plane be ordinary?
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 well-ordered 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 no-three-points-collinear 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
no-three-points-collinear 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
no-three-points-collinear 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).


