In a comment to [this question](https://mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice), Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to be shown that choice is required).


Unfortunately, I haven't been able to track down a reference, so if someone could link me to the original result (or provide a short proof) that would be great. I have roughly worked out a proof for myself, and I'd like to check it against the literature.



Subset of the plane that intersects every line exactly twice
By AC, choose a *cardinal* well-ordering of the lines in in the plane and any well-ordering of all the points.


We proceed by transfinite induction.


Suppose $A\_l$ is a set of points, no three colinear, and let $B\_l$ be the set of lines spanned by points of $A\_l$, and let $C\_l=\cup B\_l$. Suppose further that $l'\prec l$ implies $l'\in B\_l$. Note that $|A\_l| \leq |B\_l| < |l|$, so that 
$$|C\_l\cap l| = |\cup\_{l'\in B\_l} l\cap l'| < |l| .$$


* If $l\in B\_l$, let $A\_{Sl} = A\_l$.
* If $a\in l \cap A\_l$, take the minimal point $b\in l\backslash C\_l$ and let $A\_{Sl} = A\_l\cup \{b\} $.
* If $A\_l\cap l = \emptyset$, take the minimal two points $a,b\in l\backslash C\_l$, and let $A\_{Sl}=A\_l\cup\{a,b\}$.
* Otherwise if $l'\ $ is a limit ordinal, let $A\_{l'} = \bigcup\_{l\prec l'} A\_l$.


It is easy to check that $A\_{l'}$ has no three points colinear --- they'd have to all be in some $A\_l$ for $l\prec l'$. The final union $\bigcup\_l A\_l$ has exactly two points on each line $l$.


