Plane Geometry – Subset of the Plane That Intersects Every Line Exactly Twice

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In a comment to this question, 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.

Best Answer

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 $l$.

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