# Prove the existence of a set in the Euclidean plane

I got stuck on the following problem. Prove that there exists a subset $A$ of $\mathbb{R}^2$ such that every line in $\mathbb{R}^2$ goes exactly through two points in $A$. I know that I should apply the axiom of choice in some clever way but I can't think of it. Can someone help me?

• It might help to know what form of the Axiom of Choice you're supposed to use. Feb 23, 2013 at 19:14
• @GitGud: Does it matter? You could probably prove this with the assumption that every vector space has a Hamel basis, by somehow considering the lines as a vector space over some field and using the basis to generate such set. Feb 23, 2013 at 19:21
• @AsafKaragila It might matter to the OP. He could have trouble translating your answer into the AC's form he's more used to. PS: I knew you were gonna answer this question. Feb 23, 2013 at 19:21
• @GitGud: If the question is tagged [axiom-of-choice] it's rare that I won't give an answer. Combine the fact that AC is my research focus, and that I am almost all the time on the site... Feb 23, 2013 at 19:28
• @AsafKaragila Plus you're a genius, to be honest. Feb 23, 2013 at 19:28

Consider $A$ to be the set of all lines, note that its cardinality is $2^{\aleph_0}$, so we can enumerate it $A=\{L_\alpha\mid\alpha<2^{\aleph_0}\}$.

We define by transfinite induction sets $C_\alpha\subseteq\Bbb R^2$ such that for every $\beta,\alpha<2^{\aleph_0}$: $|C_\alpha|<2^{\aleph_0}$ and $|C_\alpha\cap L_\beta|\leq 2$.

If $C_\beta$ was defined for all $\beta<\alpha$, let $\gamma$ be the least ordinal such that $L_\gamma\cap\bigcup_{\beta<\alpha} C_\beta$ has at most one point. Since two distinct lines meet at at most one point the set $\bigcup_{\delta<\gamma}(L_\delta\cap L_\gamma)$ has size $<2^{\aleph_0}$ and therefore the set $L_\gamma\setminus\Big(\bigcup_{\beta<\gamma} L_\beta\, \cup\, \bigcup_{\beta<\alpha} C_\beta\Big)$ is non-empty, and we can choose $x_\alpha$ from it.

Let $C_\alpha=\bigcup_{\beta<\alpha} C_\alpha\cup\{x_\alpha\}$. And let $C=\bigcup_{\alpha<2^{\aleph_0}} C_\alpha$ be our set. As this is a homework assignment, I leave it to you to verify this part (and to formalize the above argument better).

Equally, and perhaps more easily, you could do this with Zorn's lemma, by defining the partially order set of all those subsets of the plane which meet every line in at most two points, and order it by inclusion.

Show that every chain has an upper bound (i.e. the increasing union of such sets is itself a set with this property), and conclude that the maximal element is the one you are after.

• Such sets are called "2-point sets" in the plane. It's still an open problem (people at my old university have worked on it quite a bit!) whether a Borel 2-point set exists.. Feb 23, 2013 at 19:36
• See cs.elte.hu/~vidnyanz/el1.pdf for an overview of the problem. It's quite intriguing Feb 23, 2013 at 19:37
• @Henno: Pretty cool. Feb 23, 2013 at 19:55
• This is an example of what Gowers calls "Just-do-it proofs": gowers.wordpress.com/2008/08/16/just-do-it-proofs Feb 23, 2013 at 23:48
• @AsafKaragila About your second idea to solve the problem (use of Zorn'n Lemma): Every circle is a maximal element of your partially ordered set (i.e., as you suggested: the family of those subsets of the plane which meet every line in at most two points). So if this approach gonna work then we can drop the zorn part and start with a given circle. Jun 8, 2017 at 9:25