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? 
 A: 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.
