We need to prove that there is a set $A \subset \mathbb{R}^2$ such that for every circle $C \subset \mathbb{R}^2$ we have that they intersect on exactly $3$ points,meaning $|A \cap C|=3$.
I know that $|C| = |\mathbb{R}| = \aleph$, so by AC we well-order the circles on the plane, for every ordinal $\alpha < \aleph$
1) If $\alpha = 0 $ we pick there random point from $C_0$ and put them into $A_0$ such that $|A_0 \cap C_0|=3$
2) If $\alpha $ is successor then ...
3) If $\alpha$ is a limit then ...
I am really stuck here, first time to see such thing, also in every step how i can make sure that $|A_{\alpha} \cap C_{\alpha}|=3$ ??
At stage $\alpha$ you pick points from $C_\alpha$ to make three and add them to $A$.
At stage $0$ you pick any three points from $C_0$ and add them to $A$. At stage $1$ you look how many points that are already in $A$ are on $C_1$, then pick enough points to make three. At stage $2$ you look how many points of $C_2$ are already in $A$ and pick enough points to make three and so on. When you pick points to add you have to avoid any that are on preceding circles, but that is easy. There are less than continuum many preceding circles and each one only intersects the current circle in at most two points, so you have lots of points left to choose from. What can go wrong is that at some stage you find there are already four or more points of the circle already in $A$, so you need to avoid that. Again there are less than continuum many sets of three points so you can avoid all the circles that are determined.