Let $F\subset\mathbb{Q}^2$ a closed set. Does there exists some closed and connected set $G\subset\mathbb{R}^2$ such that $F=G\cap\mathbb{Q}^2$?


For example if $F=\{a,b\}$, you can take $G$ the reunion of two lines of different irrational slopes passing through $a$ and $b$. This is a connected set and the intersection with $\mathbb{Q}^2$ is $\{a,b\}$ because the slopes are irrationnals.


But I don’t know how to prove it in general (and I don’t know if it’s true). When there are many connected components this is not clear how to connect them without adding new rational points.



Is every closed set of Q² the intersection of some connected closed set of R² with Q²
Enumerate all rational points outside your set. Then cover these points by open balls by induction as follows: the next ball is centered at the first rational point not covered so far, its radius is so small that is does not intersect $F$ and the previous balls and is chosen so that the boundary of the ball does not contain rational points. Then the complement of the union of these balls is path-connected: to connect two points, draw a segment between them and go around every ball intersected by this segment.


Note that this works for any countable set, not just $\mathbb Q^2$.


