# Is every closed set of Q² the intersection of some connected closed set of R² with Q²

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.

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I wonder whether you can do it by enumerating $F$ and then somehow building the closed set you want, connecting each new point to the set via an arc containing no rational points? But I can't see how to ensure that the union at the end is closed :-/ –  Kevin Buzzard Nov 15 '10 at 22:01
How about inducting on Cantor-Bendixon rank? After all, the base case is when the set has only isolated points, which we could seem to handle by the idea mentioned in the question, and the next level occurs when the limits of the isolated points are isoltated from each other, which would seem not too much harder... –  Joel David Hamkins Nov 15 '10 at 23:03
You can use an irrational grid and assume that F is bounded. If G is closed and bounded and the supremum of the distance between connected components of G is zero, then G is connected. Perhaps you could then join the two components at maximum distance by a line bent by less than 10 degrees say. Repeat, and check that the union of these lines does not have any unwanted limit points. –  Colin McQuillan Nov 15 '10 at 23:55
Colin, in a Cantor-set situation, however, then the closure of $F$ may have uncountably many components. e.g. $F$ = endpoints in usual Cantor set construction. –  Joel David Hamkins Nov 16 '10 at 0:24
Maybe it would help to rephrase the problem as: If $H\subset\mathbb{R}^2$ is a closed but not connected, must there be $G\subset\mathbb{R}^2$ which is closed and connected and has $G\cap\mathbb{Q}^2=H\cap\mathbb{Q}^2$? –  Aaron Meyerowitz Nov 16 '10 at 0:31

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 $\mathbb Q^2$.

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Awesome! I suspect I'll look up this answer in the future, so here are some hints for future me: (1) there are uncountably many choices of radius but only countably many rational points, so you can always choose the radius so that the boundary doesn't contain rational points, (2) yes, the straight line segment may hit infinitely many balls, but that's okay. –  Anton Geraschenko Nov 16 '10 at 20:13
Thank you for your answer, Sergei :-) @Anton : (1) You can take for the ball a square of the form $[x-\varepsilon,x+\varepsilon]\times[y-\varepsilon,y+\varepsilon]$ where $\varepsilon$ is irrationnal (2) I am not very sure but I think that if we choose the diameter of the n-th ball to be less than $2^{-n}$ then the sequence of “go around paths” will converge uniformly. –  Guillaume Brunerie Nov 16 '10 at 20:34
It truly is a brilliant proof. I'll also mention another answer of Sergei's that I thought was quite nice, another answer to a non-trivial point-set topology problem. Somehow Sergei makes it look so easy... –  Todd Trimble Nov 17 '10 at 12:54

There might be something wrong with this suggestion but isn't it all right to enumerate all points of F and then connect two "consecutive" points the way Guillaume described in his question (by drawing two lines with irrational slope)?

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I thought about the same thing, but then I realized that the set you obtain may not be closed in $\mathbb{R}^2$. –  Diego Matessi May 16 '11 at 12:42
oh yes! That's right! Thanks for your help! –  over May 17 '11 at 20:08