# Formal proof that $\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}$ is connected.

Cam anyone provide me the proof of:

that $\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}$ is connected.

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Ideally you should say what you tried. Hint: you could try showing it is path connected. – Matthew Pressland May 16 '12 at 16:19
This is answered in math.stackexchange.com/questions/16948/… in both the comments (by Jacob Schlather) and as an actual answer (by Jeremy Hurwitz.) The questions, themselves, are different, so I'm not sure if this should be closed as a duplicate or not. – Jason DeVito May 16 '12 at 16:21
I would suggest this question should be left open, as it's far more likely to come up in a search related to this problem. – Matthew Pressland May 16 '12 at 16:37
Michael Hardy's solution seems clear to me, but if for some reason you don't get it you might prefer to consider why $\mathbb{R}\setminus(\mathbb{Z}\times\mathbb{Z})$ is connected; it might be easier to visualize. – MJD May 16 '12 at 17:12
Another near duplicate? – Jyrki Lahtonen May 16 '12 at 18:11

Theorem. $\ \mathbb{R}^2-A\$ is connected for any set $A\subset\mathbb{R}^2$ of cardinality less than the continuum.

Proof. Consider any two points $u,v$ in $\mathbb{R}^2-A$. There is a foliation of continuum many paths from $u$ to $v$, which are disjoint except at $u$ and $v$. For example, one could consider all the various circle fragments containing $u$ and $v$. Since only fewer than continuum many of these paths contain points from $A$, it follows that almost all of them are contained in $\mathbb{R}^2-A$, which is therefore path-connected, and even arc-connected. QED

In particular, $\mathbb{R}^2-\mathbb{Q}\times\mathbb{Q}$ is connected.

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JDH this comment is just to point out a small typo :$A\subset\mathbb{R}^2$. +1 nice answer! – Leandro May 27 '12 at 18:18
Thanks, I have corrected. – JDH Oct 6 '13 at 20:06
Always nice to see you here! – Asaf Karagila Oct 6 '13 at 20:17
Asaf, thanks; I had been away for about a year, and my computer logged me in by accident a few days ago, and I noticed some interesting questions, so I posted. – JDH Oct 6 '13 at 20:24
I know that you've been away since September 25th of 2012. I was checking in every now and then in hopes that you'd return someday! – Asaf Karagila Oct 6 '13 at 20:29

Suppose $(x,y),(a,b)\in \mathbb{R}^2\setminus\mathbb{Q}\times\mathbb{Q}$. Then either $x$ or $y$ is irrational. Suppose $x$ is irrational. Then there's a path from $(x,y)$ to $(x,b)$ that remains in $\mathbb{R}^2\setminus\mathbb{Q}\times\mathbb{Q}$, namely, the second coordinate changes from $y$ to $b$ while $x$ stays put. Now you have to get from $(x,b)$ to $(a,b)$ along a path. If $b$ is irrational, you do it the same way except that it's the $x$-coordinate that changes, from $x$ to $a$. If $b$ is rational, then you're going to need to put the "horizontal" path elsewhere than at $b$ and use two vertical paths instead of one.

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Yeah, the Pacman path ! – Damien L Mar 9 '13 at 21:22
This argument, by the way, actually proves that $\Bbb R^2 \setminus \Bbb Q^2$ is arc-connected. And it can be made a little more uniform if you hook everything to, say, $(\sqrt 2,\sqrt 2)$. – dfeuer Oct 6 '13 at 20:27