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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… 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

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