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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.2012-05-16
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    This is answered in http://math.stackexchange.com/questions/16948/pi-1-mathbb-r2-mathbb-q2-is-uncountable 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.2012-05-16
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    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.2012-05-16
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    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.2012-05-16
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    Another near [duplicate?](http://math.stackexchange.com/q/65705/11619)2012-05-16

2 Answers 2

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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!2012-05-27
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    Always nice to see you here!2013-10-06
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    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.2013-10-06
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    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!2013-10-06
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    Wow! I'm impressed. Why don't you come to New York and we can talk math in person.2013-10-06
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    @JDH: Oh, I wish! It's just that flying to NYC costs as two conferences in Europe (of which there are plenty). And while I can't say that I'm starving in Israel as a Ph.D. student, I can't say that I'm getting enough money to afford these sort of visits...2013-10-06
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    @AsafKaragila: Stressful time right now; brain a few steps behind. Sorry.2013-10-06
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    @dfeuer: Happens to everyone. This is why I recommend a hearty glass of whisky.2013-10-06
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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 !2013-03-09
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    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, $(\sqrt 2,\sqrt 2)$.2013-10-06