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I am trying to prove that the set $\mathbb R^2 \setminus \mathbb Q^2$ is connected.

I don't know if the following is true: could it be that it is also path connected? If that is the case, maybe it's easier to prove path-connectedness and from here one would conclude connectedness. So, I pick two points $x, y \in \mathbb R^2 \setminus \mathbb Q^2$. I think I must separate in cases, so, for example, if $x=(x_1,x_2)$ and $y=(y_1,y_2)$ are such that $x_1$ is irrational and $y_1$ is irrational, how could I join these two points by union of line segments? I am clueless on how to go on.

Stefan Hamcke
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user100106
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1 Answers1

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Using your example, just join $(x_1, x_2)$ with $(x_1, y_2)$, and the latter with $(y_1, y_2)$, with straight segments.

rewritten
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    path connected implies connected... to complete the thought – Jonathan Aronson Dec 02 '13 at 17:03
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    I'm not sure about the upvotes, but the solution is wrong, when $y_2$ is rational. One correct way would be to join $(x_1, x_2)$ with $(x_1, π)$, then with $(y_1, π)$ and finally with $(y_1, y_2)$. – rewritten Jan 24 '18 at 12:57