This I find it really hard to solve. I suppose the set {$x+iy|x,y\in \mathbb{Q}$} is neither closed or open. But I just cannot seem to find a way to go forward. Can someone help me out. Thanks
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Consider paths consisting of straight lines parallel to the axes? – Mark Bennet Nov 11 '14 at 17:54
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@MarkBennet Nice idea but suppose I chose z with rational Rez and irrational Imz and $z_0$ with irrational $Rez_0$ and rational $Imz_0$ Do you think we can construct straight lines which will not have z with both rational components? – Heisenberg Nov 11 '14 at 17:58
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You should be able to join points using at most three lines. The example you give takes just two. Go along $Imz$ until you get to $Rez_0$ and then go up or down. Then you have fixed one component as irrational on both lines. – Mark Bennet Nov 11 '14 at 18:07
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Hint 1: Find a path connecting 2 points with both coordinates irrational.
Hint 2: Connect all points with one coordinate irrational to a point with both coordinates irrational.
user2345215
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Are you saying that C \ {x+iy|x,y∈Q} contains complex numbers with real and imaginary parts which are only irrational? – Heisenberg Nov 11 '14 at 18:02
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It seems to me from your answer that $z\in$ C \ {x+iy|x,y∈Q} means Rez AND Imz are both irrational – Heisenberg Nov 11 '14 at 18:05
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This is a kind of dual to a question asked just yesterday. The solution is similar: every point in your set is path-connected to the point $\sqrt 2+\sqrt 2i$, so every point is path-connected to every other point.
Indeed, if $a \notin \mathbb Q$, then we have a path $a+bi \to a+\sqrt 2i \to \sqrt 2+\sqrt 2i$; the case $b \notin \mathbb Q$ is similar.
(I suppose I shouldn't say that two points are "path-connected", because that adjective properly applies to the whole set; but you know what I mean.)
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Just the answer I was waiting for. This is brilliant and simple thanks – Heisenberg Nov 11 '14 at 18:17
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