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Consider the closed unit disk $D\subset\mathbb{C}$ with the normal topology. Are there two connected sets, $A,B\subset D$ such that $1,-1\in A$ and $i,-i\in B$ but $A\cap B=\emptyset$.

This is impossible if you ask for $A$ and $B$ to be path connected, but is it possible if you relax the condition to just connected (but not path connected). I have a feeling this might be possible using the Cantor leaky tent (also known as the Knaster–Kuratowski fan).

hjr
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    Your reference to the Cantor leaky tent made me think of something like $A=D\cap((\mathbb{Q}\times{0}) \cup ((\mathbb{R}\setminus\mathbb{Q})\times(\mathbb{R}\setminus{0})))$ and $B=D\setminus A$. I'm not sure though if $A$ and $B$ are connected. Maybe kind of related: Partitions of $\mathbb{R}^2$ into disjoint, connected, dense subsets. – jplitza Feb 14 '17 at 15:43
  • This kind of sets look promising. I need some time to convince myself that $A$ and $B$ are connected. I think we also need to modify $A$ a bit, make the line $(x,1)$ the same way you did the line $(x,0)$. – hjr Feb 15 '17 at 13:34
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    Do you mean $(0,y)$ instead of $(x,1)$? Because as $D$ is the unit disk, $D\cap(x,1)$ only contains $(0,1)$. Also, I'm starting to think the sets are connected for the same reason the leaky tent is connected. – jplitza Feb 15 '17 at 17:10
  • Sorry I didn't finish my thought, I was thinking its just easier to work in the unit square and consider the points (1/2,0), (1/2,1), (0, 1/2) and (1, 1/2), so i was answering you still thinking about that. So disregard my rambling in the previous comment. But i believe what you said works fine for the original question. – hjr Feb 15 '17 at 19:28

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