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I have a question. Could you please help me to solve this problem?

Is it possible that $\mathbb{R}^2\setminus A$ and $\mathbb{R}^2\setminus B$ are homeomorphic, when $A$ and $B$ are non-homeomorphic closed subsets of $\mathbb{R}^2$?

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fun fact: in $X= \mathbb{R}^{\mathbb{N}}$ we can take any 2 sigma-compact sets $A$ and $B$ and $X \setminus A$ will be homeomorphic to $X \setminus B$. So there we have plenty of such examples... – Henno Brandsma Mar 6 '11 at 12:16

$A= \{(0,0)\}$, $B=B[0,1] $ i.e closed unit ball is an example of this.

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This sounds wrong. Maybe I can't see the picture, but if you take $A={0}$ and $B={0,1}$, then the first space will have two connected components and the second three.

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No, what you say is true for $\mathbb{R}$, not for $\mathbb{R}^2$. – Henno Brandsma Mar 6 '11 at 12:12
Yes, I stand corrected. I should have read the question better :) – shamovic Mar 6 '11 at 17:29
But then again, take A to be the $y$ axis and take B to be the $y$-axis and the $x$-axis, won't it work the same way? – shamovic Mar 6 '11 at 17:30

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