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Suppose I have a subset $U\subset\mathbb R^2$ and a real number $r>1$ with the following properties:

  1. $U$ is compact;
  2. $U\subset rU$ (self-similarity);
  3. $0\in U$;
  4. there exists an open set $H\subset \mathbb R^2$ such that $0\in\partial H$ (boundary) and $H\cap U=\emptyset$.

Let $V:=\bigcup_{n\geq0} r^n U$.

Question: Whether exists $H_1$ open such that $0\in\partial H_1$ and $H_1\cap V=\emptyset$.

Of course, if the answer is "yes", I would like to see a way how to prove it (it needn't be a complete proof, just some crucial hint).

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Of course, $H$ (if it exists) can always be taken to be $\mathbb R^2\setminus U$. –  Hagen von Eitzen Nov 26 '12 at 18:39
    
Of course, and the question is: is still $0\in\partial H\setminus V$ and $H\setminus V$ open? –  tohecz Nov 26 '12 at 18:40
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1 Answer 1

up vote 2 down vote accepted

No.

Consider $U=\{(x,y)\mid x^2+(y-1)^2\le 1\lor x^2+(y+1)^2\le 1\}$ (union of two touching closed disks) and $r=2$. Then $V$ is dense in $\mathbb R^2$.

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damn, a good one! Now I have to find what other good properties does my set $U$ have. –  tohecz Nov 26 '12 at 18:39
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