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Let $U,V$ be two nonempty connected open sets in $\mathbb{R}^2$ and $U\subsetneqq V$.I want to ask if there must exist an open ball $B\subset V$ such that $B\not\subset U$ and $B\cap U$ is a nonempty connected open set.

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    $\begingroup$ You could reformulate as follows: does there exist an open set $U$ such that for every ball $B$ which incersects $U$, but is not contained in $U$, the set $B\cap U$ is disconnected? $\endgroup$ Nov 27 '14 at 11:49
  • $\begingroup$ @EmilJeřábek, $\ U\ $ is supposed to be connected. $\endgroup$ Nov 27 '14 at 12:53
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    $\begingroup$ Please don't modify your question, for which answers were thoughtfully provided. Best would be to ask your modified question in another post, but at least retain the text of your original question, and add your modification as an edit if you really don't want to start another post. $\endgroup$
    – Todd Trimble
    Nov 27 '14 at 15:19
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    $\begingroup$ if someone provides a nice answer to the modified question you would have a difficult dilemma to solve, namely which of the more that one deserving answers to accept. (If you start a new question you should probably link this one to the new one, or link them both ways). $\endgroup$
    – Mirko
    Nov 27 '14 at 15:31
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    $\begingroup$ OK,now the question is still the original version!I start a new question and in my new question I link this one.Here is my new question: mathoverflow.net/questions/188225/…. $\endgroup$
    – user173856
    Nov 27 '14 at 15:56
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Let $V$ be the complement of a point $a$, and $U$ the complement of a ray $r$ with end-point $a$. Every ball not containing $a$ and intersecting $r$ is in fact split by $r$.

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  • $\begingroup$ Nice! What if one assume that $\bar{U} \subseteq V$? $\endgroup$ Nov 27 '14 at 14:40
  • $\begingroup$ I am wondering about that, too. It seems rather more difficult. $\endgroup$ Nov 27 '14 at 14:45
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    $\begingroup$ @user173856, you should´t change your question after a good answer has been given to the original one. $\endgroup$ Nov 27 '14 at 15:18
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    $\begingroup$ right, I ponder how this answer relates to the current version of the question (did not see the original formulation). I like the new (current) version, but nevertheless it is a bit confusing (and time-consuming) to try to figure out how this answer relates to it. $\endgroup$
    – Mirko
    Nov 27 '14 at 15:23
  • $\begingroup$ For completeness, I note that Alexandre Eremenko’s answer to the new question also answers the $\overline U\subseteq V$ question. $\endgroup$ Nov 27 '14 at 17:05
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Choose an enumeration $\{ r_n \}_{n \in \mathbb{Z}_{\geq 0}}$ of the rational numbers, and form the subset $A \subset \mathbb{R}$ given by the union of neighborhoods of radius $2^{-n}$ around $r_n$. Let $U$ be the union of $A \times \mathbb{R}$ with a suitable open half-plane that makes the set connected. Any ball $B$ that is both disjoint from the half-plane and not contained in $U$ satisfies the property that $B \cap U$ has infinitely many connected components. Thus, we may choose $V$ to be the union of $U$ and the complement of $A \times \{0\}$.

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