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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1$\begingroup$ 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? $\endgroup$– Beni BogoselCommented Nov 27, 2014 at 11:49
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$\begingroup$ @EmilJeřábek, $\ U\ $ is supposed to be connected. $\endgroup$– Włodzimierz HolsztyńskiCommented Nov 27, 2014 at 12:53
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9$\begingroup$ 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 TrimbleCommented Nov 27, 2014 at 15:19
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1$\begingroup$ 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). $\endgroup$– MirkoCommented Nov 27, 2014 at 15:31
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1$\begingroup$ 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/…. $\endgroup$– user173856Commented Nov 27, 2014 at 15:56
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2 Answers
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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 $r$.
answered Nov 27, 2014 at 14:31
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$\begingroup$ Nice! What if one assume that $\bar{U} \subseteq V$? $\endgroup$ Commented Nov 27, 2014 at 14:40
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$\begingroup$ I am wondering about that, too. It seems rather more difficult. $\endgroup$ Commented Nov 27, 2014 at 14:45
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4$\begingroup$ @user173856, you should´t change your question after a good answer has been given to the original one. $\endgroup$ Commented Nov 27, 2014 at 15:18
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1$\begingroup$ 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. $\endgroup$– MirkoCommented Nov 27, 2014 at 15:23
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$\begingroup$ For completeness, I note that Alexandre Eremenko’s answer to the new question also answers the $\overline U\subseteq V$ question. $\endgroup$ Commented Nov 27, 2014 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 $A \times \{0\}$.
answered Nov 27, 2014 at 14:32