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Nov 27, 2014 at 15:56 comment added user173856 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/….
Nov 27, 2014 at 15:34 vote accept user173856
Nov 27, 2014 at 15:34 vote accept user173856
Nov 27, 2014 at 15:34
Nov 27, 2014 at 15:34 vote accept user173856
Nov 27, 2014 at 15:34
Nov 27, 2014 at 15:34 history edited user173856 CC BY-SA 3.0
deleted 322 characters in body
Nov 27, 2014 at 15:31 comment added Mirko 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).
Nov 27, 2014 at 15:26 history edited user173856 CC BY-SA 3.0
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Nov 27, 2014 at 15:19 comment added Todd Trimble 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.
Nov 27, 2014 at 15:13 history edited user173856 CC BY-SA 3.0
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Nov 27, 2014 at 15:12 history edited Stanley Yao Xiao CC BY-SA 3.0
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Nov 27, 2014 at 15:12 comment added user173856 Sorry,I just modified my question!
Nov 27, 2014 at 15:11 history edited user173856 CC BY-SA 3.0
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Nov 27, 2014 at 14:32 answer added S. Carnahan timeline score: 6
Nov 27, 2014 at 14:31 answer added Emil Jeřábek timeline score: 13
Nov 27, 2014 at 13:24 comment added Emil Jeřábek Sorry, you are right.
Nov 27, 2014 at 13:15 comment added Tom Leinster Oh, oops. I misread it.
Nov 27, 2014 at 13:14 comment added Joonas Ilmavirta @TomLeinster, the OP assumes $\emptyset\neq U\subset V$.
Nov 27, 2014 at 13:12 comment added Tom Leinster Surely the answer is an easy "no". Just take $U$ and $V$ to be disjoint. Am I missing something?
Nov 27, 2014 at 12:53 comment added Włodzimierz Holsztyński @EmilJeřábek, $\ U\ $ is supposed to be connected.
Nov 27, 2014 at 11:57 review Close votes
Nov 27, 2014 at 19:51
Nov 27, 2014 at 11:49 comment added Beni Bogosel 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?
Nov 27, 2014 at 11:24 history undeleted user173856
Nov 27, 2014 at 11:24 history deleted user173856 via Vote
Nov 27, 2014 at 11:23 history asked user173856 CC BY-SA 3.0