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Nov 13, 2012 at 19:27 vote accept AmirHosein Sadeghimanesh
Nov 13, 2012 at 19:23 vote accept AmirHosein Sadeghimanesh
Nov 13, 2012 at 19:27
Nov 13, 2012 at 19:22 vote accept AmirHosein Sadeghimanesh
Nov 13, 2012 at 19:23
Nov 13, 2012 at 19:22 vote accept AmirHosein Sadeghimanesh
Nov 13, 2012 at 19:22
Nov 12, 2012 at 14:03 comment added Pietro Majer by the way, the term "atomless" is better than the unclear "non-atomic" (="no atoms at all", or "not atoms only"?), which is also a funny litotes. I'll correct now.
Nov 9, 2012 at 13:13 comment added Andreas Blass As far as I can see, the revised version of the question was already answered affirmatively (before the revision) by Pietro Majer, since Lebesgue measure on a (measurable) subset of the reals is always atomless.
Nov 9, 2012 at 13:06 history edited AmirHosein Sadeghimanesh CC BY-SA 3.0
added 589 characters in body
Nov 7, 2012 at 17:19 answer added Pietro Majer timeline score: 5
Nov 7, 2012 at 17:06 comment added Davide Giraudo Not necessarily, for example $X=\Bbb N$,the set of natural numbers, with counting measure. Then \{\emptyset\}$ is open, and so is $\{A\subset \Bbb N,A\neq\emptyset\}=\{A,d(A,\emptyset)>0\}$.
Nov 7, 2012 at 17:05 comment added Davide Giraudo Not necessarily, for example $X=\Bbb N, the set of natural numbers, with counting measure. Then $\{\emptyset\}$ is open, and so is $\{A\subset \Bbb N,A\neq\emptyset\}=\{A,d(A,\emptyset)>0\}$.
Nov 7, 2012 at 16:52 history asked AmirHosein Sadeghimanesh CC BY-SA 3.0