I try to solve the following question:
Show that there is a Lebesgue null set A so that A-A is a neighbourhood of zero; where $A-A :=\{x-y|x,y\in A\}$.
I really don't no how to do this.
Thanks to everyone who can help.
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Please edit the question, what is A-A ? – seamp Oct 25 '18 at 11:57
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1@seamp $A-A={a-a',|,a,a'\in A}$. – José Carlos Santos Oct 25 '18 at 12:00
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Not knowing how to do something doesn't mean you can't try. https://math.meta.stackexchange.com/questions/27923/how-to-prevent-no-clue-questions – Calvin Khor Oct 25 '18 at 12:00
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2@CalvinKhor The OP is a new contributor. Besides, someone not familiar with the Cantor set any not get any idea at all on this question. Let us be sympathetic to new contributors. – Kavi Rama Murthy Oct 25 '18 at 12:22
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The title asks for a result that is false, so the title should be reworded a bit. – Dave L. Renfro Oct 25 '18 at 12:37
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@KaviRamaMurthy Sorry, I was terse but did not mean to not have sympathy. A course ideally provides students with enough tools to have a fighting chance, and that is what I was hoping for the OP to provide us – Calvin Khor Oct 25 '18 at 12:56
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@DaveL.Renfro You are right about the title. OP should change the title. – Kavi Rama Murthy Oct 25 '18 at 23:17
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Sorry, have changed the titel. – Oct 26 '18 at 02:11
1 Answers
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If $C$ is the Cantor set then $C-C=[-1,1]$ and $C$ has measure $0$. To prove that $C-C=[-1,1]$ use expansion of numbers to base 3. I will be glad to help if this is not clear to you.
Kavi Rama Murthy
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This is a bit late, but here is a question with several answers showing that $C+C=[0,2]$. Given the symmetry $C=1-C$, this is easily seen to be equivalent to $C-C=[-1,1]$. – Harald Hanche-Olsen Jan 02 '19 at 20:36