Re: A set containing a nonempty open interval
- From: A N Niel <anniel@xxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 26 Aug 2005 14:28:57 -0400
In article <denmfc$o5k$1@xxxxxxxxxxxxxxxxxxx>, Keith A. Lewis
<klewis@xxxxxxxxxxxxxxxx> wrote:
> "Amanda" <sca18@xxxxxxxxxxx> writes in article
> <1125067955.450481.88600@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> dated 26 Aug 2005
> 07:52:35 -0700:
> >
> >I'd like some hints on how to prove that, if a set S has positive
> >Lebesgue measure, then (A + A)/2 = {(x + y)/2 | x and y are in A}
> >contains a non-empty open interval.
>
> Unless you specify x <> y, you might get all closed intervals.
By "contains a non-empty open interval" he means it has a subset which
is a non-empty open interval.
>
> For example,
> A = [0,1] U [10,11]
> gives
> (A+A)/2 = [0,1] U [5,6] U [10,11]
> which contains no open intervals.
This contains the open interval (0,1) for example
>
> But if you stipulate x<>y you get
> (A+A)/2 = (0,1) U [5,6] U (10,11)
> which does contain open intervals.
>
> U is the union operator.
>
> --Keith Lewis klewis {at} mitre.org
> The above may not (yet) represent the opinions of my employer.
.
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