Re: A set containing a nonempty open interval

In article <denuim$rdt$2@xxxxxxxxxxxxxxxxxxx>,
klewis@xxxxxxxxxxxxxxxx (Keith A. Lewis) wrote:

> A N Niel <anniel@xxxxxxxxxxxxxxxxxxxxx> writes in article
> <260820051428574700%anniel@xxxxxxxxxxxxxxxxxxxxx> dated 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.
>
> Since every interval has a subset which is an open interval, I wonder why
> the word "open" is even in the question. Probably not important, I guess.
>
> --Keith Lewis klewis {at} mitre.org
> The above may not (yet) represent the opinions of my employer.

Non-empty closed intevals may consist of a single point, which case I
think they mean to exclude.
.

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