Re: A set containing a nonempty open interval
- From: Jim Spriggs <jim.sprigs@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 26 Aug 2005 21:59:12 +0000 (UTC)
"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
Doesn't [x, x] count as an interval?
> which is an open interval, I wonder why
> the word "open" is even in the question.
I was wondering why the word "nonempty" occurred. Maybe (x, x) is an
interval, but that seems far more perverse than [x, x].
> Probably not important, I guess.
--
I don't know who you are Sir, or where you come from,
but you've done me a power of good.
.
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