Re: A set containing a nonempty open interval
- From: "W. Dale Hall" <mailtodhall@xxxxxxxxx>
- Date: Fri, 26 Aug 2005 22:00:05 GMT
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.
The idea is that the interval sought not be degenerate, such as [1 1]. If you don't consider degenerate intervals to be intervals, then the word "open" could be omitted.
Dale. .
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