W^3
Posts:
28
Registered:
4/19/11
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Re: Measure and Density
Posted:
Feb 20, 2013 5:46 PM
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In article
<bd451ba9-b0c2-4ede-aab1-0e3b8a1fc69f@m12g2000yqp.googlegroups.com>,
Butch Malahide <fred.galvin@gmail.com> wrote:
> On Feb 19, 1:39 pm, W^3 <82nd...@comcast.net> wrote:
> > In article <Pine.NEB.4.64.1302161828050.5...@panix2.panix.com>,
> > William Elliot <ma...@panix.com> wrote:
> >
> > > Topology Q+A Board Ask An Analyst
> >
> > > How can we find a measurable dense subset S of [0,1], with m(S) < 1,
> > > and such that for any (a,b) in [0,1], we have m(S /\ (a,b)) > 0?
> >
> > > I have thought of fat Cantor sets, but I cannot see well how to
> > > do it. Any suggestions, please?
> >
> > More interesting is to require 0 < m(S /\ I) < m(I) for all nonempty
> > open intervals I contained in (0,1).
>
> Let I_0, I_1, I_2, . . . be an enumeration of the open intervals with
> rational endpoints. Construct a sequence of pairwise disjoint sets
> F_0, F_1, F_2, . . . so that F_n is a nowhere dense set of positive
> measure contained in the interval I_{floor(n/2)}. Let S be the union
> of the sets F_n where n is even.
Is it possible that there exist 0 < c < d < 1 such that cm(I) < m(S /\
I) < dm(I) for all nonempty open intervals I contained in (0,1)?
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