W^3
Posts:
29
Registered:
4/19/11


Re: Measure and Density
Posted:
Feb 20, 2013 5:46 PM


In article <bd451ba9b0c24edeaab10e3b8a1fc69f@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)?

