quasi
Posts:
11,740
Registered:
7/15/05


Re: Measure and Density
Posted:
Feb 16, 2013 10:56 PM


quasi wrote: >William Elliot wrote: >> >> [User "Herb" on forum "Ask An Analyst" asked]: >> >>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? > >Let Q denote the set of rational numbers and let > > x_1, x_2, x_3, ... > >be an enumeration of Q /\ (0,1). > >For each positive integer k, let > > a_k = max(0,x_k  1/(2^(k+1))) > > b_k = min(1,x_k + 1/(2^(k+1))) > >and define the open interval I_k by > > I_k = (a_k,b_k) > >Finally, let S be the union of the intervals > > I_1, I_2, I_3, ... > >Then S satisfies the required conditions.
Now that I see Rotwang's solution (our solutions are essentially the same with a minor difference), I realize that I should have used
a_k = max(0,x_k  1/(2^(k+2)))
b_k = min(1,x_k + 1/(2^(k+2)))
to make it more evident that m(S) < 1.
However, my original solution was not actually wrong since, for any enumeration of Q /\ (0,1), the intervals
I_1, I_2, I_3, ...
cannot be pairwise disjoint, hence it's still true that m(S) < 1.
quasi

