Let (r_n) be an enumeration of the rationals and let (a_n) be a sequence of positive numbers. For each n =1,2,3...define I_n as the open interval (r_n - a_n , r_n + a_n) and let I = Union (n = 1, oo) I_n. It's clear that I is dense in R.
Now, consider the series Sum (n=1, oo) a_n. We have 2 possibilities:
(1) If the series converges, then, according to the sub-additivity of the measure (Lebesgue measure, in this case), we have m(I) <= Sum (n=1, oo) m(I_n) = Sum (n=1, oo) (2 a_n) = 2Sum (n=1, oo) a_n < oo. Since m(R) = oo, it follows I is not all of R, but one of its proper subsets.
(2) If, on the other hand, Sum (n=1, oo) a_n diverges, then the argument given in (1) leads to no conclusion, all we get is the obvious inequality m(I) <= oo. In this case, is it possible that I is still a proper subset of R or will we always have I = R?
If lim inf a_n >0, then I = R, but if If lim inf a_n = 0, I'm not sure.
Thank you for any suggestion.
2007-12-12
06:58:23
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2 answers
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asked by
Steiner
7
in
Mathematics