Is this set all of R?

Question

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.

I think we'll come to the same conclusion if r_n is an enumeration of any countable set dense in R, like the algebraics, for example.

To ksoileu: Hey, Sum (1/n^2) has only positive terms and converges! There are infinitely many examples!

Answer 1

If liminf a_n = 0, then ℓ is NOT (necessarily) all of R.

For example, let M be the subset of N where r_n ≥ 0.

Now, take a divergent series sum{a_n} where the sub-series sum{a_j: j in M} is convergent.* Then, by the argument given for convergent series, ℓ will not cover all of the positive real numbers!

* (By this I really mean to let b_j = {a_j, if j in M, and 0 otherwise}, and have sum{b_j} converge.)


** NOTE TO KSOILEAU: It says positive NUMBERS, not positive INTEGERS. Presumably this means real numbers...!

** NOTE TO ZENTRAED: [deleted - It seems that Zentraed has deleted his comment after (or while) I responded.]

***ADDENDUM:

After a tiny bit of thought, it occurs to me that you can get a much stronger result...

Fix any є>0. Let M = {n: r_n is in (-є/6, є/6)}, and M' be the complement of M in N.

Choose b_n > 0 such that ∑b_n = є/3. Now define a_n=b_n for n in M', and a_n = є/6 for n in M. Note that ∑a_n diverges since it has infinitely many terms equal to є/6, but that it is convergent off of (-є/6, є/6) with sum ≤ є/3.

Now, the punchline: |ℓ| ≤ є! Yet ℓ covers Q, and ∑a_n diverges...

Answer 2

Maybe I missed something. How can Sum (n=1, oo) a_n converge if the a_n are all positive integers? Such a series MUST diverge...

Oh, I read integers where you said numbers! My mistake.