How to prove S is an interval?

Question

This problem is, in my opinion, interesting.

Let (a_n) be a nonincreasing sequence of nonnegative numbers such that sum(a_k, k=1..+oo) <+oo. Let S denote the set of all numbers which are sums of subseries of sum(a_k,k=1..+oo). Prove that S is an interval if, and only if, for each natural n, a_n <= sum(a_k,k=n+1..+oo).

Answer 1

I think I basically agree with Magee Coe, but I don't quite follow his explanation on the second part. So I'll do my own presentation anyway. Probably the difference is just a matter of terminology, but I think my presentation will be easier to follow.

Preliminary:
Since the a_n are a non-increasing sequence of non-negative numbers, I define the following notation: the ∞-tuple
(b_1, b_2, b_3, ....) = sum(k=1, ∞) [b_k*a_k]
where each b_k = 0 or 1

In other words, (b_1, b_2, b_3...) is the binary representation of the sum of a subseries of the a_n.

In this terminology, S = set of all real numbers x such that x is equal to one of the binary ∞-tuples described above.

First: "Suppose S is an interval of R. Then
a_n ≤ sum (k=n+1, ∞)[a_k]"
Proof: Suppose not. Then there is some a_N such that
a_N > sum (k=N+1, ∞). Then consider:
x1 = (1,1,1,...1,0,0,0...) where the last "1" is b_N-1
x2 = (1,1,1,...1,1,0,0...) where the last "1" is b_N

By construction,
x2 - x1 = a_N.
The only values of S that can "bridge the gap" between x1 and x2 are obtained by adding elements of a_n for n > N (the rest having already been used), and by our assumption, even taking all of these will not be enough to do the job.
Hence, if the "a_n ≤ sum (k=n+1, ∞)[a_k]" property is not true, S cannot be an interval.
So, if S is an interval, it must be true that
a_n ≤ sum (k=n+1, ∞)[a_k] for all n.
(This is essentially identical to Magee Coe's proof.)

Second: "If a_n ≤ sum (k=n+1, ∞)[a_k] for all n, then S is an interval."
Proof:
The smallest x = (0,0,0...) = 0 , and the largest x = (1,1,1..). For any y in between, we want to find a convergent sequence that leads to y. This is not hard:
- If a_1 ≥ y, set b_1 = 0. Otherwise, set b_1 = 1.
y_1 = (b_1, 0, 0...)
- If y_1 + a_2 ≥ y, set b_2 = 0. Otherwise, set b_2 = 1.
y_2 = (b_1, b_2, 0, 0...)
- If y_2 + a_3 ≥ y, set b_3 = 0. Otherwise, set b_3 = 1.
y_3 = (b_1, b_2, b_3, 0,0...)
...
If we consider the sequence of y_n, this is a monotonically increasing infinite sequence in a bounded interval of R. It will therefore converge to a real number within the interval. Because the a_n go to 0, this must be the value of y, for which we have constructed the binary ∞-tuple representation, and so this is an element of S.
Thus, every point y within the interval (0, ∑a_n) is included in S.

By the way:
- The statement of the question does not preclude the possibility that consecutive a_n could have equal values. This means that the binary ∞-tuple representation may not be unique. This does not matter with respect to the proof.
- The same issue applies even if the a_n are strictly decreasing, since the "≤" in the summation inequality means that there can be multiple representation. Indeed, we should be familiar with this already from our ordinary decimal number representation: 0.99999999... = 1.

Answer 2

( => )
Assume that the statement "for each natural n, a_n <= sum(a_k,k=n+1..+oo)" does not hold. Then for some N, one number in S is sum(a_k, k=1..+oo) -a_N, and the closest higher number in S is sum(a_k, k=1..+oo) -sum(a_k,k=N+1..+oo). Hence, S is not an interval.

( <= )
Now assume that the statement "for each natural n, a_n <= sum(a_k,k=n+1..+oo)" holds. Disregard the trivial case of all zeros and let x be any number between 0 and sum(a_k,k=n+1..+oo), and let epsilon>0. Construct a subseries (b_n) in the following manner:
Pick the first element from a_n that is less than or equal to x as the first element in (b_n). Choose the second as the first element from a_n that is less than or equal to x-b_1. Choose the third as the first element from a_n that is less than or equal to x-(b_1+b_2). And so on.
Hence, for every natural K, there exists an N>=K such that the following relation holds:
0 <= a_N <= x -sum(b_k,k=1,...,K-1) <= a_(N-1) <= sum(a_k,k=N,...,+oo).
For large enough K, which implies large enough N, the most right hand side will be less than epsilon, that is,
0 <= x -sum(b_k,k=1,...,K-1) <= epsilon.
By the property of the real numbers, x -sum(b_n)=0, and hence S is in the nontrivial case the interval ]0,sum(a_n)].

Answer 3

enable x be in [0, a million). Then 0 <= x < a million. Then -a million < -x <= 0 by technique of mutliplication of -a million. Then 0 < a million - x <= a million by technique of including a million. for this reason, a million-x is in [0, a million). outline the function f(x) = a million - x. that's a bijection.