Converging Uncountable Sum of Positive Reals

AI Thread Summary
The discussion centers on the possibility of a converging uncountable sum of strictly positive real numbers. Participants clarify that traditional definitions of sums apply to finite and countable cases, with integrals being limits of countable sums. A key point raised is that if a sum of positive reals converges to a finite value, the index set must be countable. The conversation also touches on the notation of ordinals, specifically correcting the reference to uncountable ordinals. Ultimately, the consensus is that a converging uncountable sum of strictly positive reals does not exist.
Dragonfall
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Does there exist a converging uncountable sum of strictly positive reals?
 
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Would that be an integral?
 
No. I mean an actual uncountable sum. An (Riemann) integral is the limit of a sequence of countable sums.
 
In the sensible way to define uncountable sums, you prove that for a sum of real terms, if it converges (to a real number) then all but countably many terms must be zero.
 
Dragonfall said:
Does there exist a converging uncountable sum of strictly positive reals?
First you will have to define what you mean by "uncountable sum"! I know a definition for finite sums and I know a definition for countable sums (the limit of the partial, finite, sums), but I do not know any definition for an uncountable sum except, possibly the integral that bpet suggested.
 
Definition Let S be an index set. Let a \colon S \to \mathbb{R} be a real function on S. Let V be a real number. Then we say V = \sum_{s\in S} a(s) iff for every \epsilon > 0 there is a finite set A_\epsilon \subseteq S such that for all finite sets A , if A_\epsilon \subseteq A \subseteq S we have \left|V - \sum_{s \in A} a(s)\right| < \epsilon .
 
I'm surprised. I thought this would have been defined at some point.

Suppose x_i is a (possibly uncountable) sequence of positive reals indexed by some ordinal L. Then their sum is \sum_{i\in L}x_i=\sup\{\sum_{i\in k}x_i:k<L\}. This takes care of limit ordinals.

So does there exist sequences x_i indexed by ordinals D\geq\epsilon_0 such that \sum_{i\in D}x_i is finite, and that each x_i is positive?
 
g_edgar said:
Definition Let S be an index set. Let a \colon S \to \mathbb{R} be a real function on S. Let V be a real number. Then we say V = \sum_{s\in S} a(s) iff for every \epsilon > 0 there is a finite set A_\epsilon \subseteq S such that for all finite sets A , if A_\epsilon \subseteq A \subseteq S we have \left|V - \sum_{s \in A} a(s)\right| < \epsilon .

I don't know what this definition is trying to achieve. I prefer mine.
 
Dragonfall said:
So does there exist sequences x_i indexed by ordinals D\geq\epsilon_0 such that \sum_{i\in D}x_i is finite, and that each x_i is positive?

If and only if D is countable.
 
  • #10
Well \epsilon_0 is the first uncountable ordinal, so why not?
 
  • #11
Strange ... \epsilon_0 is commonly used to represent a certain countable ordinal, while \omega_1 denotes the least uncountable ordinal. In any case, that notation doesn't matter. Here is a repeat of the same answer as before: If a sum of positive real terms converges to a finite value, then the index set is countable.
 
  • #12
Yes I was mistaken on the notation, it should be \omega_1.

You have yet to say why. You asserting it true doesn't constitute a proof.
 
  • #13
Dragonfall said:
Yes I was mistaken on the notation, it should be \omega_1.

You have yet to say why. You asserting it true doesn't constitute a proof.

Hint for the proof: the real line has a countable dense set, and every term of the convergent series corresponds to an interval.
 
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