Converging Uncountable Sum of Positive Reals

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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 [itex]S[/itex] be an index set. Let [itex]a \colon S \to \mathbb{R}[/itex] be a real function on [itex]S[/itex]. Let [itex]V[/itex] be a real number. Then we say [itex]V = \sum_{s\in S} a(s)[/itex] iff for every [itex]\epsilon > 0[/itex] there is a finite set [itex]A_\epsilon \subseteq S[/itex] such that for all finite sets [itex]A[/itex] , if [itex]A_\epsilon \subseteq A \subseteq S[/itex] we have [itex]\left|V - \sum_{s \in A} a(s)\right| < \epsilon[/itex] .
 
I'm surprised. I thought this would have been defined at some point.

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

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

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

If and only if [itex]D[/itex] is countable.
 
Well [tex]\epsilon_0[/tex] is the first uncountable ordinal, so why not?
 
Strange ... [itex]\epsilon_0[/itex] is commonly used to represent a certain countable ordinal, while [itex]\omega_1[/itex] 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.
 
Yes I was mistaken on the notation, it should be [tex]\omega_1[/tex].

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

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.