Converging Uncountable Sum of Positive Reals

In summary, there is no converging uncountable sum of strictly positive reals. The only way to define such a sum is through the integral, which is the limit of a sequence of countable sums. In order for an uncountable sum to converge, all but countably many terms must be zero.
  • #1
Dragonfall
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Does there exist a converging uncountable sum of strictly positive reals?
 
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  • #2
Would that be an integral?
 
  • #3
No. I mean an actual uncountable sum. An (Riemann) integral is the limit of a sequence of countable sums.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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] .
 
  • #7
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?
 
  • #8
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.
 
  • #9
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.
 
  • #10
Well [tex]\epsilon_0[/tex] is the first uncountable ordinal, so why not?
 
  • #11
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.
 
  • #12
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.
 
  • #13
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.
 

1. What is a converging uncountable sum of positive reals?

A converging uncountable sum of positive reals refers to the concept in mathematics where an infinite sum of positive real numbers is considered to converge, or have a finite value, even though the sum contains an uncountable number of terms. This is in contrast to a converging countable sum, where the sum contains a countable number of terms.

2. How is a converging uncountable sum of positive reals different from a converging countable sum?

The main difference between a converging uncountable sum of positive reals and a converging countable sum is the number of terms involved. A converging uncountable sum has an uncountable number of terms, while a converging countable sum has a countable number of terms. Additionally, a converging uncountable sum involves the sum of positive real numbers, while a converging countable sum can involve any type of number.

3. Can a converging uncountable sum of positive reals be infinite?

No, a converging uncountable sum of positive reals cannot be infinite. By definition, a converging sum has a finite value, meaning it cannot be infinite. However, it is possible for a converging uncountable sum to be equal to the infinity symbol (∞), but this is not the same as being infinite.

4. What is an example of a converging uncountable sum of positive reals?

An example of a converging uncountable sum of positive reals is the sum of all positive real numbers less than 1. This sum is equal to 1, even though it contains an uncountable number of terms. Another example is the sum of all positive real numbers less than or equal to 1, which is equal to 2.

5. How is the convergence of an uncountable sum determined?

The convergence of an uncountable sum is determined by using mathematical concepts such as the Monotone Convergence Theorem and the Cauchy Criterion. These concepts help determine if the sum has a finite value, or if it diverges (i.e. has an infinite value or does not have a defined value).

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