# Uncountable Sum

1. Aug 3, 2009

### Dragonfall

Does there exist a converging uncountable sum of strictly positive reals?

2. Aug 4, 2009

### bpet

Would that be an integral?

3. Aug 4, 2009

### Dragonfall

No. I mean an actual uncountable sum. An (Riemann) integral is the limit of a sequence of countable sums.

4. Aug 4, 2009

### g_edgar

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. Aug 4, 2009

### HallsofIvy

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. Aug 4, 2009

### g_edgar

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 $\left|V - \sum_{s \in A} a(s)\right| < \epsilon$ .

7. Aug 4, 2009

### Dragonfall

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?

8. Aug 4, 2009

### Dragonfall

I don't know what this definition is trying to achieve. I prefer mine.

9. Aug 5, 2009

### g_edgar

If and only if $D$ is countable.

10. Aug 5, 2009

### Dragonfall

Well $$\epsilon_0$$ is the first uncountable ordinal, so why not?

11. Aug 5, 2009

### g_edgar

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. Aug 5, 2009

### Dragonfall

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. Aug 6, 2009

### g_edgar

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