# Sums over arbitrary index set

1. Nov 13, 2011

### Damidami

I have read somewhere that we can extend the notion of a series of a sequence
$$\sum_{i=1}^{\infty} a_n$$
to sums over an arbitrary index set, say
$$a : I \to \mathbb{R}$$
is a family of real number indexed by I, then
$$\sum_{i \in I} a_i$$
is the sum of all the elements.

I think the text said that $$a_i \geq 0$$ in order to has sense, and that if $$a_i \neq 0$$ for a non-numerable size of elements, then the series can't converge.

1) My question is with that last sentence, why cant converge such a sequence?, for example if the family is
$$a : \mathbb{R_{>0}} \to \mathbb{R}, a(i) = \frac{1}{i}$$
how does one sum over all it's elements?

2) Other question, the fact that all the $$a_i \geq 0$$ is required because we can't asume an order on the elements of the index set, and if I have negative elements the convergence can vary according to the order in wich I made the sum?

Thanks in advance for any help in understanding this.

2. Nov 13, 2011

### micromass

Staff Emeritus
The thing is that we define the convergence slightly different.

If a series $\sum_{n=1}^{+\infty}{a_n}$ has positive terms, then we can say that the series converges if and only if

$$\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq \mathbb{N}~\text{finite}\right\}$$

is bounded. (this does not hold for negative terms).

So we mimic this definition. If $I\rightarrow \mathbb{R}:i\rightarrow u_i$ is all positive, then we say that it is "summable" if

$$\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq I\right~\text{finite}\}$$

is bounded.

It can now be proven that if $I\rightarrow \mathbb{R}:i\rightarrow u_i$ is "summable", then at most countably many terms are nonzero. Indeed, let's put

$$D_n=\{i\in I~\vert~a_i\geq 1/n\}$$

The $D_n$ is finite. Otherwise, we could find a subset of k elements with k>nM with M an upper bound of

$$\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq I\right~\text{finite}\}$$

But then

$$\sum_{i\in D_n}{a_i}>M$$

which is a contradiction. This implies that the $D_n$ are finite. And thus

$$\{i\in I~\vert~a_i\neq 0\}=\bigcup_n{D_n}$$

is countable.

So summable families coincide with convergent series.

Last edited: Nov 13, 2011
3. Nov 13, 2011

### Damidami

Hi micromass,
Thanks, it's pretty clear!
Some questions thought:
1) In all instances when you write
$$K \subseteq I$$
do we have to assume that K is finite? (for the sum to have any sense)
2) In the last step when you say
$$\displaystyle \cup_n D_n$$
is countable, you are using that a countable union of finite sets is countable?
3) In all cases we are assumming the standard topology in
$$R$$
? (the order topology when we say the sum is bounded)? Or isn't a sense of limit needed somehow to say it's summable isn't equivalent to say that it converges?

4. Nov 13, 2011

### micromass

Staff Emeritus
Yes, I'm sorry. All of it needs to be finite. I forgot that.

Yes.

Yes, we assume the standard topology on $\mathbb{R}$.

If you're acquainted with nets, then you'll see that I just described a net here.

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