Why can't sequences with non-numerable elements converge?

[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 can't 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 which I made the sum?

Thanks in advance for any help in understanding this.

 

[micromass]

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.

 

[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?

 

[micromass]

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)

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

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?

Yes.

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?

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.