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Sums over arbitrary index set

  1. Nov 13, 2011 #1
    I have read somewhere that we can extend the notion of a series of a sequence
    [tex] \sum_{i=1}^{\infty} a_n [/tex]
    to sums over an arbitrary index set, say
    [tex] a : I \to \mathbb{R} [/tex]
    is a family of real number indexed by I, then
    [tex] \sum_{i \in I} a_i [/tex]
    is the sum of all the elements.

    I think the text said that [tex]a_i \geq 0 [/tex] in order to has sense, and that if [tex]a_i \neq 0[/tex] 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
    [tex]a : \mathbb{R_{>0}} \to \mathbb{R}, a(i) = \frac{1}{i}[/tex]
    how does one sum over all it's elements?

    2) Other question, the fact that all the [tex]a_i \geq 0[/tex] 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. jcsd
  3. Nov 13, 2011 #2

    micromass

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    The thing is that we define the convergence slightly different.

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

    [tex]\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq \mathbb{N}~\text{finite}\right\}[/tex]

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

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

    [tex]\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq I\right~\text{finite}\}[/tex]

    is bounded.

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

    [tex]D_n=\{i\in I~\vert~a_i\geq 1/n\}[/tex]

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

    [tex]\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq I\right~\text{finite}\}[/tex]

    But then

    [tex]\sum_{i\in D_n}{a_i}>M[/tex]

    which is a contradiction. This implies that the [itex]D_n[/itex] are finite. And thus

    [tex]\{i\in I~\vert~a_i\neq 0\}=\bigcup_n{D_n}[/tex]

    is countable.

    So summable families coincide with convergent series.
     
    Last edited: Nov 13, 2011
  4. Nov 13, 2011 #3
    Hi micromass,
    Thanks, it's pretty clear!
    Some questions thought:
    1) In all instances when you write
    [tex]K \subseteq I[/tex]
    do we have to assume that K is finite? (for the sum to have any sense)
    2) In the last step when you say
    [tex]\displaystyle \cup_n D_n[/tex]
    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
    [tex]R[/tex]
    ? (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?
     
  5. Nov 13, 2011 #4

    micromass

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

    Yes.

    Yes, we assume the standard topology on [itex]\mathbb{R}[/itex].

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