Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook