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Let $\Omega$ be a countably infinite set. Can one use the $\sum_{x\in\Omega} E$ notation when $\Omega$ is countably infinite? I have only seen this being used when $\Omega$ is a finite set.

To make the question more concrete, suppose I have a function $f\colon\Omega\to\mathbb R_{+}$ in $\ell_1(\Omega)$.

Is it a valid notation / good practice to write \begin{align*} \sum_{x\in\Omega} f(x) \end{align*}

or one should rather let $\mu = \#(\Omega)$ and consider the measure space $(\Omega, \mathcal{P}(\Omega), \mu)$ and denote the sum as \begin{align*} \int_\Omega f\,\mathrm{d}\mu\;\;? \end{align*}

The reason I am suspecting that the first expression is frowned upon is that $\sum_{i=1}^\infty$ is defined as $\lim_{n\to\infty}\sum_{i=1}^n$ however it is not clear how to write a corresponding limit when $\Omega$ is an arbitrary countably infinite set (as opposed to $\Omega=\mathbb N$).

Thanks

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    $\begingroup$ I don't understand you qualms. By comparison $\sum_ia_i$ is only defined when the series is convergent, and it is not clear how to define it when the series is divergent; is that a reason not to ever write the sum in the convergent case? If you later want to generalise to a situation where $\Omega$ is not countable, you might want to choose a notation that can survive the generalisation unchanged, but if you have no plan to do so, why worry? $\endgroup$ Commented Oct 1, 2016 at 15:10
  • $\begingroup$ Let me clarify. For this question $\Omega$ is always a countable set, even when I use the measure theoretic language such as measure space. Furthermore, summing $f(x)$ over all $\Omega$ is always a finite number by virtue of $f$ being in $\ell_1(\Omega)$. $\endgroup$
    – Dosetsu
    Commented Oct 1, 2016 at 15:13
  • $\begingroup$ @MarcvanLeeuwen The question is whether the notation $\sum_{x\in\Omega}$ a valid / recommended notation when $\Omega$ is countably infinite. $\endgroup$
    – Dosetsu
    Commented Oct 1, 2016 at 15:14

1 Answer 1

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It turns out, if $f\in\ell^1(\Omega)$, then it makes sense to write \begin{align*} \sum_{x\in \Omega} f(x) \end{align*} since the order in which we perform the sum does not matter in this case, due unconditional convergence of the underlying series.

If $f\notin\ell^1(\Omega)$, then one has to fix an explicit bijection $b\colon\mathbb{N}\to\Omega$ and explicitly write

\begin{align*} \sum_{i=1}^\infty f(b(i)) := \lim_{n\to\infty}\sum_{i=1}^nf(b(i)) \end{align*}

since in general this sum depends on the chosen bijection. (e.g., https://en.wikipedia.org/wiki/Riemann_series_theorem)

Please let me know if this makes sense.

Thanks

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