4
$\begingroup$

Let $S\subset S_{\infty}$ be a set of permutations of $\mathbb{N}$. A real series $\sum_{n\geq0}u_{n}$ will be called $S$-conditionally convergent if it is absolutely divergent and if, for all $\sigma\in S$, $\sum_{n\geq0}u_{\sigma(n)}$ converges.

Question 1: Which sets $S$ admit $S$-conditionally convergent series?

By Riemann's rearrangement theorem, $S_{\infty}$ does not admit such a series. Similarly, for any partition $I_{1},\cdots, I_{r}$ of $\mathbb{N}$, $S_{I_{1}}\times\cdots\times S_{I_{r}}$ does not admit such a series, as the same rearrangement argument holds.

On the other hand, any finite set $S=\{\sigma_{1},\cdots,\sigma_{r}\}$ admits an $S$-conditionally convergent series. Indeed, one may find an infinite subset $I$ of $\mathbb{N}$ such that $\sigma_{1}^{-1},\cdots,\sigma_{r}^{-1}$ are monotone upon restriction to $I$, and consider a conditionally convergent series supported in $I$. There is also the notion of a convergence-preserving permutation. Let's say that $\sigma$ satisfies condition $(*)$ if the number of disjoint intervals that make up $\sigma([0,n])$ is bounded with respect to $n$. If all the elements of $S$ satisfy $(*)$ then any conditionally convergent series is $S$-conditionally convergent (see P. Schaefer, "Sum-Preserving Rearrangements of Infinite Series").

One can give a more general sufficient condition: if there is an infinite set $I\subset\mathbb{N}$ such that for every $\sigma\in S$, the permutation $\tilde{\sigma}:\mathbb{N}\cong\sigma^{-1}(I)\rightarrow I\cong\mathbb{N}$ (where the bijections $\cong$ are order-preserving) satisfies $(*)$, then a conditionally convergent series supported in $I$ is $S$-conditionally convergent. Note that under this condition, the sum of the series is also preserved, whereas a set $S$ can admit an $S$-conditionally convergent series whose sum varies depending on the permutation (again by Riemann's rearrangement theorem).

Question 2: Can a countable set $S$ admit no $S$-conditionally convergent series? Can one find such a set of the form $\{\sigma^{k};\;k\in\mathbb{Z}\}$?

$\endgroup$

1 Answer 1

10
$\begingroup$

Let me rephrase your Question 2: How many permutations of the natural numbers are needed so that every conditionally convergent series of real numbers can be rearranged to no longer converge? This very question was explored in the following paper:

Blass, Andreas; Brendle, Jörg; Brian, Will; Hamkins, Joel David; Hardy, Michael; Larson, Paul B., The rearrangement number, Trans. Am. Math. Soc. 373, No. 1, 41-69 (2020). ZBL1516.03015.

The answer: uncountably many permutations are required, the exact number being denoted by $\mathfrak{rr}$, a cardinal number called the rearrangement number, defined to be the answer to the above question. (Well, actually $\mathfrak{rr}$ is defined as the answer to a slightly different question, where the permutations are allowed either to cause divergence or to cause convergence to a different sum. But we prove in our paper that removing the bit about convergence to a different sum does not change the number of required permutations. So $\mathfrak{rr}$ is the answer to the above question, but more by a small theorem than by a definition.)

Because the number of permutations required is uncountable, your question is inherently set theoretic. In our paper, we show that the value of the cardinal $\mathfrak{rr}$ can depend sensitively on what model of set theory you are looking at. It is consistent, for example, that the cardinality of the continuum $\mathfrak{c}$ is anything you like (say $\aleph_{42}$) while the value of $\mathfrak{rr}$ is anything else you like, provided it is at least $\aleph_1$ and at most $\mathfrak{c}$. There is a whole area of combinatorial set theory studying cardinals like this. (For example, the cardinal $\mathrm{non}(\mathcal M)$, defined as the smallest size of a non-meager subset of $\mathbb R$, or the cardinality of the smallest non-Lebesgue null set, or the smallest number of null sets required to cover $\mathbb R$, etc.) A good bit of our paper consists of finding the place of $\mathfrak{rr}$ in this zoo of cardinal numbers. For example, $\mathfrak{rr} \leq \mathrm{non}(\mathcal M)$ (essentially because a "generic" permutation will disrupt the convergence of a given series), but $\mathfrak{rr}$ is $\geq$ the number of null sets required to cover $\mathbb R$ (essentially because a random series, from the appropriate probability space, will not be disturbed by a given permutation).

By the way, I should mention that the collaboration that led to the above paper began with a MathOverflow question (this one) asked by Michael Hardy.

$\endgroup$
2
  • $\begingroup$ Thank you for the answer! I'm surprised I didn't find that MO post earlier. Is there anything one can say about sets $S$ of cardinality $\geq\mathfrak{rr}$ that do admit $S$-conditionally convergent series? I'm thinking for instance self-embeddings of $S_{\infty}$ that satisfy this property. $\endgroup$
    – abeaumont
    Commented Aug 12 at 20:05
  • 1
    $\begingroup$ @abeaumont: Some of the results in our paper could be interpreted that way. For example, you can put a (not group action invariant) measure on $S_\infty$ such that every null $S$ admits an $S$-conditionally convergent series. We define a notion of "jumbling" such that every non-jumbling set $S$ admits an $S$-conditionally convergent series. My guess would be that there are other things you could say along these lines. Please post an update to your question at some point if you end up finding any interesting ones! $\endgroup$
    – Will Brian
    Commented Aug 12 at 20:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .