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Call a bijection $f : \mathbb{N} \rightarrow \mathbb{N}$ irrelevant over $\mathbb{R}$ iff for all sequences $a : \mathbb{N} \rightarrow \mathbb{R}$, if $$\sum_{i=0}^\infty a_n$$ exists, call its value $\lambda$, then $$\sum_{i=0}^\infty a_{f(n)}$$ also exists, and its limits is also $\lambda$.

(I feel slightly bad about the melodramatic title.)

Question. Which bijections $f : \mathbb{N} \rightarrow \mathbb{N}$ are irrelevant over $\mathbb{R}$?

goblin GONE
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    Possibly only those that differ from the identity in a finite amount of terms? Since given a converging, but not absolutely converging, succession, you can permute the terms to make it have pretty much whatever limit it wants... – Riccardo Orlando Apr 29 '16 at 11:53
  • It seems the relevant functions would encode the sort of permutation required to do that. – Daron Apr 29 '16 at 11:55
  • The permutation ${2,1,4,3,6,5,\cdots}$ would appear to preserve partial sums, no? At least, even partial sums are entirely unchanged while odd ones differ in one term only. Similarly, I think a case could be made for any bijection which moves every element by at most a bounded amount. – lulu Apr 29 '16 at 12:17
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    @RiccardoOrlando: however, the infinite permutation $(1,2)(3,4)(5,6)\ldots$ is also irrelevant. You are claiming that the set of irrelevant function is $S_{\infty}$, but it looks to me that any element of $S_{\omega}$ having a decomposition in disjoint cycles of finite length does the job. – Jack D'Aurizio Apr 29 '16 at 12:17
  • @lulu: what about $(1,2)(3,4,5)(6,7,8,9)\ldots$? – Jack D'Aurizio Apr 29 '16 at 12:18
  • @Jack: Not hard to prove for those having a decomposition into cycles of bounded finite length; I’m not so sure about the unbounded case. – Brian M. Scott Apr 29 '16 at 12:19
  • @JackD'Aurizio Unbounded? Not so sure. – lulu Apr 29 '16 at 12:21
  • Neither I am, but the answer is clearly some group $G$ fulfilling $S_{\infty}<G<S_{\omega}$. Interesting question. – Jack D'Aurizio Apr 29 '16 at 12:25
  • I agree...it's a very interesting question. – lulu Apr 29 '16 at 12:25
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    This seems to have been asked already: see here. – Paul McKenney Apr 29 '16 at 18:28

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