Series that converge to every real number via permutation

Question

This great answer at MathOverflow, https://mathoverflow.net/a/29488/8784, shows that the set of permutations of $\mathbb N$ is uncountable. However, I did not grasp the fact that he uses: any conditionally convergent series [and that such exists] can be rearranged to converge to any given real number $x$ proves that there is an injection $P$ from the reals to the permutations of $\mathbb N$

How did he arrive at this fact? Is there a known proof?

Are you asking why any conditionally convergent series can be rearranged to converge to any real number, or why this fact shows there is an injection from $\Bbb R$ into the permutations of $\Bbb N$? (I ask because the first seems more likely, but the second is the question you actually asked...) — David C. Ullrich, Apr 07 '16 at 16:22
Given the Riemann Rearrangement Theorem, the desired claim is clear, no? Take your favorite conditionally convergent series ($\sum (-1)^n\frac 1n$ say). As this can be rearranged to get any desired real number, there must be at least as many rearrangements (i.e. permutations) as there are real numbers. — lulu, Apr 07 '16 at 16:29
@DavidC.Ullrich I am asking the first one but I am also interested in clear explanation of the second one. — Mohammad Al-Turkistany, Apr 07 '16 at 16:29
You've been given links to the first. For the second, map $x\in\Bbb R$ to a permutation $\sigma$ such that using $\sigma$ to rearrange the series makes it converge to $x$. This map is certainly injective. — David C. Ullrich, Apr 07 '16 at 16:32

score 1 · Accepted Answer · edited Apr 13 '17 at 12:58

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As discussed in the comments, once you know that any conditionally convergent series can be rearranged to yield any desired real number, we immediately get an injection of $\mathbb R$ into the permutations on $\mathbb N$. To be somewhat explicit about it, pick a conditionally convergent series, such as $\sum (-1)^n\frac 1n$. Then, given $\alpha \in \mathbb R$ we choose a permutation of that sequence which converges to $\alpha$.

Technical note: as the above suggests, there are some issues concerning the Axiom of Choice here. More discussion of this point can be found, e.g., here

edited Apr 13 '17 at 12:58

Community

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answered Apr 07 '16 at 16:49

lulu

70,402

I was going to mention that this needed AC - decided not to mention that when I realized it doesn't! Given $\alpha\in\Bbb R$ and our conditionally convergent series there's an explicit construction of a rearrangement that converges to $\alpha$. – David C. Ullrich Apr 07 '16 at 17:05
@DavidC.Ullrich True. Though I think you need AC if you want to prove that the permutations are bijective with $2^{\mathbb N}$. Mind you, I'm not sure I've got that right. – lulu Apr 07 '16 at 17:10

Series that converge to every real number via permutation

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