Timeline for Associativity of infinite products
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 26 at 20:52 | history | undeleted |
Calum Gilhooley Eric Wofsey Moishe Kohan |
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| Jun 23 at 0:32 | history | deleted |
amWhy Harish Chandra Rajpoot Ѕᴀᴀᴅ |
via Vote | |
| Jun 22 at 20:17 | review | Low quality posts | |||
| Jun 23 at 0:32 | |||||
| Jun 13 at 20:04 | comment | added | Calum Gilhooley | I have made it clear that I don't wish to continue this conversation, so please don't force me to do so against my will, by twisting my words. In the last two sentences of my last comment but one, I addressed your criticism of my application of Theorem 2 by making a simple request, which you simply ignored (as you ignored a similar request in my very first comment). By all means ignore my comment about Theorems 1 and 2! On all previous occasions I have resisted the temptation to speculate about what is bothering you, and I wish I had done so on that occasion too, although I was trying to help. | |
| Jun 13 at 19:12 | comment | added | Hilbert Jr. | You don't seem to understand my concern: I have no doubts as to the veracity of Theorem 2, having proved it for myself in the past, what I don't understand is how you're applying it to this situation. | |
| Jun 13 at 14:45 | comment | added | Calum Gilhooley | Theorem 2 is a reason why! I think we are done here. I certainly have nothing more to add. OK, one thing: I'm probably not much happier than you are with having to take such a lot on trust from Bourbaki. I most certainly have not checked all the details of all the proofs, or even of all the definitions! But I trust Bourbaki to have got their sums right. Theorems 1 and 2, no matter what their internal complexity, can be applied to the present problem like "black boxes"; and the structure of the argument as a whole, as given in the last paragraph of my answer, is blessedly simple, is it not? | |
| Jun 13 at 14:26 | comment | added | Hilbert Jr. | Please edit your answer with a concrete proof that $(|\prod_{j\in J_n}z_j-1|)_{n\in\mathbb{N}}$ is summable (not just "we're done by Theorem 2"), because I see absolutely no reason why it should be or why Theorem 2 is applicable to it. Our original additive family is $(|z_j-1|)_{j\in\Bbb N}$ (which is summable), from which we pass to the additive family $(\sum_{j\in J_n}|z_j-1|)_{n\in\Bbb N}$ (which is summable by Theorem 2). As I've already said several times, I don't see how this last additive family relates to the multiplicative family $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$. | |
| Jun 13 at 14:11 | comment | added | Calum Gilhooley | You correctly assert that the additive family corresponding to $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$ is $(|\prod_{j\in J_n}z_j-1|)_{n\in\mathbb{N}},$ but you go astray in seeking a second proof of the summability of this family. It would be nice to have a proof which does not require wading through two volumes of Bourbaki (replete with distractions such as "commutative convergence"); indeed, I delayed posting my answer, in the hope that someone would post such a proof. Nevertheless, Theorem 2 does apply. If you maintain that it does not, please say which of its hypotheses are not satisfied. | |
| Jun 13 at 10:06 | comment | added | Hilbert Jr. | I'm saying that Theorem 2 isn't applicable to $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$ in the way you seem to apply it. In order for $(z_n)_{n\in\mathbb N}$ to be multipliable, the family $(|z_n-1|)_{n\in\mathbb N}$ needs to be summable, yes? So the additive family corresponding to $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$ would be $(|\prod_{j\in J_n}z_j-1|)_{n\in\mathbb{N}}$, no? How do you relate this to the family $(\sum_{j\in J_n}|z_j-1|)_{n\in\Bbb N}$? This family is indeed summable by Theorem 2, but it has no discernible relation to $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$ that I can see. | |
| Jun 12 at 20:41 | comment | added | Calum Gilhooley | Are you saying, then, that Theorem 2 does not imply that the family $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$ is multipliable in $\mathbb{C}^*$? (I'm sorry if I'm missing your point; but in order to try to understand, I asked you to specify, if possible, precisely which part of the last paragraph of my answer is incorrect. You haven't done that yet, so I'm trying to guess - with little sense that I'm guessing correctly.) | |
| Jun 12 at 20:10 | comment | added | Hilbert Jr. | I fully agree that $(z_j)_{j\in\Bbb N}$ absolutely multipliable $\implies(|z_j-1|)_{j\in\Bbb N}$ summable $\implies(\sum_{j\in J_n}|z_j-1|)_{n\in\Bbb N}$ summable, but this implication chain has no (discernible, to me) relation to $(\prod_{j\in J_n}z_j)_{n\in\Bbb N}$ being absolutely multipliable, since the corresponding family whose summability we'd have to investigate is $(|\prod_{j\in J_n}z_j-1|)_{n\in\Bbb N}$. | |
| Jun 12 at 17:03 | comment | added | Calum Gilhooley | Which of the logical steps in the last paragraph of my answer do you suspect may be invalid? (I'm sorry if I've made an obvious goof; but I just can't see it.) | |
| Jun 11 at 21:40 | comment | added | Hilbert Jr. | I don't see how the multipliability of $(\prod_{j\in J_n}z_j)_n$ follows from the summability of the original family. The problem is that to go from a product to a sum you need to pass from $(z_n)_n$ to $(|z_n-1|)_n$ and in this case I don't see how $(\prod_{j\in J_n}z_j)_n$ relates to $(\sum_{j\in J_n}|z_j-1|)_n$, which is indeed summable. | |
| May 20 at 21:42 | history | answered | Calum Gilhooley | CC BY-SA 4.0 |