Timeline for How many rearrangements must fail to alter the value of a sum before you conclude that none do?
Current License: CC BY-SA 3.0
19 events
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| Oct 7, 2019 at 16:16 | comment | added | Ben Burns | Now available in Transactions of the AMS: ams.org/journals/tran/0000-000-00/S0002-9947-2019-07881-9/… or DOI doi.org/10.1090/tran/7881 | |
| Dec 26, 2016 at 15:09 | comment | added | Joel David Hamkins | See Todd's MO success story post here: meta.mathoverflow.net/a/3071/1946. | |
| Dec 26, 2016 at 14:59 | comment | added | Joel David Hamkins | In the article, we treated only real series, but I suspect that many of the arguments, because of their fundamental set-theoretic character, will be more generally appliable. | |
| Dec 26, 2016 at 14:51 | comment | added | Maxime Ramzi | Is your article only about real numbers or can its results be applied to other banach spaces ? | |
| Dec 26, 2016 at 14:44 | comment | added | Todd Trimble | Thanks for the edit! I'm going to add this to the list of MO success stories: meta.mathoverflow.net/questions/617/best-of-mathoverflow | |
| Dec 26, 2016 at 14:37 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Dec 26, 2016 at 14:03 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Updated to mention and link to our article, which is now available.
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| Oct 29, 2015 at 17:09 | vote | accept | Michael Hardy | ||
| Aug 20, 2015 at 22:08 | comment | added | Joel David Hamkins | I called this cardinal the rearrangement number in my post last week plus.google.com/u/0/+JoelDavidHamkins1/posts/98mga1VW8ka, where also Yemon Choi makes a funny remark about the title of this question. | |
| Aug 14, 2015 at 20:26 | comment | added | Joel David Hamkins | Unfortunately, I can't yet determine what happens in the Sacks model, or even whether the collection of ground-model permutations still have the rearrangement property after adding one Sacks real (or one Laver real, one Cohen real etc.). | |
| Aug 14, 2015 at 16:45 | comment | added | Andreas Blass | Does "try the Sacks model" count as intuition? Actually, unless I'm overlooking something, a theorem of Zapletal ensures that, if it's consistent to have $\kappa<\mathfrak c$, then this will happen in the Sacks model. | |
| Aug 14, 2015 at 16:39 | comment | added | Joel David Hamkins | Thanks! Do you have any intuition for forcing kappa less than continuum? | |
| Aug 14, 2015 at 16:19 | comment | added | Andreas Blass | Since the forcing in your proof from MA is $\sigma$-centered, you've actually proved that $\kappa\geq\mathfrak p$. I think a slight rephrasing of the argument will give $\kappa\geq\mathfrak h$, but I don't have time to check it now. The idea is that, given a permutation $f$, there is a dense (in $([\omega]^\omega,\subseteq)$) family of sets $J$ such that, if you move the $b_k$'s to positions in $J$ then $f$ will not mess up their relative ordering. | |
| Aug 14, 2015 at 14:44 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Aug 14, 2015 at 14:18 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Aug 14, 2015 at 14:11 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Aug 14, 2015 at 14:06 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Aug 14, 2015 at 13:48 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Aug 14, 2015 at 13:33 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |