Timeline for Are there exotic polynomial bijections from $\mathbb N^d$ onto $\mathbb N$?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2021 at 23:14 | comment | added | Joel David Hamkins | @PaceNielsen I agree that that was what he probably meant, which is why I had commented, because I think despite your remark about "perfectly valid" that the phrase "so-called" is commonly taken by English speakers to carry the negative connotation (unlike its counterpart in other languages). | |
| Aug 9, 2021 at 20:56 | comment | added | Pace Nielsen | @JoelDavidHamkins I thought it was pretty clear that Roland was using "so-called" in the "commonly named" sense, rather than the "falsely or inappropriately named" sense, which is a perfectly valid form of the adjective. It is hard to read his original sentence as finding fault specifically with "Cantor" when "so-called" is modifying the entire phrase "Cantor bijection". | |
| Aug 9, 2021 at 20:31 | history | edited | Roland Bacher | CC BY-SA 4.0 |
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| Aug 9, 2021 at 20:27 | comment | added | Roland Bacher | Sorry, no skepticism was not intended. I have suppressed the ill-chosen adjectif. | |
| S Aug 9, 2021 at 19:00 | history | suggested | Buzz | CC BY-SA 4.0 |
closed bracket
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| Aug 9, 2021 at 18:46 | comment | added | Joel David Hamkins | "So-called"? Do you intend the negative connotation that this bijection should not be attributed to Cantor? (I ask because non-native speakers of English are sometimes surprised to hear that the use of "so-called" expresses skepticism.) | |
| Aug 9, 2021 at 18:18 | review | Suggested edits | |||
| S Aug 9, 2021 at 19:00 | |||||
| Aug 8, 2021 at 13:45 | vote | accept | Roland Bacher | ||
| Aug 7, 2021 at 11:42 | comment | added | user44191 | @WlodAA But in that case, with 3 variables your formula generalizes to $(x, y, z) \mapsto {x + y + z \choose 3} - {x + y \choose 2} + {x \choose 1} - {0 \choose 0}$. | |
| Aug 7, 2021 at 11:27 | answer | added | Peter Taylor | timeline score: 14 | |
| Aug 7, 2021 at 7:55 | answer | added | Aaron Meyerowitz | timeline score: 7 | |
| Aug 5, 2021 at 9:33 | history | edited | Roland Bacher | CC BY-SA 4.0 |
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| Aug 5, 2021 at 8:51 | comment | added | Wlod AA | You may be consistent: $(x,y)\longmapsto {x+y\choose 2}-{x\choose 1}+{0\choose 0}$ | |
| Aug 5, 2021 at 7:01 | comment | added | Roland Bacher | Thanks for pointing out some subtleties ignored by me of the English language. (I have changed the text in order to avoid ambguity.) | |
| Aug 5, 2021 at 6:58 | history | edited | Roland Bacher | CC BY-SA 4.0 |
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| Aug 5, 2021 at 2:47 | comment | added | Gerry Myerson | These are the "little Schroeder numbers," oeis.org/A001003 ? | |
| Aug 4, 2021 at 22:52 | comment | added | Aaron Meyerowitz | Aha, as in “To be ignorant of” | |
| Aug 4, 2021 at 22:45 | comment | added | Peter Taylor | @LSpice, the original meaning of "ignore" is "be ignorant of", and while it's fallen out of favour in English it persists in other languages with Latin influence. | |
| Aug 4, 2021 at 22:17 | comment | added | LSpice | Does "I ignore if an exotic bijection is known for $d = 3$" mean that you don't know (which is suggested by context), or you don't care (which seems to be the literal meaning)? | |
| Aug 4, 2021 at 22:17 | history | edited | LSpice | CC BY-SA 4.0 |
Semantic dots; `\mod` -> `\bmod`
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| Aug 4, 2021 at 21:38 | history | asked | Roland Bacher | CC BY-SA 4.0 |