Timeline for Is every real number in [0,1] a product of three (or more) Cantor set's numbers?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2023 at 21:11 | vote | accept | Pietro Majer | ||
| Dec 24, 2023 at 11:07 | history | edited | John Bentin | CC BY-SA 4.0 |
minor general editing
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| Dec 22, 2023 at 20:43 | history | became hot network question | |||
| Dec 22, 2023 at 20:20 | history | edited | Pietro Majer | CC BY-SA 4.0 |
added 10 characters in body
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| Dec 22, 2023 at 17:01 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
edited title
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| Dec 22, 2023 at 15:47 | answer | added | Zach Teitler | timeline score: 32 | |
| Dec 22, 2023 at 13:47 | comment | added | Zach Teitler | Yes. arxiv.org/abs/1711.08791, "Cantor Set Arithmetic" by Athreya, Reznick, and Tyson, in the Monthly in 2019: every element in $[0,1]$ can be written as $x^2y$ for some Cantor set elements $x,y$. | |
| Dec 22, 2023 at 13:42 | comment | added | Will Brian | Hmm. I could be wrong, but it looks like the numbers that are products of $2$ members of the Cantor set should be a closed nowhere dense subset of $[0,1]$. If this were true for all $n \geq 2$, then you'd have your answer. | |
| Dec 22, 2023 at 13:32 | comment | added | Joel David Hamkins | I'd suggest "every" in place of "any" in your title and questions. The word "any" is famously ambiguous between $\exists$ and $\forall$. In questions especially, it often seems more to carry the $\exists$ meaning, such as in "Was any student able to solve the problem?" or "Is any doctor on the plane?". But I believe you want the $\forall$ meaning. | |
| Dec 22, 2023 at 13:15 | comment | added | YCor | It may be useful to call it , at least the first time, "the triadic Cantor set". | |
| Dec 22, 2023 at 12:42 | history | asked | Pietro Majer | CC BY-SA 4.0 |