Timeline for Is a stronger version of the Erdős-Turan conjecture on arithmetic progessions reasonable? (And related questions.)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2018 at 14:06 | comment | added | Yaakov Baruch | ... Gowers's... | |
| Apr 12, 2018 at 13:43 | comment | added | Yaakov Baruch | @WillBrian: thank you for the pointer to Gower's sensible point. Even though it doesn't answer my question it does manage to reduce the "gap" between problem and solution in a way, by making my question definitely less interesting. | |
| Apr 12, 2018 at 13:33 | history | edited | Yaakov Baruch | CC BY-SA 3.0 |
added 17 characters in body
|
| Apr 12, 2018 at 13:31 | comment | added | Yaakov Baruch | @WillBrian: you are absolutely right! | |
| Apr 12, 2018 at 13:18 | comment | added | Will Brian | Tim Gowers's answer here seems relevant: mathoverflow.net/questions/13230/… | |
| Apr 12, 2018 at 13:13 | comment | added | Will Brian | I think that in your Q1, you mean for the implication to go the other way. Otherwise it's too easy: of course for any $k$, there are sets of arbitrarily small "size" containing length-$k$ progressions: for example, there are lots of length-$4$ progressions in the set $[10^6,10^6+3] \cup [10^{12},10^{12}+3] \cup \dots$. | |
| Apr 12, 2018 at 13:02 | history | asked | Yaakov Baruch | CC BY-SA 3.0 |