Timeline for Adding sets not containing arithmetic progressions of length three by forcing
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2015 at 4:52 | vote | accept | Mohammad Golshani | ||
| Mar 2, 2015 at 15:16 | comment | added | Mohammad Golshani | Yes, this is clear, just note that we need to meet countably many dense sets, and we can find the generic filter meeting these dense sets in the ground model, this is why I wrote the remark above. | |
| Mar 2, 2015 at 11:31 | comment | added | Peter LeFanu Lumsdaine | Just as a side remark: the machinery of forcing seems like overkill here. Suppose your question is answered positively, say by some construction sending $(s,N)$ to $(f(s),N+1)$. Then just start with $s_0 = \emptyset$, and iterate $f$ from $(s_0,0)$ to get a sequence of conditions $(s_n, n)$; now $S = \bigcup_n s_n$ gives a counterexample to Erdos-Turan. | |
| Mar 2, 2015 at 3:28 | history | edited | Mohammad Golshani |
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| Mar 1, 2015 at 18:55 | answer | added | Paul McKenney | timeline score: 10 | |
| Mar 1, 2015 at 16:18 | history | asked | Mohammad Golshani | CC BY-SA 3.0 |