Timeline for Existence of a certain set of 0/1-sequences without the Axiom of Choice
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2019 at 7:08 | vote | accept | M. Winter | ||
| Aug 18, 2019 at 2:55 | history | became hot network question | |||
| Aug 17, 2019 at 23:20 | answer | added | Andreas Blass | timeline score: 14 | |
| Aug 17, 2019 at 19:23 | comment | added | Asaf Karagila♦ | Ohhhh, I see. I misread. Well, this is not an equivalence relation, so the term selector is definitely unfit. | |
| Aug 17, 2019 at 19:18 | comment | added | M. Winter | @AsafKaragila It is self-explanatory in hindsight :P. But I would say it is something different: if you want so, my sequences are equivalent if they agree exactly on an initial segment of $\Bbb N$. Whether this is a significant difference, I don't know. | |
| Aug 17, 2019 at 19:11 | comment | added | Asaf Karagila♦ | Two sequences are equivalent if they are equal except for finitely many points? I hoped the name would be self explanatory... like "selector". | |
| Aug 17, 2019 at 19:10 | comment | added | M. Winter | @AsafKaragila Sorry, I do not know "mod-finite relation". But what I do sounds like a selector, so ... | |
| Aug 17, 2019 at 19:09 | comment | added | Asaf Karagila♦ | Are you trying to say that this is a selector for "half" of the mod-finite relation? | |
| Aug 17, 2019 at 19:07 | comment | added | M. Winter | @Todd Yeah, had the same feeling. Especially, as $\mathcal X$ contains always either $x$ or its complement (the sequence with entries $1-x_i$). Is it easy to make this feeling more concrete? | |
| Aug 17, 2019 at 19:03 | comment | added | Todd Trimble | This looks very much like existence of a nonprincipal ultrafilter on $\mathbb{N}$, which cannot be proven in ZF. (But is far weaker than AC of course.) | |
| Aug 17, 2019 at 18:43 | history | asked | M. Winter | CC BY-SA 4.0 |