Timeline for Posets such that the collection of principal down-sets does not have property ${\bf B}$
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2023 at 20:13 | comment | added | Joel David Hamkins | I agree with Andrej that this usage is often bad, since people sometimes use it to mean that all three are different, but that isn't actually what it says. A similar problem arises with $p\iff q\iff r$, which people use to mean that all three are equivalent, but one cannot put brackets in to make it have this meaning. | |
| Aug 9, 2023 at 19:05 | answer | added | KP Hart | timeline score: 3 | |
| Aug 7, 2023 at 21:23 | vote | accept | Dominic van der Zypen | ||
| Aug 7, 2023 at 17:23 | answer | added | KP Hart | timeline score: 4 | |
| Jul 21, 2023 at 18:14 | comment | added | Dominic van der Zypen | You're welcome. - Why do you think it should be avoided? | |
| Jul 21, 2023 at 17:00 | comment | added | Andrej Bauer | Thanks for the explanation. (I think $x \neq y \neq z$ is better avoided in general.) | |
| Jul 21, 2023 at 16:42 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
added 37 characters in body
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| Jul 21, 2023 at 16:28 | comment | added | Dominic van der Zypen | Oh right... and your answer to my previous question implies that it is also consistent that for all posets, the hypergraph of principal down-sets has property ${\bf B}$? | |
| Jul 21, 2023 at 16:23 | comment | added | Joel David Hamkins | But now the question is answered by the remark you made before, concerning a model of ZF with a non-Ramsey function. | |
| Jul 21, 2023 at 16:16 | comment | added | Dominic van der Zypen | @AndrejBauer I would take $x \neq y \neq z$ to mean $[x\neq y \text{ and } y \neq z]$, so that $x = z$ would be possible. However, in the special case of this question, if $S\cap e\neq \emptyset$ and $e\setminus S\neq \emptyset$ we cannot have $S\cap e = e\setminus S$. | |
| Jul 21, 2023 at 16:14 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
added 24 characters in body
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| Jul 21, 2023 at 16:13 | comment | added | Dominic van der Zypen | @JoelDavidHamkins - correct, will reformulate the question | |
| Jul 21, 2023 at 16:06 | comment | added | Andrej Bauer | What does $x \neq y \neq z$ mean? | |
| Jul 21, 2023 at 15:58 | comment | added | Joel David Hamkins | But if you are still asking for a poset without property B, there seems no AC angle any longer. I'm not yet sure how to do it even in ZFC, but you ask about ZF. Indeed, I suspect that ZFC proves that every poset has property B, so we shouldn't expect ZF to prove there is a poset without B. Rather, the question should be: is it consistent with ZF that there is a poset without property B? | |
| Jul 21, 2023 at 15:56 | comment | added | Dominic van der Zypen | Thanks @JoelDavidHamkins for spotting this, and apologies for the wrong part in the question. I hope that I got it right this time and that the bit after the question which is supposed to have a clarifying function is not confusing (or wrong). | |
| Jul 21, 2023 at 15:55 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
removed wrong part and added (hopfefully) clarifying bit after question
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| Jul 21, 2023 at 15:53 | comment | added | Joel David Hamkins | Yes, that is right. | |
| Jul 21, 2023 at 15:52 | comment | added | Dominic van der Zypen | If I get it right this time: In ${\sf (ZFC)}$, the poset $([\omega]^\omega, \subseteq)$ does have property ${\bf B}$, and if I understand your answer to my previous question correctly, it is consistent in ${\sf (ZF)}$ that $([\omega]^\omega, \subseteq)$ does not have property ${\bf B}$. | |
| Jul 21, 2023 at 15:49 | comment | added | Dominic van der Zypen | You are right! I got it exactly wrong. I am wondering whether there is a poset $(P, \leq)$ such that no matter how you color $P$, you will get a mono-chromatic down-set (with more than $1$ element). Will modify the question! | |
| Jul 21, 2023 at 15:36 | comment | added | Joel David Hamkins | More specifically, one uses AC to pick representatives from each finite-difference equivalence class, and then let $S$ be the family of sets that have even-difference from their chosen representative. This shows that $[\omega]^\omega$ fulfills property B, not that it omits it. Right? | |
| Jul 21, 2023 at 15:03 | comment | added | Joel David Hamkins | I'm confused. In your blog post (and here), you seem to say that the existence of a non-Ramsey function $c:[\omega]^\omega\to 2$ is equivalent to the question whether the poset $\langle[\omega]^\omega,\subset\rangle$ does not have property B. But isn't this backwards? A non-Ramsey function shows that the poset does have property B, since the coloring provides the set $S$, the non-Ramseyness means that no set is homogeneous, which makes it fulfill property B. Please help me out. | |
| Jul 21, 2023 at 13:21 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |