Timeline for The partial preorder on $\mathbb N$ generated by the finite axioms of choice
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2018 at 9:38 | vote | accept | Taras Banakh | ||
| Jul 5, 2018 at 9:38 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 63 characters in body
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| Jul 5, 2018 at 9:14 | history | edited | Taras Banakh | CC BY-SA 4.0 |
deleted 9 characters in body
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| Jul 5, 2018 at 9:12 | comment | added | Asaf Karagila♦ | That also works, good. I will edit my answer accordingly. | |
| Jul 5, 2018 at 9:11 | comment | added | Taras Banakh | @AsafKaragila Ok. I will change notation from $AC(n)$ to $AC_n$, which is more near to that from Jech who denote it by $C_n$ (but $C_n$ is ovecharged, for example it can denote the cyclic group of order $n$). | |
| Jul 5, 2018 at 8:53 | comment | added | Asaf Karagila♦ | Let me make a notational remark. Often $\mathsf{AC}(X)$ denotes the axiom of choice restricts to subsets of $X$. In which case $\mathsf{AC}(n)$ is a theorem of ZF for any finite $n$. I have a strong preference to the convention $\mathsf{AC}^A_B(C)$ where $B$ is the size of the family, $A$ is the size of the members in the family, and $C$ is the superset of the sets in the family, along with the understanding that omitting any of the parameters implies a universal quantifier. | |
| Jul 5, 2018 at 8:50 | answer | added | Asaf Karagila♦ | timeline score: 12 | |
| Jul 5, 2018 at 8:48 | comment | added | Emil Jeřábek | “Theorem 3” in Mathieu Baillif’s answer to his own question gives a complete description of the preorder, doesn’t it? | |
| Jul 5, 2018 at 8:28 | history | edited | Taras Banakh | CC BY-SA 4.0 |
put more info in the question
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| Jul 5, 2018 at 8:16 | comment | added | Taras Banakh | @AsafKaragila Yes, I have seen this question, but my interest to this question was motivated by a yesterday discussion with one of my coauthors who currently reads some book in Set Theory (probably Jech's "Axiom of Choice") and told me about $AC(2)=> AC(4)$ so I started to think about this question (without extensive search in the literature) and hoped that asking this question on mathoverflow will give me more quick and complete answer than my own browsing through internet. | |
| Jul 5, 2018 at 8:07 | comment | added | Asaf Karagila♦ | Have you seen mathoverflow.net/questions/202586/… by the way? | |
| Jul 5, 2018 at 8:05 | history | edited | Taras Banakh | CC BY-SA 4.0 |
edited title
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| Jul 5, 2018 at 8:05 | comment | added | Asaf Karagila♦ | I had a conversation with Lorenz Halbeisen about this once. I don't remember the conclusion, though. There's some information in Jech "The Axiom of Choice" (Ch. 7), but I don't remember if there were any significant developments after that. | |
| Jul 5, 2018 at 8:05 | comment | added | Taras Banakh | @AsafKaragila Oh, sorry, Asaf! I had in mind preorder (and wrote so in the body of the question). Now I will fix the title. | |
| Jul 5, 2018 at 8:03 | comment | added | Asaf Karagila♦ | It's not a partial order, since $\sf AC(2)\iff AC(4)$. (You are right to say preorder, though.) | |
| Jul 5, 2018 at 7:55 | history | asked | Taras Banakh | CC BY-SA 4.0 |