Timeline for About the existence of a particular kind of "splitting" function on atomless complete Boolean algebras
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2018 at 2:18 | vote | accept | Zoorado | ||
| Sep 29, 2018 at 3:55 | |||||
| Sep 29, 2018 at 2:18 | vote | accept | Zoorado | ||
| Sep 29, 2018 at 2:18 | |||||
| Sep 29, 2018 at 2:18 | vote | accept | Zoorado | ||
| Sep 29, 2018 at 2:18 | |||||
| Sep 28, 2018 at 15:43 | answer | added | Adam | timeline score: 1 | |
| Sep 28, 2018 at 11:13 | answer | added | KP Hart | timeline score: 1 | |
| Sep 28, 2018 at 6:17 | history | edited | Zoorado | CC BY-SA 4.0 |
deleted 14 characters in body
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| Sep 28, 2018 at 5:41 | comment | added | YCor | Thanks, you're right. The sup of $c_i$ for $c\ge b$ is the same as the sup for $c$ such that $b\vee c\neq 1$, but that's it. | |
| Sep 28, 2018 at 1:51 | comment | added | Zoorado | For any fixed $b \neq 0$ in the domain, $b_i$ and $(\neg b)_i$ are both defined but $(b \vee \neg b)_i$ is not. Same holds if you replace $\neg b$ with any $c \geq \neg b$ in the previous sentence. | |
| Sep 27, 2018 at 17:07 | comment | added | Zoorado | Sorry, I don't quite get why the supremum does not depend on $b$. $b$ varies over the domain of $f$. | |
| Sep 27, 2018 at 16:54 | comment | added | YCor | Given 4, your condition 5 is a bit hopeless: the supremum $\sup\{c_i:c\ge b\}$, when $b$ is fixed, does not depend on $b$. More precisely, it implies that $b_{1-i}$ does not depend on $b$, so has to be constant. | |
| Sep 27, 2018 at 14:14 | history | edited | YCor | CC BY-SA 4.0 |
edited body; edited title
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| Sep 27, 2018 at 12:44 | history | asked | Zoorado | CC BY-SA 4.0 |