Timeline for C.c.-ness of a forcing notion based on an atomless complete Boolean algebra
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2018 at 5:51 | comment | added | Monroe Eskew | @Jonathan if you allow 0 to be in the descending sequence in the definition of closure, then the notion trivializes. | |
| Oct 15, 2018 at 19:22 | comment | added | Jonathan Schilhan | @MonroeEskew To me $0$ is a lower bound of this sequence but anyway it doesn't matter. Closedness is just not a property of the boolean algebra. | |
| Oct 15, 2018 at 0:50 | history | edited | Zoorado | CC BY-SA 4.0 |
added 19 characters in body
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| Oct 15, 2018 at 0:44 | comment | added | Zoorado | I have made some edits to my question. | |
| Oct 15, 2018 at 0:36 | history | edited | Zoorado | CC BY-SA 4.0 |
See body
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| Oct 15, 2018 at 0:24 | comment | added | Zoorado | @Andreas Blass, in some sense, yes it is intentional. I wanted to simplify the definition and saw no reason this additional condition will give me trouble that is not already there. | |
| Oct 15, 2018 at 0:15 | history | edited | Zoorado | CC BY-SA 4.0 |
edited body
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| Oct 15, 2018 at 0:15 | comment | added | Andreas Blass | In the earlier question that you linked to in your comment here, I don't see any requirement that the pieces $b_0,b_1$ into which $f$ splits an element $b$ must have join $b$. But in the present question, requirement 2 includes $a_0\lor a_1=a$. Is that difference intentional? | |
| Oct 15, 2018 at 0:13 | comment | added | Zoorado | @Monroe Eskew, given $p(a) = (a_0, a_1)$, $p(a_0)$ will be $(a_0, 0)$. | |
| Oct 15, 2018 at 0:09 | history | edited | Zoorado | CC BY-SA 4.0 |
Edit: rephrasing and adding the requirement that $a_1 \wedge b_0 = 0$ for 2 points in domain of which join is not $1$.
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| Oct 15, 2018 at 0:03 | comment | added | Zoorado | @Andrea Blass, I think there is some bad sentence structure to 2. I will edit it. My forcing is supposed to add a function of this kind: mathoverflow.net/questions/311559/… | |
| Oct 14, 2018 at 23:33 | comment | added | Andreas Blass | Trying to be more specific about my previous comment: Suppose that $a,b,c$ are pairwise disjoint elements of $B$ whose join is $1$, and let $x$ be any element of $B$. Then there's a condition $p$ with domain $\{a,b,c\}$ that sends $a$ to $(a\land x,a\land\neg x)$ and similarly for $b$ and $c$. It seems to me that this $p$ will be an atom in the separative quotient of your forcing; all its extensions in your forcing just add more pairs $(m\land x,m\land\neg x)$ for various $m$ and the same $x$, so they're all compatible. What am I missing? | |
| Oct 14, 2018 at 23:26 | comment | added | Andreas Blass | In requirement 2, you say that no two elements of the domain of a condition $p$ should have join equal to 1, but you apparently permit three (or more) elements of that domain to have join 1. That doesn't fit my attempt to understand what this forcing is supposed to do, so I second @MonroeEskew's request for information about the intention behind the forcing. | |
| Oct 14, 2018 at 19:21 | comment | added | Monroe Eskew | Can you say a bit about what you want your forcing to do? | |
| Oct 14, 2018 at 19:12 | comment | added | Monroe Eskew | Maybe you intend this, but if $p(a) = (a_0,a_1)$, then there is no $q \leq p$ such that $a_0 \in dom(q)$. | |
| Oct 14, 2018 at 19:07 | comment | added | Monroe Eskew | @Jonathan, incorrect. In fact quite the opposite is true. A complete boolean algebra is never countably closed. This is because we can form a countable maximal antichain by taking joins of parts of any maximal antichain. If $\{ b_n : n < \omega \}$ is such a partition, then let $a_n = \bigvee_{m\geq n} b_m$. Then the sequence of $a_n$'s gives a descending chain with no lower bound. | |
| Oct 14, 2018 at 18:05 | comment | added | Jonathan Schilhan | If your Boolean algebra is complete, then it is $\lambda$ closed for any $\lambda$. | |
| Oct 14, 2018 at 17:58 | history | edited | Zoorado | CC BY-SA 4.0 |
edited title
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| Oct 14, 2018 at 16:29 | history | asked | Zoorado | CC BY-SA 4.0 |