Timeline for Complete Boolean algebra not isomorphic to a $\sigma$-algebra
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2014 at 19:45 | comment | added | Joseph Van Name | One can strengthen these distributivity conditions to obtain necessary and sufficient conditions for whether every quotient $B/I$ of a $\sigma$-complete Boolean algebra $B$ by a $\sigma$-complete ultrafilter $I$ can be represented as an algebra of sets, and these algebras can be put into a one-to-one correspondence with the Lindelof $P$-spaces. These results generalize to any Boolean algebra which is endowed with a notion of which least upper bounds are important and which least upper bounds are unimportant. | |
| Feb 22, 2014 at 19:35 | comment | added | Joseph Van Name | The distributivity law mentioned in this problem can be generalized to necessary and sufficient conditions. A $\sigma$-complete Boolean algebra $B$ is representable as a $\sigma$-algebra if and only if whenever $I$ is an index set, $R_{i}\subseteq B$ is a countable set with $\bigvee R_{i}=1$ for $i\in I$, and whenever $x_{i}\in R_{i}$ for $i\in I$ there is a countable $J\subseteq I$ with $\bigwedge_{i\in I}x_{i}=\bigwedge_{j\in J}x_{j}$ (i.e. you have some compactness), then $$\bigvee\{\bigwedge_{i\in I}x_{i}|x_{i}\in R_{i}\,\textrm{for}\,i\in I\}=1.$$ | |
| Feb 22, 2014 at 19:19 | comment | added | Bjørn Kjos-Hanssen | Here is the usual recursion theoretical comment: I gather that if $A_{\langle\sigma,n\rangle,\tau}$ says $\sigma$ is a prefix of $\tau$ and $\tau$ belongs to or strongly avoids a certain dense set $D_n$, then François' last equation says there is an $f$ that given $\sigma$ and $n$ finds such a $\tau$; if the sequence $D_n$ is the collection of $\Sigma^0_1$ sets then $f$ computes a 1-generic $G$. Interesting... | |
| Feb 22, 2014 at 1:54 | comment | added | François G. Dorais | Note that it is consistent with ZFC that there is a ccc algebra that doesn't add reals (a Souslin algebra) and so this answer does not subsume all of Joseph's answer. | |
| Feb 22, 2014 at 1:47 | history | edited | François G. Dorais | CC BY-SA 3.0 |
added 31 characters in body
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| Feb 22, 2014 at 1:35 | history | edited | François G. Dorais | CC BY-SA 3.0 |
added 49 characters in body
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| Feb 22, 2014 at 1:27 | history | answered | François G. Dorais | CC BY-SA 3.0 |