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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| Jan 10, 2018 at 23:28 | comment | added | Joel David Hamkins | Simon, your algebra is what is what is known as the random algebra, corresponding to the forcing to add a random real. This is different from the Cohen algebra, which corresponds to the forcing to add a Cohen real. The Cohen algebra is the unique atomless complete Boolean algebra with a countable dense set. Your algebra does not have a countable dense set. | |
| Jan 10, 2018 at 22:19 | comment | added | Simon Henry | @JoelDavidHamkins : My above comment, is actually a question on whether my terminology is correct or if "cohen algebra" really means the thing I'm talking about in my answer as you said. | |
| Jan 10, 2018 at 22:16 | comment | added | Simon Henry | That was a long time ago, but I think that the example I gave in my answer is actually not the Cohen algebra (but I might be wrong on the terminology). From my understanding, the Cohen algebra corresponds to the double negation topology on $[0,1]$ and it admit no absolutely continuous valuation, while the example I mentioned might not be countably generated, but admit a completely additive valuation given by the Lebesgue measure. Through the frame <-> locales duality, the Cohen algebra is the space of "Bair generic reals" while my example is the space of "Lebesgue generic reals". | |
| Feb 21, 2014 at 19:46 | comment | added | Bjørn Kjos-Hanssen | Oh okay, so if we think in terms of propositional logic with variables $p_1,p_2,\ldots$ then an atom would correspond to a complete truth assignment on all the variables, but (1) in the Cohen algebra a complete truth assignment gets identified with 0, and (2) if $\mathbb P(p_n=\text{True})=1/2$ then any particular truth assignment has probability 0, i.e., is Lebesgue negligible, and that's the connection with @Simon Henry's answer. Thanks. | |
| Feb 21, 2014 at 19:38 | comment | added | Joel David Hamkins | I added more details. The completion of an atomless Boolean algebra has no atoms, since the original algebra is dense in it. | |
| Feb 21, 2014 at 19:37 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Improved the exposition
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| Feb 21, 2014 at 19:21 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |