Timeline for Representation of free Boolean sigma-algebras
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2017 at 8:59 | comment | added | Robert Furber | Borel sets modulo meagre sets and sets with the Baire property modulo meagre sets are isomorphic (because every set with the Baire property symmetrically differs from an open set by a meagre set, and every Borel set has the Baire property). The relationship between Borel sets and sets with the Baire property is the "Baire category" analogue of the relationship between Borel sets and Lebesgue measurable sets in measure theory. | |
| Sep 9, 2017 at 8:57 | comment | added | Robert Furber | Spectrum is another name for Stone space, and the name I more usually use (I must have lapsed into it without realizing that you might not know what it meant). So the spectrum of $B$ is a complicated non-metrizable space not homeomorphic to the Cantor space. By "we get $B$ back again", I mean that the map taking an $B$ to its corresponding equivalence class of Baire sets modulo meagre sets is an isomorphism (exactly the statement of Loomis-Sikorski). | |
| Sep 7, 2017 at 10:49 | comment | added | user111723 | Sorry, Robert, by spectrum, do you mean the Cantor space without any reference to its metrizability? Also, I am not sure to fully understand what you mean when you say: " if we take the σ-algebra generated by the clopens (or equivalently the σ-algebra of Baire sets) modulo meagre sets, we get B back again". Finally, what is the difference between Borel sets mod meagre sets and sets with BP mod meagre sets? Thank you in advance. | |
| Sep 5, 2017 at 22:25 | comment | added | Robert Furber | In fact, for the spectrum of $B$, the two algebras differ - if we take the $\sigma$-algebra generated by the clopens (or equivalently the $\sigma$-algebra of Baire sets) modulo meagre sets, we get $B$ back again, and if we take Borel sets modulo meagre sets (or we could also use sets with the Baire property (a very different thing from Baire sets) modulo meagre sets), we get the completion of $B$. | |
| Sep 5, 2017 at 22:24 | comment | added | Robert Furber | Yes. You can find a proof of this in Halmos's Lectures on Boolean Algebras, section 13, theorem 4. But note that it is only because $2^\omega$ is metrizable that the $\sigma$-algebra generated by the clopen sets is the Borel $\sigma$-algebra. | |
| Sep 4, 2017 at 16:47 | comment | added | user111723 | Thank you for the clarification. It makes more sense now. So, if I understand correctly, every free Boolean algebra on a countable set of free generators is isomorphic the field of all open-closed sets of the Cantor space $2^\omega$ (which is countable, atomless, but also incomplete). The completion this field of open-closed sets, i.e. the regular open algebra of $2^\omega$, is in turn isomorphic to the Borel algebra of $2^\omega$ mod meagre sets in $2^\omega$ or, equivalently, the $\sigma$-algebra generated by the open-closed sets of $2^\omega$ mod meagre sets. | |
| Sep 2, 2017 at 20:56 | history | answered | Robert Furber | CC BY-SA 3.0 |