Timeline for Is every element of $\omega_1$ the rank of some Borel set?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2020 at 14:53 | vote | accept | Hannes Jakob | ||
| Nov 12, 2020 at 14:35 | answer | added | Alessandro Codenotti | timeline score: 18 | |
| Nov 12, 2020 at 14:16 | comment | added | Alessandro Codenotti | In $\mathsf{ZFC}$ the standard way of showing that the Borel hierarchy has length $\omega_1$ is through the construction of so called "universal sets" for the various levels of the Borel hierarchy, see chapter 22 of Kechris Classical Descriptive Set Theory | |
| Nov 12, 2020 at 14:07 | comment | added | Alessandro Codenotti | In $\mathsf{ZF}$ it is consistent that every set of Reals is the countable union of countable sets, each such set is a countable union of singletons, so every set in $\Bbb R$ is $F_{\sigma\sigma}$ and we get $P(\Bbb R)=\mathbf{\Sigma^0_4}=\mathbf{\Pi}^0_4$.It is also possible to have longer Borel hierarchies in $\mathsf{ZF}$, for example of length $\omega_2$ | |
| Nov 12, 2020 at 14:01 | comment | added | YCor | Actually that $B_{\omega_1}$ is a $\sigma$-algebra seems to use some countable choice (DC or AC, I'm not sure): indeed according to A. Caicedo's answer to the linked question ZF is consistent with "$B_{\omega_1}$ is not a $\sigma$-algebra". | |
| Nov 12, 2020 at 13:58 | comment | added | YCor | This is equivalent to asking whether no $B_\alpha$ equals $B_{\omega_1}$ for any $\alpha<\omega_1$. The answer is known to be positive (at least for ZFC) but I'm not aware of the argument. | |
| Nov 12, 2020 at 13:58 | comment | added | Wojowu | See this thread on Math.SE | |
| Nov 12, 2020 at 13:56 | history | edited | YCor | CC BY-SA 4.0 |
added tags, formatting
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| Nov 12, 2020 at 13:47 | history | asked | Hannes Jakob | CC BY-SA 4.0 |