Timeline for How to prove that the Lebesgue $\sigma$-Algebra is not countably generated?
Current License: CC BY-SA 4.0
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| Jul 12, 2023 at 10:32 | comment | added | Joel David Hamkins | The basic lemma is that if you have $\kappa$ many subsets of a set, with $\kappa\geq 2$, then the $\sigma$-algebra they generate has size at most $\kappa^\omega$. The proof is to perform the transfinite process I described. At each countable stage, you'll have at most $\kappa^\omega$ many, and by stage $\omega_1$ you have a $\sigma$-algebra. Finally, since $\omega_1\cdot\kappa^\omega=\kappa^\omega$, this means the generated algebra has size at most $\kappa^\omega$ altogether. | |
| Jul 12, 2023 at 2:25 | comment | added | Joel David Hamkins | Basically, the countably generated case turns into the continuum generated case after one step. | |
| Jul 12, 2023 at 0:32 | comment | added | Joel David Hamkins | @DieterKadelka It seems to me that even for the countably generated case, you still need to do the calculation that the generated $\sigma$-algebra has size at most continuum. And that argument amounts to the same as the argument I gave, since it blows up to size continuum at the first step, and you still need to go out to $\omega_1$. So I don't think the countably generated case is any easier than the continuum-generated case. Right? | |
| Jul 11, 2023 at 21:50 | comment | added | Dieter Kadelka | The last part of your answer alone proves in an elementary way that $\cal{L}(\mathbb{R})$ is not countably generated. Of course you prove much more. | |
| Jul 11, 2023 at 20:46 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jul 11, 2023 at 20:24 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jul 11, 2023 at 20:07 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jul 11, 2023 at 20:03 | vote | accept | Joris Wk | ||
| Jul 11, 2023 at 19:58 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |