Let $\mathscr{C}\subset \mathscr{P}(\Omega)$ be a class of subsets of a nonempty set $\Omega$ containing $\Omega$ and $\varnothing$. Define $\mathscr{C}_0=\mathscr{C}$ and for each $n\geq 1$ define $$ \mathscr{C}_{n+1}=\mathscr{C}_n\cup \{A^c:~ A\in \mathscr{C}_n\}\cup \{\bigcup_{i=1}^\infty A_{i}: ~\{A_{i}\}_{i=1}^\infty\subset \mathscr{C}_n\}. $$

Question: I am looking for an example of $\Omega$ such that $\mathscr{A}(\mathscr{C})=\bigcup_{i=1}^\infty \mathscr{C}_n$ do not equal to $\sigma(\mathscr{C})$ the sigma álgebra generated by $\mathscr{C}$.

I first try to pick a $\Omega=\mathbb{R}$ and consider $\mathscr{C}$ the class given by the finite and cofinite sets, however applying the above procedure we get the the class of the contable and co-countable subsets, which is a sigma álgebra....

I know that must be examples that $\mathscr{A}(\mathscr{C})=\bigcup_{i=1}^\infty \mathscr{C}_n$ do not equal to $\sigma(\mathscr{C})$ because we need of an ordinal argument in the construction...

  • $\begingroup$ How do you get any infinite but not cofinite set in your example? If $\mathscr C_0$ consists of the finite and cofinite subsets, then $\mathscr C_1$ will be exactly the same as $\mathscr C_0$, and this continues ad infinitum to every $\mathscr C_\alpha$. Note that you're only taking finite unions at each step. $\endgroup$ – hmakholm left over Monica Aug 20 '18 at 15:40
  • $\begingroup$ My apologies, I made a Typo! $m$ must be $\infty$. $\endgroup$ – Eduardo Aug 20 '18 at 15:44
  • 2
    $\begingroup$ Taking $\mathscr C_0$ to be (say) all the open intervals of $\mathbb R$ will yield an example because we know the Borel hierarchy keeps growing all the way to $\omega_1$. But I can't offhand reproduce an argument for this fact ... $\endgroup$ – hmakholm left over Monica Aug 20 '18 at 15:58
  • $\begingroup$ @HenningMakholm Once you have, in $\mathbb R$, Borel sets $B_n$ of level $n$ for all finite $n$, you can get one of level $\omega$ by putting a shrunk copy of $B_n$ in the interval $(n,n+1)$ and taking their union. It's not as good as going all the way up to $\omega_1$ (which you could do by iterating the same idea), but it's enough to answer the original question. $\endgroup$ – Andreas Blass Aug 20 '18 at 21:31
  • $\begingroup$ @AndreasBlass: Indeed. Unfortunately, having missed out on learning descriptive set theory properly, I can't even convince myself that the finite levels all bring something new ... $\endgroup$ – hmakholm left over Monica Aug 20 '18 at 21:51

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