Suppose that $X$ is a set and $\mathcal{F}$ is a $\sigma$-algebra on $X$ such that for some uncountable family $\mathcal{A}\subset \mathcal{F}$ we have
- $X = \bigcup \mathcal{A}$,
- for every countable subfamily $\mathcal A_0 \subset \mathcal{A}$ the union $\bigcup \mathcal{A}_0$ is not $X$.
Is it possible to show (in ZFC) that $\sigma(\mathcal{A})$, the $\sigma$-algebra generated by $\mathcal{A}$ is strictly smaller than $\mathcal{F}$?
Some relevant references would be also appreciated.
No. Consider the $\sigma$-algebra $\mathcal P_{< \omega_1}(\omega_1)$ on $\omega_1$ and $\mathcal{A} = \mathcal P_{< \omega_1}(\omega_1)$. Clearly $\bigcup A = \omega_1$, $\sigma(\mathcal{A}) = \mathcal P_{< \omega_1}(\omega_1)$ and for any countable $\mathcal A_0 \subseteq \mathcal A$, we have that $\bigcup \mathcal A_0$ is a countable union of countable subsets of $\omega_1$ and hence not cofinal in (in particular not equal to) $\omega_1$.