# $\sigma$-field and Uncountable ordinal

I have been trying to get my head around this question. Any help greatly appreciated.

Let $\mathscr{C}$ be any class of subsets $\Omega$ with $\emptyset,\Omega\in \mathscr{C}$. Define $\mathscr{C}_0=\mathscr{C}$ and for any ordinal $\alpha>0$ write inductively, $$\mathscr{C}_\alpha=\Big(\bigcup\left\{ \mathscr{C}_\beta:\beta<\alpha\right\} \Big)^{\prime}$$ where $\mathscr{D}^{\prime}$ denotes the class of all countable unions of differences of sets in $\mathscr{D}$. Let $\mathscr{S}=\bigcup\left\{\mathscr{C}_\alpha:\alpha<\omega_1 \right\}$ where $\omega_1$ is the first uncountable ordinal and let $\mathscr{F}$ be the minimal $\sigma$-field over $\mathscr{C}$.

Show $\mathscr{S}\subset \mathscr{F}$ and $\mathscr{C}\subset\mathscr{C}_\alpha$. Also, $\mathscr{C}_\alpha$ increases with $\alpha$.

• The least uncountable ordinal is usually denoted by $\omega_1$. – Asaf Karagila Jun 13 '14 at 22:52
• This is the notation of the book I am using, but thanks anyway. – user64066 Jun 13 '14 at 22:54
• The essential property to use is that the first uncountable $\omega_1$ does not have countable cofinality. That is, for any countable family of ordinals smaller than $\omega_1$, there is an ordinal smaller than $\omega_1$ larger than all of them. – Michael Greinecker Jun 13 '14 at 22:55
• Does this count as proof of $\mathscr{S}\subset\mathscr{F}$? Let $\Lambda$ be index set of $\alpha$ where $\mathscr{C}_\alpha\not\subset\mathscr{F}$. Let $\alpha_0=\inf\Lambda$, so $A\in\mathscr{F}$ where $A=\cup_{i=1}D_i$, but $A\not\in\mathscr{F}$. But then for some $D_i\in\mathscr{C}_\beta$ with $\beta<\alpha_0$ we have $\mathscr{C}_\beta\not\subset\mathscr{F}$. This is contradiction thus, $\mathscr{C}_\alpha\subset\mathscr{F}$ for all $\alpha$ which implies $\mathscr{S}\subset\mathscr{F}$. – user64066 Jun 13 '14 at 23:18