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Let $Y$ be a collection of subsets of the set X. Show that for each $A \in \sigma(Y)$ there is a countable subfamily $B_0 \subset Y$ such that $A\in \sigma(B_0)$

My try: I look at $\cup B_i$ where $B_i$ is a countable subfamily of $Y$. And I want to show that $Y\subset \cup B_i \subset\sigma(Y)$. Both the $\cup B_i \subset\sigma(Y)$ and $Y\subset \cup B_i$ feels intuitive, but how do I write it out rigorously?

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  • $\begingroup$ Could you clarify what is a countable subfamily? Does this mean a collection of a countable number of sets or does this mean a collection of countable sets? $\endgroup$ – user100109 Oct 10 '13 at 20:52
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The way to prove it is to consider the set $\mathcal{C}$ of all $X\in\sigma(Y)$ such that for some countable $B_0\subseteq Y$, $X\in\sigma(B_0)$. Clearly, $\mathcal{C}$ is closed under complements and contains every element of $Y$. If you can show that $\mathcal{C}$ is closed under countable unions, which follows from the fact that countable unions of countable sets are countable, you have established that $\mathcal{C}$ is a $\sigma$-algebra satisfying $$Y\subseteq\mathcal{C}\subseteq\sigma(Y),$$ and hence $\mathcal{C}=\sigma(Y)$.

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  • $\begingroup$ Thanks! can you please expand on why $\mathcal{C}$ is closed under complements and contains every element in $Y$? $\endgroup$ – Johan Feb 1 '13 at 12:10
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    $\begingroup$ @Johan If $X$ is in $\sigma(B_0)$, then $X^c\in\sigma(B_0)$ because $\sigma(B_0)$ is a $\sigma$-algebra. For each $X\in Y$, we have $X=\sigma(\{X\})$, so $X$ is generated by a countable subfamily of $Y$. $\endgroup$ – Michael Greinecker Feb 1 '13 at 17:48
  • $\begingroup$ @MichaelGreinecker How does the fact that countable unions of countable sets is countable help me here? $\endgroup$ – Eran Oct 18 '18 at 16:26
  • $\begingroup$ @Eran Take a countable family $\langle E_n\rangle$ in $\mathcal{C}$. For each set $E_n$ in your countable family, there is a countable family $\mathcal{C}_n\subseteq Y$ such that $E_n\in\sigma(\mathcal{C}_n)$. Then $\bigcup_n\mathcal{C}_n$ is countable and $\bigcup_n E_n\in\sigma(\bigcup_n\mathcal{C}_n)$. $\endgroup$ – Michael Greinecker Oct 18 '18 at 18:58

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