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This set$$\{E\subset(0,1]:E\text{ is countable or } E^c\text{ is countalbe}\}$$is a $\sigma$-algebra.I strongly believe it can't be generated by some countable collection of sets. But I don't know how to prove it rigorously. Thanks!

By Nate Eldredge's hints, I write the details here:

For hint 1: for the each set $A_i\in \mathcal{A}$ with $A_i^c$ is countable, we can replace $A_i$ by $A_i^c$. For the new set $\mathcal A'$, we have $\sigma(\mathcal A)=\sigma(\mathcal A')$. So "without loss of generality, we can assume all the $A_i$ are countable."

For hint 2: Because every $A_i\in \mathcal{A}$ is countable, so $\mathcal A$ can only cover countable many point in $(0,1]$. There must be some point $x\in (0,1]$, and $x\notin A_i,i=1,2,...$. Let $B=\{x\}$.

For hint 3: Let $A=\bigcup_{i=1}^\infty A_i$, and $$\mathcal F=\{E\subset(0,1]:E\subset A\text{ or }E^c\subset A\}$$ We can easily check $\mathcal F$ is a $\sigma$-algebra, and $\mathcal A\subset\mathcal F$. Hence $\sigma(\mathcal A)\subset\mathcal F$ but $B=\{x\}\notin\mathcal F$.

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Suppose it could be generated by some countable collection of sets $\mathcal{A} = \{A_1, A_2, \dots\}$. Now try to find a countable set $B$ which is not in $\sigma(\mathcal{A})$. Hint 1: Without loss of generality, we can assume all the $A_i$ are countable. Hint 2: $B$ could be a singleton. Hint 3: Try looking for a $\sigma$-algebra $\mathcal{F}$ such that $\mathcal{A} \subset \mathcal{F}$ (and hence $\sigma(\mathcal{A}) \subset \mathcal{F}$) but $B \notin \mathcal{F}$.

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  • $\begingroup$ Yes, I get it. Thank you very much! $\endgroup$ – Danielsen Nov 22 '12 at 4:28
  • $\begingroup$ @Danielsen, would you mind presenting your solution? I'm curious about the details, but unfortunately couldn't follow Nate Eldredge's hints. $\endgroup$ – Kuba Helsztyński Nov 22 '12 at 15:33

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