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Let $\Sigma$ be a countable $\sigma$-algebra.

Show that there is a sequence $A_1,A_2,...$ of disjoint elements of $\Sigma$ such that every $B$ in $\Sigma$ is a countable union of elements in $\{A_n\}$.

My attempt:

Since $\Sigma$ is countable, we have $\Sigma = \{B_1,B_2,B_3,...\}$

Then I took:

$A_1 = B_1$

$A_2 = B_2 - B_1$

$A_3 = B_3 - (B_2 \cup B_1) $

$A_n = B_n - \bigcup_{i=1}^{n-1}{B_i} $

The sequence $\{A_n\}$ is disjoint and $\{A_n\}\subset \Sigma$.

I'm having trouble showing that every $B$ in $\Sigma$ is a countable union of $\{A_n\}$.

How do I show that every $B$ in $\Sigma$ is a countable union of $\{A_n\}$?

Thanks in advance!

Santos
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    There is no such thing as a countable $\sigma$-algebra: http://math.stackexchange.com/questions/320035/if-s-is-an-infinite-sigma-algebra-on-x-then-s-is-not-countable –  Aug 23 '15 at 23:19
  • There's such a thing as an at most countable $\sigma$-algebra! Well, we know that it has to be finite. So I suppose that it would be more accurate to say that there exists a finite $\sigma$-algebra... Is this question moot then? Nothing useful to glean? – Clark Zinzow Aug 23 '15 at 23:27
  • Byron Schmuland is right: there is no such thing as a countable σ-algebra (in the sense of infinity countable $\sigma$-algebra). Then, either the question above is void or we must read it as "Let $\Sigma$ be an at most countable $\sigma$-algebra", and then the question becomes actually about finite $\sigma$-algebras. – Ramiro Aug 24 '15 at 00:30
  • Well, I think there is a mistake in my homework. Thanks for helping me. – Santos Aug 24 '15 at 00:56
  • @Santos countably-generated $\sigma-$algebra? – BCLC Apr 24 '18 at 07:47
  • @ClarkZinzow same question – BCLC Apr 24 '18 at 07:49
  • @Ramiro same question – BCLC Apr 24 '18 at 07:49

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