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Let $E_1,...,E_n$ be non-empty pair disjoint sets such that $\Omega=\cup_k E_k$. Find the sigma-algebra defined on $\Omega$ generated by the sets $E_k$ and determine the cardinality.

I was thinking about the simplest sigma-algebra: $S=\lbrace{\Omega, \emptyset \rbrace}$ seeing that $\Omega$ being the union it contains all the sets $E_k$. The cardinality would depend on the cardinality of the set $\Omega$. Can someone confirm that or am I missing something?

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It would also have to contain e.g. $E_1 \cup E_2$, which is not in $\{\Omega,\emptyset\}$ if $n > 2$. Moreover, the sets $E_i$ themselves have to be in there. – fuglede Apr 28 '14 at 10:04
This doesn't follow directly from $\Omega=\cup_k E_k$? – user73793 Apr 28 '14 at 10:14
I think you are mistaking $E_i\subset\Omega\,(\,\in S)$ with $E_i\in S$. $E_i\in S$ means that $E_i$ is itself an element of the family of sets $S$. It does not mean that $E_i$ is contained in an element of $S$. Hopes this clears things up. – Ian Apr 28 '14 at 10:21
Now I understand the issue. Thanks for your clarifications. – user73793 Apr 28 '14 at 10:23
up vote 1 down vote accepted

$$\mathcal{F}=\left\{ A_{1}\cup\cdots\cup A_{n}\mid A_{i}\in\left\{ E_{i},\emptyset\right\} \text{ for }i=1,\dots,n\right\} $$

Cardinality $2^{n}$.

It contains every set $E_i$ and is closed under complements ($A\in\mathcal F\Rightarrow A^c\in\mathcal F$) and countable unions.

Any $\sigma$-algebra containing the sets $E_{i}$ will contain any set in $\mathcal{F}$.

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This is clear. Thanks – user73793 Apr 28 '14 at 10:46

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