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Let $X$ is a nonempty set with $m$ members . How many $\sigma$ -algebra can we make on this set?

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    @DavideGiraudo I think that would be a good idea. I also think that this question is applicable to a greater audience so perhaps some simpler notation and explanation may be in order. Thank you for responding :)2015-04-14

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Short answer. You can make as much $\sigma$-algebras as partitions on $X$.

Formal verification. Let $\mathcal A$ be the collection of all the algebras over $X$, and let $\Pi$ be all the partitions of $X$ (which consist of non-empty sets). There is a bijective correspondence between $\mathcal A$ and $\Pi$. Indeed, for a partition $P\in \Pi$, consider $\mathfrak A_{\Pi}$ the algebra generated by $\{A_1,\ldots,A_k\}$, the elements of $P$, i.e. $\mathfrak A_{\Pi}$ consists of the set $\bigcup_{j\in J}A_j$, where $J\subset \{1,\ldots,k\}$. To see that this correspondence is bijective, given an algebra $\mathfrak A$, you can define for all $x\in X$ the set $A_x:=\bigcap_{A\in \mathfrak A,x\in A}A$ (it is a finite intersection), and that will give you a unique partition.

Indeed, define the equivalence relation $x\sim y$ if and only if $A_x=A_y$. It gives a partition, and it is the unique one. If $P=\{S_1,\ldots,S_m\}$ works, then $A_x=S_{i(x)}$ for some $i(x)$, and you can check that this partition consists of the equivalence classes of $\sim$.

So the problem is to enumerate the number of partitions of a set which has $m$ elements. You will have to use Bell's numbers.

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    What about when X is countably infinite? Does it still work? That intersection of elements of the sigma-algebra would no longer be finite, correct?2015-09-11