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I had some doubt with my proof, but I'll list the question here along with the proof:

Claim: Show that the cardinality of a finite $\sigma$-algebra $\mathfrak{M}$ on a set $X$ is $2^n$ for $n \in \mathbb{N}$. Describe the exponent $n$ in terms of $\sigma$-algebra.

Proof: For each member in $\mathfrak{M}$, we can pair it with its complement by properties of $\mathfrak{M}$. This pairing gives a specific partition. Using this idea, we can pick any combination of members of $X$ such that these sets of members of $X$ forms a partition of $X$. Just observing the collection of partitions of $X$, we consider the partition which has the most cells by superimposing all of the partition on top of each other. So suppose there are $n$ cells for a specific partition of $X$. Let $S$ be the set that contains these $n$ cells. So $\mathcal{P}(S) = 2^n$, where $n$ is the maximum number of cells one can achieve through all possible partition of $X$. From $\mathcal{P}(S)$, we get the other partition of $X$. Hence, $S$ generates $\mathfrak{M}$.

On a similar note, there's a question that is similar to the one I am posing:

If we are given any infinite ($|\mathfrak{M}| =$ is infinite) $\sigma$-algebra $\mathfrak{M}$ on set $X$, then there is a subset with cardinality of the real numbers $2^{\aleph_{0}}$.

Proof: I don't think my argument would work in the infinite case, but I'll give it a go. So I thought that $\mathfrak{M}$ has infinitely many partition of $X$, so if we were to countably infinitely take intersection of the partitions of $X$, we get a countably infinite cells of a partition of $X$. Using the argument by the previous problem, we take the power set of the natural number, which gives us $2^{\aleph_{0}}$. For some reason, I am unsure how this infinite case would work out.

-Thanks in advance

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There is a countably infinite Boolean algebra; see Examples 5 and 6 here. Thus, you can’t hope to show that an infinite B.a. has a subset of cardinality $2^\omega$. –  Brian M. Scott Feb 10 '12 at 6:59
    
The first part looks okay, though I’d express it differently: $n$ is the number of atoms of $\mathfrak{M}$. (And now I’ve a better idea of what you were doing in the second half of the other question.) –  Brian M. Scott Feb 10 '12 at 7:01

1 Answer 1

up vote 1 down vote accepted

For the infinite case, let us pick a countable subset $T$ of $X$ such that $\mathfrak{M}$ induces an infinite sigma algebra over $T$ (that is, define $\mathfrak{M}(T) = {A\cap T: A\in\mathfrak{M}}$). If we manage to show that $\mathfrak{M}(T)$ contains a subset of cardinality $2^{\aleph_0}$, we are done. Hence it is enough to show that an arbitrary infinite sigma algebra over $\mathbb{N}$ contains a subset of cardinality $2^{\aleph_0}$. Let's work with that.

We need more assumptions on $\mathfrak{\mathbb{N}}$ (see Brian's comments). So assume that

$(*)$ there exists $\mathfrak{A} = \{A_k\}_{k\in\mathbb{N}}$ a sequence of nonempty pairwise disjoint subsets of $\mathfrak{M}(\mathbb{N})$.

This sequence is in bijective correspondence with $\mathbb{N}$, hence the set of all possible finite and countable unions of elements of the sequence $\mathfrak{A}$ are in bijective correspondence with the powerset of $\mathbb{N}$, that is, $\mathfrak{M}(\mathbb{N})$ is at least of size $2^{\aleph_0}$. On the other hand, $\mathfrak{M}(\mathbb{N})\subset \mathcal{P}(\mathbb{N})$.

So, given $X$ and an infinite sigma algebra on $X$, there exists (countable) $T\subset X$ such that if the the induced sigma algebra on $T$ satisfies $(*)$, then $\mathfrak{M}(T)$ has cardinality $2^{\aleph_0}$.

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Suppose that $\mathfrak{M}(\mathbb{N})$ is the set of all subsets of $\mathbb{N}$ whose indicator functions are periodic; then $\mathfrak{M}(\mathbb{N})$ is countably infinite and contains no infinite family of pairwise disjoint subsets of $\mathbb{N}$. –  Brian M. Scott Feb 10 '12 at 7:35
    
@Brian: Thanks - I just learned something new. I was being naive. –  William Feb 10 '12 at 7:54
    
@Brian M. Scott: I don't get the issue. Every $\sigma$-algebra on $\mathbb{N}$ is atomic. If there are only finitely many atoms, the $\sigma$-algebra is finite too. So for an infinite $\sigma$-algebra on $\mathbb{N}$, the partition into atoms gives a countable infinite set of pairwise disjoint measurable sets. –  Michael Greinecker Feb 18 '12 at 16:46
    
@BrianM.Scott: This example doesn't work; $\mathfrak{M}(\mathbb{N})$ has the following infinite family of disjoint subsets: $1+2\mathbb{N}$, $2+4\mathbb{N}$, $4+8\mathbb{N}$, $8+16\mathbb{N}$, etc. It's also not a $\sigma$-algebra. –  you-sir-33433 Oct 4 '13 at 0:46

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