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

  • $\begingroup$ 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 $2^\omega$. $\endgroup$ – Brian M. Scott Feb 10 '12 at 6:59
  • $\begingroup$ 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.) $\endgroup$ – Brian M. Scott Feb 10 '12 at 7:01

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 $2^{\aleph_0}$.

  • $\begingroup$ 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 $\mathbb{N}$. $\endgroup$ – Brian M. Scott Feb 10 '12 at 7:35
  • $\begingroup$ @Brian: Thanks - I just learned something new. I was being naive. $\endgroup$ – user2093 Feb 10 '12 at 7:54
  • $\begingroup$ @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. $\endgroup$ – Michael Greinecker Feb 18 '12 at 16:46
  • $\begingroup$ @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. $\endgroup$ – Slade Oct 4 '13 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.