10
$\begingroup$

I am trying to solve Exercise 3 a) given here. The problem states:

Let $\mathcal{M}$ be an infinite $\sigma$-algebra. Prove that $\mathcal{M}$ contains an infinite sequence of nonempty, disjoint sets. (Hint: if $\mathcal{M}$ contains an infinite sequence of strictly nested sets, then we’re done, so assume that no such sequence exists. Next, use this assumption to find a nonempty set in $\mathcal{M}$ with no nonempty proper subsets in $\mathcal{M}$. Finally, show that this can be done infinitely many times.)

Here's my idea: Let's say $\mathcal{M}$ is $\sigma$-algebra on the set $X$. If $\mathcal{M}$ does not contain an infinite sequence of strictly nested sets, then given any $A\in\mathcal{M}$, every strict chain starting with $A$ must terminate after finite time: i.e. $A\subsetneq E_{1}\subsetneq E_{2}\subsetneq … \subsetneq E_{k}$ and there is no set $X\neq B\in\mathcal{M}$ such that $E_{k}\subsetneq B$. But then the complement $E_k^{c}$ has no nonempty proper subsets in $\mathcal{M}$. But I am having trouble showing that the process can repeated infinitely many times.

I have been stuck with this for the whole day, and it is driving me crazy! Thanks for the help!

$\endgroup$
4
$\begingroup$

Here is a hint: can you show that $\mathcal{M}_1 := \{Y \cap E_k\ | Y \in \mathcal{M}\}$ is an infinite $\sigma$-algebra (and a subset of $\mathcal{M}$) with the same property? Now repeat this process, obtaining $\mathcal{M}_n$ for each $n\in \mathbb{N}$ and finding a set $A_n$ disjoint from each previous set $A_i$ ($i<n$) at every step in the process.

$\endgroup$
  • $\begingroup$ Is $\mathcal{M}_1$ a $\sigma$-algebra though? It doesn't seem to contain $X$ itself. Or maybe it is a $\sigma$-algebra on another set? $\endgroup$ – Prism Dec 16 '13 at 0:57
  • $\begingroup$ It will be a $\sigma$-algebra on $E_k$. $\endgroup$ – universalset Dec 16 '13 at 1:00
  • $\begingroup$ Ah that makes sense :) Thank you! So I can see that $\mathcal{M}_1$ will be a $\sigma$-algebra on $E_k$. I have to think why it is infinite though. But once it is infinite, I can see that the process you describe will lead to infinitely many desired sets. +1! $\endgroup$ – Prism Dec 16 '13 at 1:05
  • $\begingroup$ If $Y\cap E_k=\emptyset$, then $Y\subseteq E_{k}^{c}$, but we know $E_{k}^{c}$ has no nonempty proper subsets. So $Y=\emptyset$. Does this show $\mathcal{M}_{1}$ infinite? $\endgroup$ – Prism Dec 16 '13 at 1:19
  • $\begingroup$ Basically. Indeed, if $Y \neq X$, then $Y \cap E_k = Y$ or $Y = (Y\cap E_k) \cup E_k^c$. So the map $Y \rightarrow Y \cap E_k$ is $2$ to $1$, and so its range must be infinite. $\endgroup$ – universalset Dec 16 '13 at 1:34
4
$\begingroup$

I found a very interesting answer to the question using Hausdorff Maximal Principle here in page 2. Let me take a screenshot for now. I will come back and latexify this when I have the time. enter image description here

$\endgroup$

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.