I realize this question has been answered but I wonder if this argument would work.


If the infinite Sigma algebra X contains an infinite sequence of strictly nested nonempty sets, then we are done.

Otherwise, take the set of all maximal strict nests in X, then the smallest element of these nests are disjoint from each other. If there are infinitely many then we are done.

Otherwise, delete those small elements from each set in X and repeat the process above. Since the process can be repeated infinitely many times and the resulting smallest set in each nest form a disjoint set, the proof is finished.

  • $\begingroup$ Please make the question self contained. $\endgroup$ – Joshua Ruiter Apr 10 '17 at 3:50
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    $\begingroup$ Easier solution: $\emptyset,\emptyset,\emptyset,\dots$ is an infinite sequence of disjoint sets. However, if you require the sets to be nonempty, the statement is false: there are finite $\sigma$-algebras. $\endgroup$ – bof Apr 10 '17 at 5:15
  • $\begingroup$ Your "proof" is more like a "hand-waving" level intuition. It seems a good intuition, though. $\endgroup$ – Will M. Apr 12 '17 at 19:03
  • $\begingroup$ @WillM. Would putting it into symbolic form make it rigorous? Is this a common requirement for all proofs? Thanks $\endgroup$ – Zee Apr 12 '17 at 19:51
  • $\begingroup$ The answer below is a formalisation of your intuition. The point is that you made hard claims (minimal sets, etc) that are far from being obvious. $\endgroup$ – Will M. Apr 12 '17 at 19:52

EDIT: your argument does indeed work, although it could be written a bit clearer (e.g. justify why smallest elements of nests exist, and what you mean by "repeat the process infinitely many times" - this isn't hard and is a bit tedious, but it makes for a more complete proof).

Here's another nice way to attack the problem:

Call our $\sigma$-algebra $\mathbb{S}$. Say that a set $A$ is splitting if $A\in\mathbb{S}$ and $$\{A\cap B: B\in\mathbb{S}\}$$ is infinite. (Note that this is equivalent to $\{B: B\in\mathbb{S}, B\subseteq A\}$ being infinite.) That is, $A$ is splitting if it has infinitely many "pieces" in the $\sigma$-algebra. Note that most natural $\sigma$-algebras contain all singletons; for such a $\sigma$-algebra, "splitting" is the same as "infinite and in the $\sigma$-algebra." Note that since the $\sigma$-algebra is finite, the whole underlying set is splitting (finite $\sigma$-algebras, by contrast, have no splitting sets).

Now the crucial fact is the following:

$(*)\quad$ If $A\in\mathbb{S}$ is splitting, then $A$ can be partitioned as $A=B\sqcup C$ where $B, C\in\mathbb{S}$ are nonempty and $C$ is splitting.

(Of course this partition will not in general be unique, and of course both pieces can be splitting.) Note that if $(*)$ is true, we're done: we build a sequence of strictly decreasing sets in $\mathbb{S}$ $X=A_0\supsetneq A_1\supsetneq A_2\supsetneq ...$ (here "$X$" denotes the whole space), and so the sets $B_i=A_i\setminus A_{i+1}$ are disjoint elements of $\mathbb{S}$.

So can you prove $(*)$? HINT: first show that any splitting $A$ has a proper subset which is an element of $\mathbb{S}$; now show that if $D\cup E$ is splitting and $D, E\in\mathbb{S}$, then at least one of $D$ and $E$ is splitting.

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  • $\begingroup$ The reason a smallest element exists is because I assumed so in the first sentence (that any strict nest is finite). Since otherwise , we can easily use that to construct an infinite disjoint set. Thank you very much for the answer. $\endgroup$ – Zee Apr 12 '17 at 19:45
  • $\begingroup$ @Zee Ah, d'oy! Of course you're right. Edited! $\endgroup$ – Noah Schweber Apr 12 '17 at 19:47
  • $\begingroup$ How do you prove that any splitting set has a proper subset which is an element of $\mathbb{S}$ ? $\endgroup$ – Gabriel Romon Aug 2 '18 at 5:38
  • $\begingroup$ @GabrielRomon That follows directly from the definition: $A$ is a splitting set iff the set $\{A\cap B:B\in\mathbb{S}\}$ is infinite. As long as this latter set has at least two elements, that means that there must be some $B$ with $A\cap B\not=A$; but then $A\cap B$ is a proper subset of $A$ which is an element of $\mathbb{S}$. $\endgroup$ – Noah Schweber Aug 2 '18 at 22:34
  • $\begingroup$ Note that $\mathbb{S}$ having infinitely many sets does not mean that $\{A\cap B: B\in\mathbb{S}\}$ is infinite: e.g. if $A=\emptyset$, then this set is just $\{\emptyset\}$ (since $A\cap B_1=A\cap B_2$ for all $B_1,B_2\in\mathbb{S}$). I'm really counting the intersections as sets themselves, and not paying attention to how they got there (I'm not counting the $B$s). $\endgroup$ – Noah Schweber Aug 2 '18 at 22:36

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