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Problem Statement: Does there exist an infinite $\sigma-$algebra which has only countably many members?

Solution: Denote our $\sigma-$algebra by $M$. If $A$ is a finite collection of disjoint members of $M$, i.e. $A=\{m_1,...,m_n\}$ w/ $m_i\cap m_j=\varnothing$, then for $M$ to be infinite we WLOG may find some $m'\in M$ such that $m'\not\subset \cup A$. Since $M$ is closed under set difference, union, and intersection we may form a novel collection of disjoint members of $M$:

$A'=\{m_1-m',m_1\cap m',...,m_n-m',m_n\cap m',m'-\cup A\}$

which due to the properties of $m'$ has at least one more member than $A$.

Consequently, there exists a countably infinite set $S$ of disjoint members of $M$. If we take $P(S)$, the power set of $S$, then $P$ will be uncountably infinite. But each member $x$ of $P$ will correspond to a unique member of $M$, namely $\cup x$, the union of all members of $x$. Thus $M$ contains uncountably many terms.

Shaun
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Derek H.
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  • @RobArthan Your comment is self-referential. I think you linked the wrong question. – Robert Shore Feb 13 '21 at 22:57
  • @RobertShore: thanks. The possible duplicate is https://math.stackexchange.com/questions/320035/if-s-is-an-infinite-sigma-algebra-on-x-then-s-is-not-countable – Rob Arthan Feb 13 '21 at 23:04

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I see two minor things wrong with this argument as it stands. First, the claim that there exists without loss of generality some $m' \in M$ that is not a subset of $\bigcup A$. This is false: consider the $\sigma$-algebra of open sets on $\mathbb{R} \setminus \{0\}$, and let $A = \{(-\infty,0), (0,\infty)\}$. $A$ is a finite collection of disjoint sets in the $\sigma$-algebra, but it also covers the whole space, so every member of that $\sigma$-algebra is a subset of $\bigcup A$. I suspect you want just $m' \notin A$.

Also, there's a minor issue at the end: if $\emptyset$ is in the $\sigma$-algebra, then the unions aren't unique. That's an easy enough fix, though -- you might as well just exclude $\emptyset$ from $S$.

Reese Johnston
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  • Thanks for your response! I actually meant the 'WLOG' to include cases such as the one you have illustrated. My thinking was such: In general we may break at least one of these finitely many disjoint sets up into further subsets, since otherwise it's necessarily a finite sigma-algebra, and due to the closure of our sigma algebra we may thus enumerate another disjoint set within the algebra. – Derek H. Feb 13 '21 at 23:46
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    That makes sense, and the argument's essentially correct - it just doesn't seem to be quite an appropriate use of "WLOG". Generally, "WLOG" is reserved for situations where, if the thing you want isn't true, a minor transformation can make it true. For example, if talking about a set of two real numbers "a" and "b", if the names have no meaning we can say "WLOG a < b", because if that's not true we can just rename the numbers so that it is. – Reese Johnston Feb 14 '21 at 02:07