# If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable

I am going over a tutorial in my real analysis course and there is an exercise that I don't understand a part of his solution.

The exercise is:

Let $S$ be an infinite $\sigma$ algebra on $X$ .Prove that $S$ is not countable.

The proof given is:

Assume by negation $S=\{A_{i}\}_{i=1}^{\infty}$. For each $x\in X$ define $B_{x}:=\cap_{x\in A_{i}}A_{i}$. Note that $B_{x}\in S$ since this is a countable intersection.

Claim: If $B_{x}\cap B_{y}\neq\emptyset$ then $B_{x}=B_{y}$.

Proof:

If $z\in B_{x}\cap B_{y}$ then $B_{z}\subseteq B_{x}\cap B_{y}$. If $x\not\in B_{z}$then $B_{x}\setminus B_{z}$ is a set in $S$ containing $x$ and is strictly contained in $B_{x}$, in contradiction to the definition of $B_{x}$.

Hence $B_{z}=B_{x}$and similarly $B_{z}=B_{y}$ hence $B_{x}=B_{y}$.

Now consider $\{B_{x}\}_{x\in X}$ - If there is a finite sets of the form $B_{x}$ then $S$ is a union of a finite number of disjoint sets hence it is finite. We conclude there is an infinite number of sets of the form $B_{x}$. $\cup_{i\in A}B_{x_{i}}$ where $A\subseteq\mathbb{N}$ is of at least cardinality $\aleph$.

There are couple of things I don't understand in this proof:

1. Why the fact that we found a set ($B_{x}\setminus B_{z}$) in $S$ containing $x$ and is strictly contained in $B_{x}$ a contradiction ?

2. Why if there are only a finite number of different sets of the form $B_{x}$ then $S$ is a union of a finite number of disjoint sets and is finite ?

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 You may find it helpful to think about the case where the $\sigma$-algebra $S$ separates points; i.e. for every $x,y \in X$ there exists $A \in S$ with $x \in A$, $y \notin A$. In this case, $B_x = \{x\}$, and the problem reduces to showing that the discrete $\sigma$-algebra on an infinite set is uncountable. – Nate Eldredge Mar 4 at 2:36

1. Because $B_x$ is supposed to be the intersection of all measurable sets containing $x$, but you've found a measurable set containing $x$ strictly inside $B_x$.
2. Because for any measurable set $T$, we have $T=\bigcup_{x\in T}B_x$. Thus, if there are $n$ distinct sets of the form $B_x$, then there are at most $2^n$ elements of $S$.
 Thanks Zev! ${}$ – Belgi Mar 4 at 1:44 No problem, glad to help! – Zev Chonoles♦ Mar 4 at 1:45
Show that if a $\sigma$-algebra is infinite, that it contains a countably infinite collection of disjoint subsets. An immediate consequence is that the $\sigma$-algebra is uncountable.