Why aren't there infinitely countable sigma-algebras?

Why aren't there infinitely countable sigma-algebras?

This is a supplement to one of my analysis discussion sessions.

We will now prove a fascinating result in measure theory: there are no \sigma-algebras that are infinitely countable. This means that any \sigma-algebra S is either finite (and is therefore just an algebra) or very 'BIG' in cardinality, in the sense that it is uncountable. The idea behind this proof is simple. We will take a \sigma-algebra S on a set X, and show that the collection of `smallest' non-trivial sets belonging to S are in one-to-one correspondence with a power set of a certain infinite set.

Let's assume by contradiction that S is an infinitely countable \sigma-algebra defined on a set X. The set X has to be infinite as well (otherwise any \sigma-algebra defined on it is smaller in cardinality than its power set that is finite as well.) Define a function

f:X\rightarrow S

mapping

x\mapsto \bigcap_{x\in A \in S} A

Namely, the function f maps x to the smallest set of the \sigma-algebra S that contains x. Note that this function is well-defined (and maps into S) exactly because of the assumption that S is countable (and therefore any countable intersection remains in S by the definition of a \sigma-algebra.) Since S is a collection of sets, the image of X under f is a subset of the power set of X. In fact, it turns out that f(X) provides us with a partition of X.

Let's try to see why. Consider the images of two points x,y under f and assume by contradiction they have a non-trivial intersection f(x) \cap f(y) \neq \emptyset. If x \notin f(y) then f(x) \setminus f(y) \in S is a smaller set in the \sigma-algebra containing x which is a contradiction to the definition of f. Therefore x \in f(y) and by the same argument y \in f(x).  But then since f maps a point to the smallest set in the \sigma-algebra containing it, x \in f(x) \subseteq f(y) and y \in f(y) \subseteq f(x) and thus we conclude that f(x)=f(y).

We may conclude that f(X) is a partition of X. Each set A in S can be written as the union of such images,

A = \bigcup_{x\in A} f(x)

Therefore, the partition f(X) cannot be finite, as otherwise S would be finite as well. But now if the partition f(X) is infinite, one can form all the sets in S by taking all the possible (disjoint) unions of sets in f(X). This means that the cardinality of S is equal to the cardinality of the power set of f(X), namely

|S|=|P(f(X))|

which is of course uncountable.

This is a contradiction, and therefore the \sigma-algebra S cannot be infinitely countable.