Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable

The following question is from Folland Real Analysis, chapter 1 problem 3.

Let $\mathcal{M}$ be an infinite $\sigma$-algebra. Prove that

a. $\mathcal{M}$ contains an infinite sequence of disjoint sets.

b. $\text{card}(\mathcal{M}) \ge \mathfrak{c}$.

This is the problem I'm totally stuck at. First, I think there is a missing condition in (a). For (a) to be meaningful, (a) should be corrected : "M contains an infinite collection of disjoint and nonempty sets." But I can't find a way to construct such a collection of sets.

Could anyone show me how to solve it?

Solution 1:

Let $X$ be the whole space. First we show that

there is $E\in\mathcal{M}$ such that the restriction of $\mathcal{M}$ to $E^c$ is still infinite.

If no such $E$ existed, then pick any $\emptyset\neq E\in\mathcal{M}$. The restriction of $\mathcal{M}$ to $E^c$ is finite. But the restriction to $E$ must also be finite because otherwise we could take $E^c$ for the role of $E$. Notice that $\mathcal{M}$ would be generated by the two finite, and disjoint, restrictions and that would imply it is itself finite.

Now apply induction to define the infinite sequence. Pick the first $E_0$ with that property, $E_1$ with the same property from the restriction of the $\sigma$-algebra to $E^c$, $E_2$ from the restriction of the $\sigma$-algebra to $E^c\setminus E_1$, and so on ...

For part (b) consider the (uncountable many different) arbitrary unions of elements of the sequence. They must all be elements of $\mathcal{M}$.