I am new to real analysis and was looking at this problem. I have been looking at the numerous answers to this question and I feel like I am close to understanding; however, I am not quite understanding this last, bolded part of this Proof (The first answer at the link).
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 ...
I think that it is saying let $E_0$ be the first element so that the restriction of $E_0^c$ and $\mathcal{M}$ is infinite. But then I don't understand what $E_1$ is.
Can I say that $\mathcal{M}_1=\{F \cap E_0^c : F\in \mathcal{M}\}$. Then we have $\mathcal{M}_1$ is still a sigma algebra since it is closed under intersections and compliments. Then Take $E_1$ so that the restriction of $E_1^c$ and $\mathcal{M}_1$ is infinite.
I feel like this maybe isn't the right way to do it. I am a bit concerned that I can’t say that the $\mathcal{M}_1$ is a sigma algebra.
If someone could explicitly write for me what the $E_i$ are I would really appreciate the clarification. Thank you.
From the comments above.
Yes, your understanding is correct! You choose $E_0$ to be a nonempty member of $\mathcal{M}$ such that $\mathcal{M}_1 := \{ F \cap E_0^c : F \in \mathcal{M} \}$ is infinite. You can check that $\mathcal{M}_1$ is indeed a $\sigma$-algebra, and now you can choose $E_1$ to be a nonempty member of $\mathcal{M}_1$ such that $\mathcal{M}_2 := \{ F_1 \cap E_1^c : F_1 \in \mathcal{M}_1 \}$ is infinite. And you can proceed like this inductively to complete the proof.