Sigma algebra is uncountable, self-contained way of seeing?

Question

Suppose $\mathcal{A}$ is a $\sigma$-algebra with the property that whenever $X \in \mathcal{A}$ and $X$ is nonempty, there exists $Y$, $Z \in \mathcal{A}$ with $Y \cap Z = \emptyset$, $Y \cup Z = X$, and neither $Y$ nor $Z$ is empty. How do I see that $\mathcal{A}$ is uncountable?

There are "proofs" on this website, but they are unclear or defer to stuff that I don't follow, like "$\mathbb{F}_2$-vector spaces," or are just incomplete. Is it possible anybody could supply a clean and self-contained argument, or even proof?

Answer 1

Let $\mathcal A'=\{X\in \mathcal{A}:X\ne\emptyset\}.$ Choose $Y_0\in\mathcal A'.$ (If $\mathcal A=\{\emptyset\}$ we've got a problem.) Choose disjoint sets $X_1,Y_1\in\mathcal A'$ with $X_1,Y_1\subset Y_0;$ then choose disjoint sets $X_2,Y_2\in\mathcal A'$ with $X_2,Y_2\subset Y_1;$ and so on. In this way we get an infinite sequence $X_1,X_2,X_3,\dots$ of pairwise disjoint nonempty elements of $\mathcal A.$

For each set $I\subseteq\mathbb N$ let $X_I=\bigcup_{i\in I}X_i.$ The $2^{\aleph_0}$ sets $X_I$ are distinct elements of $\mathcal A.$

Answer 2

It suffices to exhibit a countably infinite $S\subset A$ where the members of $S$ are pair-wise disjoint non-empty members of $A.$ Because the set $T=\mathbb P(S)$ of all subsets of $S$ has the cardinal of $\mathbb R,$ and for $t\in T,$ the function $\psi :T\to A,$ where $\psi (t)=\cup t,$ is injective.

So take $X_1\in A$ with $X_1\ne \phi.$ For $n\in \mathbb N,$ let $X_n\supsetneqq X_{n+1}$ where $\phi \ne X_{n+1}\in A.$ And now let $Y_n=X_n$ \ $X_{n+1}$ for $n\in \mathbb N.$ And let $S=\{Y_n:n\in \mathbb N\}.$