# Why does a countable partition $\subset A$ of $X$ (with $A$ an infinite $\sigma$-algebra) exist?

The problem I'm trying to solve is as follows:

Let $X$ be a space and let $A$ be a $\sigma$-algebra on $X$ that contains infinitely many elements. An infinite partition is a countably infinite sequence on non-empty and pairwise disjoint sets $C_1, C_2, ...$, all in $A$, such that their union is $X$. Show that such a partition exists.

I tried to show this using the method described in this question, namely:

For the function $f:X\rightarrow A$ defined as $f(x) := \bigcap_{x\in S \in A} S$ we have $f(x) \cap f(y) = \varnothing$ for all $x \neq y$. Then, the set $f(X) := \{f(x) | x \in X\}$ partitions $X$.

However, this method results in a countable partition iff $X$ is countable. Does anyone know another method that will definitely result in a countable partition?

HINT: start by picking countably many distinct elements $a_1, a_2, a_3, ...$ of $X$, and let $D_i=\{a_i\}$. Clearly each $D_i$ is in $A$; as you observe, however, there's no reason to believe that $\bigcup D_i=X$.
But this isn't a problem. Think about the "leftover" set $E=X\setminus(\bigcup D_i)$. Given that $A$ is a $\sigma$-algebra, what can you say about the set $E$? From that, what can you say about the sequence $(E, D_1, D_2, D_3, ...)$?
• Why can you find each $D_i$ in $A$? – Sliem el Ela Sep 16 '18 at 19:31