A $\sigma$-algebra $F$ is said to be generated by a partition if there
is some partition $\{B_i\}$ of $\Omega$ so that every set $A$ in $F$ is a
union of some parts in the partition, and every such union in in $F$.
Show that any $\sigma$-algebra on a countable set $\Omega$ is generated
by a partition of $\Omega$.
Proof: Let $\Omega$ be a countable set and $F$ be a $\sigma$-algebra on $\Omega$, then $F$ $\subset$ $2^\Omega$. Let $x\in\Omega$, consider $\cap_i A_i$ such that $x\in A_i$ for all $i$, and $A_i \in F$. Note there maybe uncountably many $A_i \in F$ such that $x \in A_i$, but since $\Omega$ is countable, $\cap_i A_i$ can be written as an intersection of sets in $F$, so $\cap A_i \in F$.
I know that I need to work work with the intersections and see if the sigma algebra can or cannot distinguish $x,y \in \Omega$. But my question is how do I prove the intersections of all sets $A_i$ containing $x$ can be written as intersections of sets in $F$?
Best Answer
You are referring to the proof in https://math.stackexchange.com/a/932678/250955.
For every $x,y \in \Omega$, let $L_{x,y}$ be any set from $F$ that contain $x$ but does not contain $y$, if such set exists, and $\Omega$ otherwise. I guess you want to prove that $$\bigcap_{y \in \Omega} L_{x,y} = \bigcap_{\substack{A \in F \\ x \in A}} A.$$ Indeed,