A σ-algebra may have no non-empty atoms at all? How to prove the fact that a $\sigma-$algebra may have no non-empty atoms at all?

Definition: An atom of a $\sigma$-algebra $\mathscr{A}$ is a non-void set $\emptyset \neq A \in \mathscr{A}$ that contains no other set of
$\mathscr{A}$.

 A: I'm copying this very nice and elementary (but somewhat buried) answer from the following thread: https://mathoverflow.net/questions/22477/sigma-algebra-without-atoms
"A good source for results on atomless $\sigma$-algberas is chapter 3 of Borel Spaces (1981) by Rao and Rao (Diss. Math.). Here are two remarks, taken from this text:

*

*There is no atomless countably generated $\sigma$-algebra.
Let $\mathcal{C}$ be a countable set of generators for a $\sigma$-algebra on $X$. W.l.o.g., we can assume that $\mathcal{C}$ is closed under complements. Then for every $x\in X$, the set  $A(x)=\bigcap \{C:x\in C,C\in\mathcal{C}\}$ is measurable in $\sigma(\mathcal{C})$ as a countable intersection of measurable sets. Since one cannot separate points by $\sigma(\mathcal{C})$ that one cannot separate by $\mathcal{C}$, for each $x$, $A(x)$ is an atom of $\sigma(\mathcal{C})$


*On every uncountable set, there exists an atomless $\sigma$-algebra that separates points.
Let $X$ be an uncountable set. Clearly, it suffices to show that an atomless $\sigma$-algebra that separates points exists on a set with the same cardinality as $X$. The set $Y$ of all finite subsets of $X$ has the same cardinality as $X$, so we work with $Y$. We verify that the $\sigma$-algebra on $Y$ generated by elements of the form $G_x=\{F:x\in F\}$ for some $x\in X$ does the job.
It is obvious that this $\sigma$-algebra separates the elements of $Y$. So if there were any atom, it would be a singleton $\{F\}$ and it would be generated by countably many of the $G_x$. So let $C$ be a countable subset of $X$ such that $\{F\}\in\sigma\{G_c:c\in C\}$. For the reason pointed out in 1., $\{F\}$ would be the intersection of all elements in $\{G_c:c\in C\}\cup\{G_c^C:c\in C\}$ that contain $F$. Now $F\in G_c$ only if $c\in F$. So let $x\notin F\cup C$. Then $F\cup\{x\}\in G_c$ for all $c\in F$ and $F\cup\{x\}\in G_c^C$ for $c\notin F$. So the intersection doesn't contain only $F$, which is a contradiction."
Supplied by https://mathoverflow.net/users/35357/michael-greinecker
