Let $X$ be a set, and let $\mathcal{A} = (A_n)_{n=1}^{\infty}$ be a sequence of disjoint, nonempty
subsets whose union is $X$. Then the set $\mathcal{M}$ of
all finite or countable unions of elements of $\mathcal{A}$ together with $\emptyset$ is a $\sigma$-algebra. A $\sigma$-algebra of this form is called atomic. Then the Borel $\sigma$-algebra $\mathcal{B}_{\mathbb{R}}$ on $\mathbb{R}$ is not atomic.
$\text{Proof.}$
Suppose for sake of contradiction that $\mathcal{B}_{\mathbb{R}}$ is the collection of all finite or countable unions of sets in $\mathcal{A} = (A_n)_{n=1}^{\infty} $, where the $A_i$ are mutually disjoint, non-empty subsets of $\mathbb{R}$ whose union is $\mathbb{R}$. In particular it follows from this that each set in $\mathcal{A}$ is itself a Borel-set in $\mathbb{R}$. Now let
$\mathcal{U}= \left\{\left\{p\right\}:0<p<1\right\}$. Then each singleton in $\mathcal{U}$ is a Borel-set, being closed with respect to the standard topology on $\mathbb{R}$. Hence, for each $p\in (0,1)$, it follows that $\left\{p \right\}$ is a union of a finite or countable sub-collection of the $A_i$. But since the $A_i$ are each nonempty, $\left\{p\right\}$ cannot be a union of more than one $A_i$, since otherwise $\left\{p\right\}$ would have more than one element. Thus, for each $p\in (0,1)$, we can injectively associate a set $A_i\in\mathcal{A}$ with $\left\{p\right\} = A_i$. But this is absurd because $\mathcal{U}$ is uncountable and $\mathcal{A}$ is countable.
I'd appreciate if anyone here could check for the accuracy of the above proof. Thanks.
Best Answer
The idea of your proof is fine, but I think it could be rearranged to be a little simpler.
Suppose such $\mathcal{A}$ existed. Consider two cases: either every $A_i$ is a singleton, or some $A_i$ contains at least two points. The first case is impossible since $\mathbb{R}$ is uncountable. In the second case, if $A_i$ contains two points $x,y$, then $\{x\}$ is a Borel set which, as you argued, cannot be written as a union of elements of $\mathcal{A}$.