Why is $\{\{c\}:c \in \mathbb{R} \}$ not a generator of the Borel $\sigma$ field in $\mathbb{R}$?

Question

Why is $\{\{c\}:c \in \mathbb{R}\}$ not a generator of the Borel $\sigma$ field in $\mathbb{R}$? For any singleton set $\{c\}$ we can write it as a sequential intersection of $\bigcap(c-\frac{1}{n},c+\frac{1}{n})$. Is it because the unions or intersections are not countable? Please explain.

Answer 1

The set $\mathcal{A}$ of all subsets of $\Bbb R$ that are either countable or have a countable complement, forms a $\sigma$-algebra.

It clearly contains $\mathcal{S}:=\{\{c\}: c \in \Bbb R\}$, all elements being finite.

So $\sigma(\mathcal{S}) \subseteq \mathcal{A}$ by minimality and in fact we can easily show equality. But the main point is that $\mathcal{A} \neq \text{Bor}(\Bbb R)$, as e.g. $(0,1)$ is not countable nor has it a countable complement. So $(0,1) \notin \sigma(\mathcal{S})$.

It is true that singleton sets are Borel, so all members of $\sigma(\mathcal{S})$ are Borel too. But we can only "reach" countable sets with countable unions, and then complements give us the other type, and no more.