Show $\mathcal{B}( \mathbb R)$ contains all countable subsets of $\mathbb R$

Question

We defined $\mathcal{B}(\mathbb R):=\sigma(\{[a,b[\}),$ where $a < b \in \mathbb R.$

I have been able to prove:

i) for any $b \in \mathbb R$, $\{b\} \in\mathcal{B}(\mathbb R)$

and subsequently ii) $[a,b] \in \mathcal{B}(\mathbb R)$ for any $a, b \in \mathbb R$

iii) $]a,b[ \in \mathcal{B}(\mathbb R)$ for any $a, b \in \mathbb R$

Now for my problem, how can I prove that $\mathcal{B}(\mathbb R)$ contains all countable subsets of $\mathbb R$?

My idea was to take any countable subset $C$ of $\mathbb R$,

it can be written as $C = \bigcup_{n \in \mathbb N}\{a_{n}\}$. Since we were able to prove (i), $C$ is immediately as the countable union $\in \mathcal{B}(\mathbb R)$. Is this ok? Does not seem clean enough to me?

Answer 1

The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $\mathcal{B}( \mathbb R)$, every countable subset of $\Bbb R$ is a countable union of singletons.

Answer 2

Your answer is entirely correct, since $\sigma$-algebras are closed under countable unions and singeltons are in this $\sigma$-algebra:

Indeed, if $a\in \mathbb{R}$, then $$\{a\} = \bigcap_{n=1}^\infty{[a,a + 1/n[} = \bigcap_{n=1}^\infty{[a,a + 1/n]}$$