Question: How do I show that $\mathbb{N}, \mathbb{Q}, \mathbb{R}\backslash\mathbb{Q} \in \mathcal{B}(\mathbb{R})$? That is, I want to show that $\mathbb{N}, \mathbb{Q}$ and $\mathbb{R}\backslash\mathbb{Q}$ are Borel subsets of $\mathbb{R}$.
I was introduced to Borel subsets by the following definition:
Let $(S,d)$ be a metric space. Let $\mathcal{B}(S)$ be the $\sigma$-algebra generated by the open sets in $S$. Thus $$\mathcal{B}(S) = \sigma\{\text{open sets in S}\}.$$ The $\sigma$-algebra $\mathcal{B}(S)$ is called the Borel $\sigma$-algebra of S. The sets of $\mathcal{B}(S)$ are called the Borel subsets of S.
I'm not sure how I can use this definition to show that for example $\mathbb{N}$ is a Borel subset of $\mathbb{R}$. $\mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra that contains the open sets in $\mathbb{R}$, but I don't know how that will really get me somewhere.
Thanks in advance!
Note that $\mathbb N$ and $\mathbb Q$ are both countable unions of singletons. A $\sigma$-algebra is closed under countable unions, so for proving that both sets are elements of $\mathcal B(\mathbb R)$ it is enough to prove that every singleton will be an element of $\mathcal B(\mathbb R)$.
For every $x\in\mathbb R$ we have: $$\{x\}=\left[\left(-\infty,x\right)\cup\left(x,\infty\right)\right]^{\complement}$$ So it is the complement of an open set. Every open set is contained in $\mathcal B(\mathbb R)$ and as a $\sigma$-algebra $\mathcal B(\mathbb R)$ is closed under complements, so we are ready with $\mathbb N$ and $\mathbb Q$ now.
Then - again because $\mathcal B(\mathbb R)$ is closed under complements - also $\mathbb R\setminus\mathbb N$ and $\mathbb R\setminus\mathbb Q$ are Borel subsets of $\mathbb R$.