Prove that $\{t\} ∈ \mathscr B(\mathbb R)$ for any $t ∈ \mathbb R$, where $\mathscr B(\mathbb R)$ denotes the Borel σ-algebra of $\mathbb R$

Question

So I understand that the borel σ-algebra of the reals is the σ-algebra generated by the open subsets of the reals.

Would it be correct to say a single point is a closed set. A σ-algebra is closed under complements. Therefore for any {t}∈ R, {t} ∈ B(R)?

However, I am having issues describing this. Any tips helpful.

Answer 1

Your intuition is wrong, the reason that $\{t\}\in \mathcal{B}(\mathbb{R})$ is not because $t$ is in an open subset of $\mathbb{R}$. Rather, it is because $\{t\}$ is a closed set and thus the complement, $\mathbb{R}\backslash\{t\}$, of an open subset of $\mathbb{R}$. Since the Borel $\sigma$ algebra is generated by open sets and closed under complements, $\{t\}\in\mathcal{B}(\mathbb{R})$.