Why is $\{ x\} \in \mathcal{B}(\mathbb{R}^n)$

Question

I am confused how the Borel $\sigma$-algebra on $\mathbb{R}^n$ , $B(\mathbb{R}^n)$, holds every $\{ x \}$, where $x \in \mathbb{R}^n$.

The proof in Klenke's Probability Theory goes like this:

Given that $(\mathbb{R}^n, \tau)$ is a topological space, the Borel $\sigma$-algebra is the $\sigma$-algebra that is generated by the open sets of $\tau$. So, if $C \subset \mathbb{R}^n$ is a closed set, then $C ^c \in \tau$ is also in $B(\mathbb{R}^n)$ and hence $C$ is a Borel set, since $B$ is closed under complements. --(jump)--> In particular, this implies that {$ x$} $\in B(\mathbb{R}^n)$ for every $x \in \mathbb{R}^n$.

I do not understand how he is making this jump.

Going from top to bottom: If $C$ is a closet set, which is a subset of $\mathbb{R}^n$, then yes, the complement of $C$, which is $C^c$ by definition is in the topology $\tau$. Since $B$ is closed under complements, $C$ is a part of $B$. I do not see how this necessitates $C$ = {$x$}. What if the singleton set {$x$} is not a part of $B$? How do I prove this contradiction?

Answer 1

We are assuming $\tau$ is the usual topology in $\mathbb{R}^n$. Then for any singleton $x$, you should show that the set $U = \mathbb{R}^n \setminus \{ x \}$ is open, hence in the Borel sets. Then the complement $U^c = \{ x \}$ is in the Borel sets since it is closed under complements

Answer 2

I have realized that I completely missed Klenke's portion of the "usual" class of open sets defined by the metric. Applying the definition of the usual metric,

$$\tau = \left\{ \bigcup _{(x,r)\in F} B_r (x): F\subset \Omega \times (0,\infty) \right\}$$

Where $B_r(y)$ is the set of all points that are within a distance $r$ of $y$. Setting $r$ to be $d(x,y)/2$ and performing a union, we see that $x$ is never present inside the unions. The complement of that set is the singleton $\{ x\}$.