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

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?

• Every closed set belongs to the Borel sigma algebra. Singletons are closed in $\mathbb{R^n}$, and so in particular they are Borel sets.
– Mark
Aug 24, 2022 at 21:02
• Can you see that $\{x\}$ is closed in the usual topology? Aug 24, 2022 at 21:02
• @Mark, thank you for your response. But how are singleton's closed under $R^n$? What necessitates the complement of a singleton to be in the topology? Aug 24, 2022 at 21:05
• @megamence Well, try to show that $\mathbb{R^n}\setminus\{x\}$ is open. Given any $y\in\mathbb{R^n}\setminus\{x\}$ you need to find an open ball around $y$ which doesn't contain the point $x$.
– Mark
Aug 24, 2022 at 21:07
• @BrianMoehring, after seeing the term "usual" topology in the comment, I realized I wasn't reading closely enough. All good now! Aug 24, 2022 at 21:21

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