Show that $\{x\}\in \mathbb{B}(X)$ for every $x\in X$

Question

Let $(X, \tau)$ be a Hausdorff space, and let $\mathbb{B}(X)$ be the Borel $\sigma$ algebra on $X$. The question is,

Is it true that, if $x\in X$, then $\{x\}\in \mathbb{B}(X)$?

The reason why I ask is because of the previous post I made; the answer shows that one can determine a Radon measure at $\{x\}$, but I need to verify that $\{x\}\in \mathbb{B}(X)$.

Answer 1

Since $X$ is Hausdorff $\{x\}$ (a singleton) is closed and $U = X \setminus \{x\}$ is open. Since $B(X)$ is a $\sigma$-algebra, it's closed under taking the complement: $$ B(X) \ni X \setminus U = X \setminus ( X \setminus \left\{ x\right\} ) = \left\{ x \right\} $$