I am reading the book Brownian Motion by Yuval peres and Peter Morters. I am having trouble understanding the following remark. In perticular, How the set on the right equals the set on the left. The remark is as follows,
Suppose $H$ is a closed set, for example a singleton. Then the first hitting time $T=\inf \{t \geqslant 0: B(t) \in H\}$ of the set $H$ is a stopping time with respect to $\left(\mathcal{F}^{0}(t): t \geqslant 0\right)$. Indeed, we note that $$ \{T \leqslant t\}=\bigcap_{n=1}^{\infty} \bigcup_{s \in \mathbb{Q} \cap(0, t)} \bigcup_{x \in \mathbb{Q}^{d} \cap H}\left\{B(s) \in \mathcal{B}\left(x, \frac{1}{n}\right)\right\} \in \mathcal{F}^{0}(t) . $$
Here, $\mathcal{F}^{0}(t) = \sigma\left(B(s): 0 \leq s \leq t \right)$
I don't get why it is true because, if H a closed set which does not intersect with $\mathbb{Q}^d$ then the intersection on RHS would be just empty.
As stated in the comments, the identity is not correct as written. The argument should just be that there exists a countable subset $A \subseteq H$ which is dense in $H$, and then $$ \{T \leqslant t\}=\bigcap_{n=1}^{\infty} \bigcup_{s \in \mathbb{Q} \cap(0, t)} \bigcup_{x \in A}\left\{B(s) \in \mathcal{B}\left(x, \frac{1}{n}\right)\right\} \in \mathcal{F}^{0}(t) . $$